Label encoding across multiple columns in scikit-learn

Question

I'm trying to use scikit-learn's LabelEncoder to encode a pandas DataFrame of string labels. As the dataframe has many (50+) columns, I want to avoid creating a LabelEncoder object for each column; I'd rather just have one big LabelEncoder objects that works across all my columns of data.

Throwing the entire DataFrame into LabelEncoder creates the below error. Please bear in mind that I'm using dummy data here; in actuality I'm dealing with about 50 columns of string labeled data, so need a solution that doesn't reference any columns by name.

import pandas
from sklearn import preprocessing 

df = pandas.DataFrame({
    'pets': ['cat', 'dog', 'cat', 'monkey', 'dog', 'dog'], 
    'owner': ['Champ', 'Ron', 'Brick', 'Champ', 'Veronica', 'Ron'], 
    'location': ['San_Diego', 'New_York', 'New_York', 'San_Diego', 'San_Diego', 
                 'New_York']
})

le = preprocessing.LabelEncoder()

le.fit(df)

Traceback (most recent call last): File "", line 1, in File "/Users/bbalin/anaconda/lib/python2.7/site-packages/sklearn/preprocessing/label.py", line 103, in fit y = column_or_1d(y, warn=True) File "/Users/bbalin/anaconda/lib/python2.7/site-packages/sklearn/utils/validation.py", line 306, in column_or_1d raise ValueError("bad input shape {0}".format(shape)) ValueError: bad input shape (6, 3)

Any thoughts on how to get around this problem?

Coinage

To simplify encoding a multi-column dataframe of string data. I'm piclking the encoding object(s), so want to avoid having to pickle/unpickle 50 separate objects. Also, I wonder if there's a way to have the encoder simplify the data, ie just returning one row with an identifier for every unique combination of variables in each column. – Coinage

There is a simple way to do this all in pandas by passing a dictionary of dictionaries to the replace method. See this answer below – Glaab

As of scikit-learn 0.20, there is no need to implement a custom class to label encode multiple columns. You can simply use OrdinalEncoder – Berniecebernier

Answer 1

You can easily do this though,

df.apply(LabelEncoder().fit_transform)

EDIT2:

In scikit-learn 0.20, the recommended way is

OneHotEncoder().fit_transform(df)

as the OneHotEncoder now supports string input. Applying OneHotEncoder only to certain columns is possible with the ColumnTransformer.

EDIT:

Since this original answer is over a year ago, and generated many upvotes (including a bounty), I should probably extend this further.

For inverse_transform and transform, you have to do a little bit of hack.

from collections import defaultdict
d = defaultdict(LabelEncoder)

With this, you now retain all columns LabelEncoder as dictionary.

# Encoding the variable
fit = df.apply(lambda x: d[x.name].fit_transform(x))

# Inverse the encoded
fit.apply(lambda x: d[x.name].inverse_transform(x))

# Using the dictionary to label future data
df.apply(lambda x: d[x.name].transform(x))

MOAR EDIT:

Using Neuraxle's FlattenForEach step, it's possible to do this as well to use the same LabelEncoder on all the flattened data at once:

FlattenForEach(LabelEncoder(), then_unflatten=True).fit_transform(df)

For using separate LabelEncoders depending for your columns of data, or if only some of your columns of data needs to be label-encoded and not others, then using a ColumnTransformer is a solution that allows for more control on your column selection and your LabelEncoder instances.

Resurrection

This is Amazing, but in this case how can we apply inverse transform ? – Ofeliaofella

But if I want to use this solution in a pipeline e.g. separate fit and transform (fit on train, and then use on test-set --> re-use the learnt dictionary) is this supported with df.apply(LabelEncoder().fit_transform)? – Lockwood

@GeorgHeiler: No, since there is no assigned variable that retains the object. Dictionary plays that role. – Resurrection

How can this be made to work with LabelBinarizer instead and re-use the dictionary for a test set? I tried d = defaultdict(LabelBinarizer) and then fit = df.apply(lambda x: d[x.name].fit_transform(x)) but an exception is raised: Exception: Data must be 1-dimensional. I'm not sure how I expect the resulting DataFrame to look like... maybe each column should hold the binarized vectors. – Mumbletypeg

Nice solution. How to transform in certain column only? – Unripe

Seeing the following: TypeError: ("'>' not supported between instances of 'int' and 'str'", 'occurred at index 670') – Lepidolite

@There, Use astype(str) in transform like fit = df.apply(lambda x: d[x.name].fit_transform(x.astype(str))) – Influence

if i want to inverse the encode juste for one column, how do i do it ? – Hardship

I tried the method, but I got an error as "ValueError: If using all scalar values, you must pass an index". Meanwhile, I'm also a little confused about what exactly the return will be? It looks like a dataframe, but what the column names? Since after encoding, each column will turn to multiple columns, is it correct? Thanks for any help. – Hass

