Remap values in pandas column with a dict, preserve NaNs

Question

I have a dictionary which looks like this: di = {1: "A", 2: "B"}

I would like to apply it to the col1 column of a dataframe similar to:

     col1   col2
0       w      a
1       1      2
2       2    NaN

to get:

     col1   col2
0       w      a
1       A      2
2       B    NaN

How can I best do this?

Wigan

Answer 1

You can use .replace. For example:

>>> df = pd.DataFrame({'col2': {0: 'a', 1: 2, 2: np.nan}, 'col1': {0: 'w', 1: 1, 2: 2}})
>>> di = {1: "A", 2: "B"}
>>> df
  col1 col2
0    w    a
1    1    2
2    2  NaN
>>> df.replace({"col1": di})
  col1 col2
0    w    a
1    A    2
2    B  NaN

or directly on the Series, i.e. df["col1"].replace(di, inplace=True).

Licence

Answer 2

map can be much faster than replace

If your dictionary has more than a couple of keys, using map can be much faster than replace. There are two versions of this approach, depending on whether your dictionary exhaustively maps all possible values (and also whether you want non-matches to keep their values or be converted to NaNs):

Exhaustive Mapping

In this case, the form is very simple:

df['col1'].map(di)       # note: if the dictionary does not exhaustively map all
                         # entries then non-matched entries are changed to NaNs

Although map most commonly takes a function as its argument, it can alternatively take a dictionary or series: Documentation for Pandas.series.map

Non-Exhaustive Mapping

If you have a non-exhaustive mapping and wish to retain the existing variables for non-matches, you can add fillna:

df['col1'].map(di).fillna(df['col1'])

as in @jpp's answer here: Replace values in a pandas series via dictionary efficiently

Benchmarks

Using the following data with pandas version 0.23.1:

di = {1: "A", 2: "B", 3: "C", 4: "D", 5: "E", 6: "F", 7: "G", 8: "H" }
df = pd.DataFrame({ 'col1': np.random.choice( range(1,9), 100000 ) })

and testing with %timeit, it appears that map is approximately 10x faster than replace.

Note that your speedup with map will vary with your data. The largest speedup appears to be with large dictionaries and exhaustive replaces. See @jpp answer (linked above) for more extensive benchmarks and discussion.

Martellato

The last block of code for this answer is certainly not the most elegant, but this answer deserves some credit. It is orders of magnitude faster for large dictionaries and doesn't use up all of my RAM. It remapped a 10,000 line file using a dictionary that had about 9 million entries in half a minute. The df.replace function, while tidy and useful for small dicts, crashed after running for 20 minutes or so. – Cold

Related: Replace values in a pandas series via dictionary efficiently – Libeler

.map is certainly the better way. .map on a table with millions of entries runs in a matter of seconds while .replace was running for more than an hour. .map is the recommended way! – Botel

In some cases (like working with an index) it may be necessary to first cast as series, as the .fillna method for an index requires a scalar value. Would look something like this: pd.Series(my_index).map(di).fillna(pd.Series(my_index)) – Neckline

.map with a dictionary works extremely well,except for nan, because nandoesn't work at all as a key in a dictionary. Indeed nan doesn't even compare equal to itself. An alternative is to use .map with dict.get with a defaultvalue, for instance .map(lambda x: d.get(x, nan)) which will map all keys not found, including nan, to the default value. – Cavin

Answer 3

There is a bit of ambiguity in your question. There are at least three two interpretations:

1. the keys in di refer to index values
2. the keys in di refer to df['col1'] values
3. the keys in di refer to index locations (not the OP's question, but thrown in for fun.)

Below is a solution for each case.