Actually now @ogrisel's answer from below should be the accepted answer since it's an easier solution. – Neville

@YunzhaoXing: It will return dataframe with each column values tranformed, not multiple columns – Resurrection

@IbD You can select the column and added apply func like above – Resurrection

Your "easily can do this though" code does not work. – Distant

Added an edit to address @Unripe 's relevant comment just above. – Losel

Is there any possible way to see a dict of each column's mapping information all at once? – Astrolabe

For certain columns only: fit[[col1,col2,col3]] = df[[col1,col2,col3]].apply(lambda x: d[x.name].fit_transform(x)) – Orangery

Answer 2

As mentioned by larsmans, LabelEncoder() only takes a 1-d array as an argument. That said, it is quite easy to roll your own label encoder that operates on multiple columns of your choosing, and returns a transformed dataframe. My code here is based in part on Zac Stewart's excellent blog post found here.

Creating a custom encoder involves simply creating a class that responds to the fit(), transform(), and fit_transform() methods. In your case, a good start might be something like this:

import pandas as pd
from sklearn.preprocessing import LabelEncoder
from sklearn.pipeline import Pipeline

# Create some toy data in a Pandas dataframe
fruit_data = pd.DataFrame({
    'fruit':  ['apple','orange','pear','orange'],
    'color':  ['red','orange','green','green'],
    'weight': [5,6,3,4]
})

class MultiColumnLabelEncoder:
    def __init__(self,columns = None):
        self.columns = columns # array of column names to encode

    def fit(self,X,y=None):
        return self # not relevant here

    def transform(self,X):
        '''
        Transforms columns of X specified in self.columns using
        LabelEncoder(). If no columns specified, transforms all
        columns in X.
        '''
        output = X.copy()
        if self.columns is not None:
            for col in self.columns:
                output[col] = LabelEncoder().fit_transform(output[col])
        else:
            for colname,col in output.iteritems():
                output[colname] = LabelEncoder().fit_transform(col)
        return output

    def fit_transform(self,X,y=None):
        return self.fit(X,y).transform(X)

Suppose we want to encode our two categorical attributes (fruit and color), while leaving the numeric attribute weight alone. We could do this as follows:

MultiColumnLabelEncoder(columns = ['fruit','color']).fit_transform(fruit_data)

Which transforms our fruit_data dataset from

to

Passing it a dataframe consisting entirely of categorical variables and omitting the columns parameter will result in every column being encoded (which I believe is what you were originally looking for):

MultiColumnLabelEncoder().fit_transform(fruit_data.drop('weight',axis=1))

This transforms

to

.

Note that it'll probably choke when it tries to encode attributes that are already numeric (add some code to handle this if you like).

Another nice feature about this is that we can use this custom transformer in a pipeline:

encoding_pipeline = Pipeline([
    ('encoding',MultiColumnLabelEncoder(columns=['fruit','color']))
    # add more pipeline steps as needed
])
encoding_pipeline.fit_transform(fruit_data)

Simmer

Just realized the data implies that an orange is colored green. Oops. ;) – Simmer

this is a good way to transform data once, but what if i want to reuse this transform on a validation set. you would have to fit_transform again and issues could arise such as my new data set not having all the categories for all variables. eg., say the color green does not show up in my new data set. this will mess up the encoding. – Galiot

Agreed with @Ben. This doesn't actually mimic sklearn at all beyond the method names. If you tried to put this in a Pipeline, it wouldn't work – Macassar

@PriceHardman: What if I encode my training data (only input vectors) first and then next time I encode my test data; Will the encoding be came? if not, then what can we do? – Tetanize

To make sure the label encoding is consistent across both the train and test sets, you'll want to perform the encoding on your entire dataset (train + test). This can be done either before you split them into train and test, or you can combine them, perform the encoding, and split them back out again. – Simmer

@Simmer that's fine in terms of validating the model, what if the values at inference time are not present in the train or test dataset? This could happen with high cardinality categorical variables for example. – Weslee

Only one showing pipeline application (unlike other answers) – Hodson

@Weslee if the values don't appear in the training set you're in a problem either way. The key to solve this is to use the fit piece and train the encoder on the training set. I used a similar approach in this code – Theomorphic

Thanks. This saved me quite some time as a beginner. Even made this into a snippet. Due credit has been given to you as well. :) Snippet – Orelee

if i want to focus only on dtype = 'object' and reverse encoding after modeling for one column (ex : taget column : Y) how do i do it ? – Hardship

How about going reverse ? decoding back to the original ? – Sorrells

You can't inverse transform with this. – Vally

I don't know why, but this work extremely slow on my 6.5 million dataset where I'm trying to encode 8 features. Do anyone have any clue regarding this? – Glosso