---

Case 1: If the keys of di are meant to refer to index values, then you could use the update method:

df['col1'].update(pd.Series(di))

For example,

import pandas as pd
import numpy as np

df = pd.DataFrame({'col1':['w', 10, 20],
                   'col2': ['a', 30, np.nan]},
                  index=[1,2,0])
#   col1 col2
# 1    w    a
# 2   10   30
# 0   20  NaN

di = {0: "A", 2: "B"}

# The value at the 0-index is mapped to 'A', the value at the 2-index is mapped to 'B'
df['col1'].update(pd.Series(di))
print(df)

yields

  col1 col2
1    w    a
2    B   30
0    A  NaN

I've modified the values from your original post so it is clearer what update is doing. Note how the keys in di are associated with index values. The order of the index values -- that is, the index locations -- does not matter.

---

Case 2: If the keys in di refer to df['col1'] values, then @DanAllan and @DSM show how to achieve this with replace:

import pandas as pd
import numpy as np

df = pd.DataFrame({'col1':['w', 10, 20],
                   'col2': ['a', 30, np.nan]},
                  index=[1,2,0])
print(df)
#   col1 col2
# 1    w    a
# 2   10   30
# 0   20  NaN

di = {10: "A", 20: "B"}

# The values 10 and 20 are replaced by 'A' and 'B'
df['col1'].replace(di, inplace=True)
print(df)

yields

  col1 col2
1    w    a
2    A   30
0    B  NaN

Note how in this case the keys in di were changed to match values in df['col1'].

---

Case 3: If the keys in di refer to index locations, then you could use

df['col1'].put(di.keys(), di.values())

since

df = pd.DataFrame({'col1':['w', 10, 20],
                   'col2': ['a', 30, np.nan]},
                  index=[1,2,0])
di = {0: "A", 2: "B"}

# The values at the 0 and 2 index locations are replaced by 'A' and 'B'
df['col1'].put(di.keys(), di.values())
print(df)

yields

  col1 col2
1    A    a
2   10   30
0    B  NaN

Here, the first and third rows were altered, because the keys in di are 0 and 2, which with Python's 0-based indexing refer to the first and third locations.

Concupiscence

Answer 4

DSM has the accepted answer, but the coding doesn't seem to work for everyone. Here is one that works with the current version of pandas (0.23.4 as of 8/2018):

import pandas as pd

df = pd.DataFrame({'col1': [1, 2, 2, 3, 1],
            'col2': ['negative', 'positive', 'neutral', 'neutral', 'positive']})

conversion_dict = {'negative': -1, 'neutral': 0, 'positive': 1}
df['converted_column'] = df['col2'].replace(conversion_dict)

print(df.head())

You'll see it looks like:

   col1      col2  converted_column
0     1  negative                -1
1     2  positive                 1
2     2   neutral                 0
3     3   neutral                 0
4     1  positive                 1

The docs for pandas.DataFrame.replace are here.

Hessian

Answer 5

Given map is faster than replace (@JohnE's solution) you need to be careful with Non-Exhaustive mappings where you intend to map specific values to NaN. The proper method in this case requires that you mask the Series when you .fillna, else you undo the mapping to NaN.

import pandas as pd
import numpy as np

d = {'m': 'Male', 'f': 'Female', 'missing': np.NaN}
df = pd.DataFrame({'gender': ['m', 'f', 'missing', 'Male', 'U']})

---

keep_nan = [k for k,v in d.items() if pd.isnull(v)]
s = df['gender']

df['mapped'] = s.map(d).fillna(s.mask(s.isin(keep_nan)))

---

    gender  mapped
0        m    Male
1        f  Female
2  missing     NaN
3     Male    Male
4        U       U

Phaeton

could also do an update using the results of the map also i think. not sure which is faster – Public

Answer 6

TL;DR: Use map+fillna for large di and use replace for small di

---

1. An alternative: np.select()

If the remapping dictionary is not too large, another option is numpy.select. The syntax of np.select requires separate arrays/lists of conditions and replacement values, so the keys and values of di should be separated.

import numpy as np
df['col1'] = np.select((df[['col1']].values == list(di)).T, di.values(), df['col1'])

N.B. If the remapping dictionary di is very large, this may run into memory issues because as you can see from the line of code above, a boolean array of shape (len(df), len(di)) is required to evaluate the conditions.