Answer 3

Since scikit-learn 0.20 you can use sklearn.compose.ColumnTransformer and sklearn.preprocessing.OneHotEncoder:

If you only have categorical variables, OneHotEncoder directly:

from sklearn.preprocessing import OneHotEncoder

OneHotEncoder(handle_unknown='ignore').fit_transform(df)

If you have heterogeneously typed features:

from sklearn.compose import make_column_transformer
from sklearn.preprocessing import RobustScaler
from sklearn.preprocessing import OneHotEncoder

categorical_columns = ['pets', 'owner', 'location']
numerical_columns = ['age', 'weigth', 'height']
column_trans = make_column_transformer(
    (categorical_columns, OneHotEncoder(handle_unknown='ignore'),
    (numerical_columns, RobustScaler())
column_trans.fit_transform(df)

More options in the documentation: http://scikit-learn.org/stable/modules/compose.html#columntransformer-for-heterogeneous-data

Frederico

inverse_transform() is not supported on ColumnTransformer though. At least, not for the moment: github.com/scikit-learn/scikit-learn/issues/11463. That's a big disadvantage for my application, and will probably also be for others. – Hiroshima

Although ColumnTransformer is a great suggestion, this code does not run (imbalanced parentheses, column_transformer does not/no longer works that way) – Helfrich

I've proposed an edit to the original answer to fix the code – Helfrich

Answer 4

We don't need a LabelEncoder.

You can convert the columns to categoricals and then get their codes. I used a dictionary comprehension below to apply this process to every column and wrap the result back into a dataframe of the same shape with identical indices and column names.

>>> pd.DataFrame({col: df[col].astype('category').cat.codes for col in df}, index=df.index)
   location  owner  pets
0         1      1     0
1         0      2     1
2         0      0     0
3         1      1     2
4         1      3     1
5         0      2     1

To create a mapping dictionary, you can just enumerate the categories using a dictionary comprehension:

>>> {col: {n: cat for n, cat in enumerate(df[col].astype('category').cat.categories)} 
     for col in df}

{'location': {0: 'New_York', 1: 'San_Diego'},
 'owner': {0: 'Brick', 1: 'Champ', 2: 'Ron', 3: 'Veronica'},
 'pets': {0: 'cat', 1: 'dog', 2: 'monkey'}}

Backstitch

If i want to go back (reverse) for one column (example target variable : Y) how do i do it ? – Hardship

Nice one Alexander! I like this a lot!! – Selfcontrol

Answer 5

This is a year-and-a-half after the fact, but I too, needed to be able to .transform() multiple pandas dataframe columns at once (and be able to .inverse_transform() them as well). This expands upon the excellent suggestion of @PriceHardman above:

class MultiColumnLabelEncoder(LabelEncoder):
    """
    Wraps sklearn LabelEncoder functionality for use on multiple columns of a
    pandas dataframe.

    """
    def __init__(self, columns=None):
        self.columns = columns

    def fit(self, dframe):
        """
        Fit label encoder to pandas columns.

        Access individual column classes via indexig `self.all_classes_`

        Access individual column encoders via indexing
        `self.all_encoders_`
        """
        # if columns are provided, iterate through and get `classes_`
        if self.columns is not None:
            # ndarray to hold LabelEncoder().classes_ for each
            # column; should match the shape of specified `columns`
            self.all_classes_ = np.ndarray(shape=self.columns.shape,
                                           dtype=object)
            self.all_encoders_ = np.ndarray(shape=self.columns.shape,
                                            dtype=object)
            for idx, column in enumerate(self.columns):
                # fit LabelEncoder to get `classes_` for the column
                le = LabelEncoder()
                le.fit(dframe.loc[:, column].values)
                # append the `classes_` to our ndarray container
                self.all_classes_[idx] = (column,
                                          np.array(le.classes_.tolist(),
                                                  dtype=object))
                # append this column's encoder
                self.all_encoders_[idx] = le
        else:
            # no columns specified; assume all are to be encoded
            self.columns = dframe.iloc[:, :].columns
            self.all_classes_ = np.ndarray(shape=self.columns.shape,
                                           dtype=object)
            for idx, column in enumerate(self.columns):
                le = LabelEncoder()
                le.fit(dframe.loc[:, column].values)
                self.all_classes_[idx] = (column,
                                          np.array(le.classes_.tolist(),
                                                  dtype=object))
                self.all_encoders_[idx] = le
        return self

    def fit_transform(self, dframe):
        """
        Fit label encoder and return encoded labels.

        Access individual column classes via indexing
        `self.all_classes_`

        Access individual column encoders via indexing
        `self.all_encoders_`

        Access individual column encoded labels via indexing
        `self.all_labels_`
        """
        # if columns are provided, iterate through and get `classes_`
        if self.columns is not None:
            # ndarray to hold LabelEncoder().classes_ for each
            # column; should match the shape of specified `columns`
            self.all_classes_ = np.ndarray(shape=self.columns.shape,
                                           dtype=object)
            self.all_encoders_ = np.ndarray(shape=self.columns.shape,
                                            dtype=object)
            self.all_labels_ = np.ndarray(shape=self.columns.shape,
                                          dtype=object)
            for idx, column in enumerate(self.columns):
                # instantiate LabelEncoder
                le = LabelEncoder()
                # fit and transform labels in the column
                dframe.loc[:, column] =\
                    le.fit_transform(dframe.loc[:, column].values)
                # append the `classes_` to our ndarray container
                self.all_classes_[idx] = (column,
                                          np.array(le.classes_.tolist(),
                                                  dtype=object))
                self.all_encoders_[idx] = le
                self.all_labels_[idx] = le
        else:
            # no columns specified; assume all are to be encoded
            self.columns = dframe.iloc[:, :].columns
            self.all_classes_ = np.ndarray(shape=self.columns.shape,
                                           dtype=object)
            for idx, column in enumerate(self.columns):
                le = LabelEncoder()
                dframe.loc[:, column] = le.fit_transform(
                        dframe.loc[:, column].values)
                self.all_classes_[idx] = (column,
                                          np.array(le.classes_.tolist(),
                                                  dtype=object))
                self.all_encoders_[idx] = le
        return dframe.loc[:, self.columns].values

    def transform(self, dframe):
        """
        Transform labels to normalized encoding.
        """
        if self.columns is not None:
            for idx, column in enumerate(self.columns):
                dframe.loc[:, column] = self.all_encoders_[
                    idx].transform(dframe.loc[:, column].values)
        else:
            self.columns = dframe.iloc[:, :].columns
            for idx, column in enumerate(self.columns):
                dframe.loc[:, column] = self.all_encoders_[idx]\
                    .transform(dframe.loc[:, column].values)
        return dframe.loc[:, self.columns].values

    def inverse_transform(self, dframe):
        """
        Transform labels back to original encoding.
        """
        if self.columns is not None:
            for idx, column in enumerate(self.columns):
                dframe.loc[:, column] = self.all_encoders_[idx]\
                    .inverse_transform(dframe.loc[:, column].values)
        else:
            self.columns = dframe.iloc[:, :].columns
            for idx, column in enumerate(self.columns):
                dframe.loc[:, column] = self.all_encoders_[idx]\
                    .inverse_transform(dframe.loc[:, column].values)
        return dframe.loc[:, self.columns].values

Example:

If df and df_copy() are mixed-type pandas dataframes, you can apply the MultiColumnLabelEncoder() to the dtype=object columns in the following way:

# get `object` columns
df_object_columns = df.iloc[:, :].select_dtypes(include=['object']).columns
df_copy_object_columns = df_copy.iloc[:, :].select_dtypes(include=['object']).columns

# instantiate `MultiColumnLabelEncoder`
mcle = MultiColumnLabelEncoder(columns=object_columns)

# fit to `df` data
mcle.fit(df)

# transform the `df` data
mcle.transform(df)

# returns output like below
array([[1, 0, 0, ..., 1, 1, 0],
       [0, 5, 1, ..., 1, 1, 2],
       [1, 1, 1, ..., 1, 1, 2],
       ..., 
       [3, 5, 1, ..., 1, 1, 2],

# transform `df_copy` data
mcle.transform(df_copy)

# returns output like below (assuming the respective columns 
# of `df_copy` contain the same unique values as that particular 
# column in `df`
array([[1, 0, 0, ..., 1, 1, 0],
       [0, 5, 1, ..., 1, 1, 2],
       [1, 1, 1, ..., 1, 1, 2],
       ..., 
       [3, 5, 1, ..., 1, 1, 2],

# inverse `df` data
mcle.inverse_transform(df)

# outputs data like below
array([['August', 'Friday', '2013', ..., 'N', 'N', 'CA'],
       ['April', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['August', 'Monday', '2014', ..., 'N', 'N', 'NJ'],
       ..., 
       ['February', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['April', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['March', 'Tuesday', '2013', ..., 'N', 'N', 'NJ']], dtype=object)

# inverse `df_copy` data
mcle.inverse_transform(df_copy)

# outputs data like below
array([['August', 'Friday', '2013', ..., 'N', 'N', 'CA'],
       ['April', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['August', 'Monday', '2014', ..., 'N', 'N', 'NJ'],
       ..., 
       ['February', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['April', 'Tuesday', '2014', ..., 'N', 'N', 'NJ'],
       ['March', 'Tuesday', '2013', ..., 'N', 'N', 'NJ']], dtype=object)

You can access individual column classes, column labels, and column encoders used to fit each column via indexing:

mcle.all_classes_
mcle.all_encoders_
mcle.all_labels_

Criseldacrisey

Hi Jason, mcle.all_labels_ does not appear to work (Python 3.5, Conda 4.3.29, Sklearn 0.18.1, Pandas 0.20.1. I get: AttributeError: 'MultiColumnLabelEncoder' object has no attribute 'all_labels_' – Wergild

@Wergild Hi, sorry, I didn't see this until today :/ but if I had to guess, I would say that you just used the fit method from above which won't actually produce any labels until you apply it (transform / fit_transform) to the data. – Criseldacrisey

I think you need to put a better example - I couldn't rerun all your codes. – Sorrells

Answer 6

this does not directly answer your question (for which Naputipulu Jon and PriceHardman have fantastic replies)

However, for the purpose of a few classification tasks etc. you could use

pandas.get_dummies(input_df) 

this can input dataframe with categorical data and return a dataframe with binary values. variable values are encoded into column names in the resulting dataframe. more

Wheatworm

Answer 7

It is possible to do this all in pandas directly and is well-suited for a unique ability of the replace method.

First, let's make a dictionary of dictionaries mapping the columns and their values to their new replacement values.

transform_dict = {}
for col in df.columns:
    cats = pd.Categorical(df[col]).categories
    d = {}
    for i, cat in enumerate(cats):
        d[cat] = i
    transform_dict[col] = d

transform_dict
{'location': {'New_York': 0, 'San_Diego': 1},
 'owner': {'Brick': 0, 'Champ': 1, 'Ron': 2, 'Veronica': 3},
 'pets': {'cat': 0, 'dog': 1, 'monkey': 2}}

Since this will always be a one to one mapping, we can invert the inner dictionary to get a mapping of the new values back to the original.

inverse_transform_dict = {}
for col, d in transform_dict.items():
    inverse_transform_dict[col] = {v:k for k, v in d.items()}

inverse_transform_dict
{'location': {0: 'New_York', 1: 'San_Diego'},
 'owner': {0: 'Brick', 1: 'Champ', 2: 'Ron', 3: 'Veronica'},
 'pets': {0: 'cat', 1: 'dog', 2: 'monkey'}}

Now, we can use the unique ability of the replace method to take a nested list of dictionaries and use the outer keys as the columns, and the inner keys as the values we would like to replace.

df.replace(transform_dict)
   location  owner  pets
0         1      1     0
1         0      2     1
2         0      0     0
3         1      1     2
4         1      3     1
5         0      2     1

We can easily go back to the original by again chaining the replace method

df.replace(transform_dict).replace(inverse_transform_dict)
    location     owner    pets
0  San_Diego     Champ     cat
1   New_York       Ron     dog
2   New_York     Brick     cat
3  San_Diego     Champ  monkey
4  San_Diego  Veronica     dog
5   New_York       Ron     dog

Glaab

Answer 8

No, LabelEncoder does not do this. It takes 1-d arrays of class labels and produces 1-d arrays. It's designed to handle class labels in classification problems, not arbitrary data, and any attempt to force it into other uses will require code to transform the actual problem to the problem it solves (and the solution back to the original space).

Anglophile

Ok, given this, what is your suggestion on the best way I can encode string labels by an entire DataFrame at a time? – Coinage

@Coinage Look at the LabelEncoder code and adapt it. I don't use Pandas myself, so I don't know how hard that will be. – Anglophile

I'll let other pandas folks take a crack at this question too -- I'm sure I'm not the only person with this challenge, so I hope there might be a pre-built solution out there. – Coinage

Answer 9

Assuming you are simply trying to get a sklearn.preprocessing.LabelEncoder() object that can be used to represent your columns, all you have to do is:

le.fit(df.columns)

In the above code you will have a unique number corresponding to each column. More precisely, you will have a 1:1 mapping of df.columns to le.transform(df.columns.get_values()). To get a column's encoding, simply pass it to le.transform(...). As an example, the following will get the encoding for each column:

le.transform(df.columns.get_values())

Assuming you want to create a sklearn.preprocessing.LabelEncoder() object for all of your row labels you can do the following:

le.fit([y for x in df.get_values() for y in x])

In this case, you most likely have non-unique row labels (as shown in your question). To see what classes the encoder created you can do le.classes_. You'll note that this should have the same elements as in set(y for x in df.get_values() for y in x). Once again to convert a row label to an encoded label use le.transform(...). As an example, if you want to retrieve the label for the first column in the df.columns array and the first row, you could do this:

le.transform([df.get_value(0, df.columns[0])])

The question you had in your comment is a bit more complicated, but can still be accomplished:

le.fit([str(z) for z in set((x[0], y) for x in df.iteritems() for y in x[1])])

The above code does the following:

1. Make a unique combination of all of the pairs of (column, row)
2. Represent each pair as a string version of the tuple. This is a workaround to overcome the LabelEncoder class not supporting tuples as a class name.
3. Fits the new items to the LabelEncoder.

Now to use this new model it's a bit more complicated. Assuming we want to extract the representation for the same item we looked up in the previous example (the first column in df.columns and the first row), we can do this:

le.transform([str((df.columns[0], df.get_value(0, df.columns[0])))])

Remember that each lookup is now a string representation of a tuple that contains the (column, row).

Coverdale

Answer 10

Here is the script

from sklearn.preprocessing import LabelEncoder
le = LabelEncoder()
col_list = df.select_dtypes(include = "object").columns
for colsn in col_list:
    df[colsn] = le.fit_transform(df[colsn].astype(str))

Albuquerque

Answer 11

If you have numerical and categorical both type of data in dataframe You can use : here X is my dataframe having categorical and numerical both variables

from sklearn import preprocessing
le = preprocessing.LabelEncoder()

for i in range(0,X.shape[1]):
    if X.dtypes[i]=='object':
        X[X.columns[i]] = le.fit_transform(X[X.columns[i]])

Note: This technique is good if you are not interested in converting them back.

Rollo

Answer 12

After lots of search and experimentation with some answers here and elsewhere, I think your answer is here:

pd.DataFrame(columns=df.columns, data=LabelEncoder().fit_transform(df.values.flatten()).reshape(df.shape))

This will preserve category names across columns:

import pandas as pd
from sklearn.preprocessing import LabelEncoder

df = pd.DataFrame([['A','B','C','D','E','F','G','I','K','H'],
                   ['A','E','H','F','G','I','K','','',''],
                   ['A','C','I','F','H','G','','','','']], 
                  columns=['A1', 'A2', 'A3','A4', 'A5', 'A6', 'A7', 'A8', 'A9', 'A10'])

pd.DataFrame(columns=df.columns, data=LabelEncoder().fit_transform(df.values.flatten()).reshape(df.shape))

    A1  A2  A3  A4  A5  A6  A7  A8  A9  A10
0   1   2   3   4   5   6   7   9   10  8
1   1   5   8   6   7   9   10  0   0   0
2   1   3   9   6   8   7   0   0   0   0

Paleoecology

Answer 13

I checked the source code (https://github.com/scikit-learn/scikit-learn/blob/master/sklearn/preprocessing/label.py) of LabelEncoder. It was based on a set of numpy transformation, which one of those is np.unique(). And this function only takes 1-d array input. (correct me if I am wrong).

Very Rough ideas... first, identify which columns needed LabelEncoder, then loop through each column.

def cat_var(df): 
    """Identify categorical features. 

    Parameters
    ----------
    df: original df after missing operations 

    Returns
    -------
    cat_var_df: summary df with col index and col name for all categorical vars
    """
    col_type = df.dtypes
    col_names = list(df)

    cat_var_index = [i for i, x in enumerate(col_type) if x=='object']
    cat_var_name = [x for i, x in enumerate(col_names) if i in cat_var_index]

    cat_var_df = pd.DataFrame({'cat_ind': cat_var_index, 
                               'cat_name': cat_var_name})

    return cat_var_df



from sklearn.preprocessing import LabelEncoder 

def column_encoder(df, cat_var_list):
    """Encoding categorical feature in the dataframe

    Parameters
    ----------
    df: input dataframe 
    cat_var_list: categorical feature index and name, from cat_var function

    Return
    ------
    df: new dataframe where categorical features are encoded
    label_list: classes_ attribute for all encoded features 
    """

    label_list = []
    cat_var_df = cat_var(df)
    cat_list = cat_var_df.loc[:, 'cat_name']

    for index, cat_feature in enumerate(cat_list): 

        le = LabelEncoder()

        le.fit(df.loc[:, cat_feature])    
        label_list.append(list(le.classes_))

        df.loc[:, cat_feature] = le.transform(df.loc[:, cat_feature])

    return df, label_list

The returned df would be the one after encoding, and label_list will show you what all those values means in the corresponding column. This is a snippet from a data process script I wrote for work. Let me know if you think there could be any further improvement.

EDIT: Just want to mention here that the methods above work with data frame with no missing the best. Not sure how it is working toward data frame contains missing data. (I had a deal with missing procedure before execute above methods)

Krypton

Answer 14

A short way to LabelEncoder() multiple columns with a dict():

from sklearn.preprocessing import LabelEncoder
le_dict = {col: LabelEncoder() for col in columns }
for col in columns:
    le_dict[col].fit_transform(df[col])

and you can use this le_dict to labelEncode any other column:

le_dict[col].transform(df_another[col])

Antagonistic

Answer 15

Instead of LabelEncoder we can use OrdinalEncoder from scikit learn, which allows multi-column encoding.

Encode categorical features as an integer array. The input to this transformer should be an array-like of integers or strings, denoting the values taken on by categorical (discrete) features. The features are converted to ordinal integers. This results in a single column of integers (0 to n_categories - 1) per feature.

>>> from sklearn.preprocessing import OrdinalEncoder
>>> enc = OrdinalEncoder()
>>> X = [['Male', 1], ['Female', 3], ['Female', 2]]
>>> enc.fit(X)
OrdinalEncoder()
>>> enc.categories_
[array(['Female', 'Male'], dtype=object), array([1, 2, 3], dtype=object)]
>>> enc.transform([['Female', 3], ['Male', 1]])
array([[0., 2.],
       [1., 0.]])

Both the description and example were copied from its documentation page which you can find here:

https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.OrdinalEncoder.html#sklearn.preprocessing.OrdinalEncoder

Dutiable

This should be the way to go as of scikit-learn 0.20. Thank you for suggesting it! – Berniecebernier

Answer 16

Using Neuraxle

TLDR; You here can use the FlattenForEach wrapper class to simply transform your df like: FlattenForEach(LabelEncoder(), then_unflatten=True).fit_transform(df).

With this method, your label encoder will be able to fit and transform within a regular scikit-learn Pipeline. Let's simply import:

from sklearn.preprocessing import LabelEncoder
from neuraxle.steps.column_transformer import ColumnTransformer
from neuraxle.steps.loop import FlattenForEach

Same shared encoder for columns:

Here is how one shared LabelEncoder will be applied on all the data to encode it:

    p = FlattenForEach(LabelEncoder(), then_unflatten=True)

Result:

    p, predicted_output = p.fit_transform(df.values)
    expected_output = np.array([
        [6, 7, 6, 8, 7, 7],
        [1, 3, 0, 1, 5, 3],
        [4, 2, 2, 4, 4, 2]
    ]).transpose()
    assert np.array_equal(predicted_output, expected_output)

Different encoders per column:

And here is how a first standalone LabelEncoder will be applied on the pets, and a second will be shared for the columns owner and location. So to be precise, we here have a mix of different and shared label encoders:

    p = ColumnTransformer([
        # A different encoder will be used for column 0 with name "pets":
        (0, FlattenForEach(LabelEncoder(), then_unflatten=True)),
        # A shared encoder will be used for column 1 and 2, "owner" and "location":
        ([1, 2], FlattenForEach(LabelEncoder(), then_unflatten=True)),
    ], n_dimension=2)

Result:

    p, predicted_output = p.fit_transform(df.values)
    expected_output = np.array([
        [0, 1, 0, 2, 1, 1],
        [1, 3, 0, 1, 5, 3],
        [4, 2, 2, 4, 4, 2]
    ]).transpose()
    assert np.array_equal(predicted_output, expected_output)

Losel

Answer 17

Following up on the comments raised on the solution of @PriceHardman I would propose the following version of the class:

class LabelEncodingColoumns(BaseEstimator, TransformerMixin):
def __init__(self, cols=None):
    pdu._is_cols_input_valid(cols)
    self.cols = cols
    self.les = {col: LabelEncoder() for col in cols}
    self._is_fitted = False

def transform(self, df, **transform_params):
    """
    Scaling ``cols`` of ``df`` using the fitting

    Parameters
    ----------
    df : DataFrame
        DataFrame to be preprocessed
    """
    if not self._is_fitted:
        raise NotFittedError("Fitting was not preformed")
    pdu._is_cols_subset_of_df_cols(self.cols, df)

    df = df.copy()

    label_enc_dict = {}
    for col in self.cols:
        label_enc_dict[col] = self.les[col].transform(df[col])

    labelenc_cols = pd.DataFrame(label_enc_dict,
        # The index of the resulting DataFrame should be assigned and
        # equal to the one of the original DataFrame. Otherwise, upon
        # concatenation NaNs will be introduced.
        index=df.index
    )

    for col in self.cols:
        df[col] = labelenc_cols[col]
    return df

def fit(self, df, y=None, **fit_params):
    """
    Fitting the preprocessing

    Parameters
    ----------
    df : DataFrame
        Data to use for fitting.
        In many cases, should be ``X_train``.
    """
    pdu._is_cols_subset_of_df_cols(self.cols, df)
    for col in self.cols:
        self.les[col].fit(df[col])
    self._is_fitted = True
    return self

This class fits the encoder on the training set and uses the fitted version when transforming. Initial version of the code can be found here.

Theomorphic

Answer 18

if we have single column to do the label encoding and its inverse transform its easy how to do it when there are multiple columns in python

def stringtocategory(dataset):
    '''
    @author puja.sharma
    @see The function label encodes the object type columns and gives label      encoded and inverse tranform of the label encoded data
    @param dataset dataframe on whoes column the label encoding has to be done
    @return label encoded and inverse tranform of the label encoded data.
   ''' 
   data_original = dataset[:]
   data_tranformed = dataset[:]
   for y in dataset.columns:
       #check the dtype of the column object type contains strings or chars
       if (dataset[y].dtype == object):
          print("The string type features are  : " + y)
          le = preprocessing.LabelEncoder()
          le.fit(dataset[y].unique())
          #label encoded data
          data_tranformed[y] = le.transform(dataset[y])
          #inverse label transform  data
          data_original[y] = le.inverse_transform(data_tranformed[y])
   return data_tranformed,data_original

Acrospire

Answer 19

Mainly used @Alexander answer but had to make some changes -

cols_need_mapped = ['col1', 'col2']

mapper = {col: {cat: n for n, cat in enumerate(df[col].astype('category').cat.categories)} 
     for col in df[cols_need_mapped]}

for c in cols_need_mapped :
    df[c] = df[c].map(mapper[c])

Then to re-use in the future you can just save the output to a json document and when you need it you read it in and use the .map() function like I did above.

Wilcox

Answer 20

import pandas as pd
from sklearn.preprocessing import LabelEncoder

train=pd.read_csv('.../train.csv')

#X=train.loc[:,['waterpoint_type_group','status','waterpoint_type','source_class']].values
# Create a label encoder object 
def MultiLabelEncoder(columnlist,dataframe):
    for i in columnlist:

        labelencoder_X=LabelEncoder()
        dataframe[i]=labelencoder_X.fit_transform(dataframe[i])
columnlist=['waterpoint_type_group','status','waterpoint_type','source_class','source_type']
MultiLabelEncoder(columnlist,train)

Here i am reading a csv from location and in function i am passing the column list i want to labelencode and the dataframe I want to apply this.

Stancil

Answer 21

If you have all the features of type object then the first answer written above works well https://mcmap.net/q/98616/-label-encoding-across-multiple-columns-in-scikit-learn.

But, Suppose when we have mixed type columns. Then we can fetch the list of features names of type object type programmatically and then Label Encode them.

#Fetch features of type Object
objFeatures = dataframe.select_dtypes(include="object").columns

#Iterate a loop for features of type object
from sklearn import preprocessing
le = preprocessing.LabelEncoder()

for feat in objFeatures:
    dataframe[feat] = le.fit_transform(dataframe[feat].astype(str))
 

dataframe.info()

Dodgem

Answer 22

The problem is the shape of the data (pd dataframe) you are passing to the fit function. You've got to pass 1d list.

Rohde

Answer 23

How about this?

def MultiColumnLabelEncode(choice, columns, X):
    LabelEncoders = []
    if choice == 'encode':
        for i in enumerate(columns):
            LabelEncoders.append(LabelEncoder())
        i=0    
        for cols in columns:
            X[:, cols] = LabelEncoders[i].fit_transform(X[:, cols])
            i += 1
    elif choice == 'decode': 
        for cols in columns:
            X[:, cols] = LabelEncoders[i].inverse_transform(X[:, cols])
            i += 1
    else:
        print('Please select correct parameter "choice". Available parameters: encode/decode')

It is not the most efficient, however it works and it is super simple.

Cleome

Answer 24

Here is my solution to your problem. In order to convert your data-frame column containing text to encoded values just use my function text_to_numbers it returns a dictonary of LE. Key is the column name that column LabelEncoder() as a value.

def text_to_numbers(df):
        le_dict = dict()
        for i in df.columns:
            if df[i].dtype not in ["float64", "bool", "int64"]:
                le_dict[i] = preprocessing.LabelEncoder()
                df[i] = le_dict[i].fit_transform(df[i])
    
        return df, le_dict

The function below will make it possible to retain an original unencoded dataframe.

 def numbers_to_text(df, le_dict):
        for i in le_dict.keys():
            df[i] = le_dict[i].inverse_transform(df[i])
    
        return df

Churning

Answer 25

Here is my solution to transform multiple columns in one-go, along with the accurate inverse_transformation

from sklearn import preprocessing
columns = ['buying','maint','lug_boot','safety','cls']  # columns names where transform is required
for X in columns:
  exec(f'le_{X} = preprocessing.LabelEncoder()')  #create label encoder with name "le_X", where X is column name
  exec(f'df.{X} = le_{X}.fit_transform(df.{X})')  #execute fit transform for column X with respective lable encoder "le_X", where X is column name
df.head()  # to display transformed results

for X in columns:
  exec(f'df.{X} = le_{X}.inverse_transform(df.{X})')  #execute inverse_transform for column X with respective lable encoder "le_X", where X is column name
df.head() # to display Inverse transformed results of df

Agnew