2. map+fillna vs replace. Which is better?

If we look at the source code, if a dictionary is passed to it, map is an optimized method that calls a Cython-optimized take_nd() function to make replacements and fillna() calls where() (another optimized method) to fill values. On the other hand, replace() is implemented in Python and uses a loop over the dictionary. So if the dictionary is large, replace can potentially be thousands of times slower than map+fillna. Let's illustrate the difference by the following example where a single value (0) is replaced in the column (one using a dictionary of length 1000 (di1) and another using a dictionary of length 1 (di2)).

df = pd.DataFrame({'col1': range(1000)})
di1 = {k: k+1 for k in range(-1000, 1)}
di2 = {0: 1}

%timeit df['col1'].map(di1).fillna(df['col1'])
# 1.19 ms ± 6.77 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

%timeit df['col1'].replace(di1)
# 41.4 ms ± 400 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df['col1'].map(di2).fillna(df['col1'])
# 691 µs ± 27.9 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

%timeit df['col1'].replace(di2)
# 157 µs ± 3.34 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

As you can see, if len(di)==1000, replace is 35 times slower, but if len(di)==1, it's 4.5 times faster. This gap gets worse as the size of the remapping dictionary di increases.

In fact, if we look at the performance plots, we can make the following observations. The plots were drawn with particular parameters fixed in each graph. You can use the code down below to change the size of the dataframe to see for different parameters but it will produce very similar plots.

• For a given dataframe, map+fillna makes replacements in almost constant time regardless of the size of the remapping dictionary whereas replace does worse as the size of the remapping dictionary increases (top-left plot).
• The percentage of values being replaced in the dataframe has very little impact on the runtime difference. The impact of the length of di completely trumps whatever impact it has (top-right plot).
• For a given remapping dictionary, map+fillna performs better than replace as the size of the dataframe increases (bottom-left plot).
• Again, if di is large, the size of the dataframe does not matter; map+fillna is much faster than replace (bottom-right plot).

Code used to produce the plots:

import numpy as np
import pandas as pd
from perfplot import plot
import matplotlib.pyplot as plt

kernels = [lambda df,di: df['col1'].replace(di), 
           lambda df,di: df['col1'].map(di).fillna(df['col1'])]
labels = ["replace", "map+fillna"]


# first plot
N, m = 100000, 20
plot(
    setup=lambda n: (pd.DataFrame({'col1': np.resize(np.arange(m*n), N)}), 
                     {k: (k+1)/2 for k in range(n)}),
    kernels=kernels, labels=labels,
    n_range=range(1, 21),
    xlabel='Length of replacement dictionary',
    title=f'Remapping values in a column (len(df)={N:,}, {100//m}% replaced)',
    equality_check=pd.Series.equals)
_, xmax = plt.xlim()
plt.xlim((0.5, xmax+1))
plt.xticks(np.arange(1, xmax+1, 2));


# second plot
N, m = 100000, 1000
di = {k: (k+1)/2 for k in range(m)}
plot(
    setup=lambda n: pd.DataFrame({'col1': np.resize(np.arange((n-100)*m//100, n*m//100), N)}),
    kernels=kernels, labels=labels,
    n_range=[1, 5, 10, 15, 25, 40, 55, 75, 100],
    xlabel='Percentage of values replaced',
    title=f'Remapping values in a column (len(df)={N:,}, len(di)={m})',
    equality_check=pd.Series.equals);


# third plot
m, n = 10, 0.01
di = {k: (k+1)/2 for k in range(m)}
plot(
    setup=lambda N: pd.DataFrame({'col1': np.resize(np.arange((n-1)*m, n*m), N)}),
    kernels=kernels, labels=labels,
    n_range=[2**k for k in range(6, 21)], 
    xlabel='Length of dataframe',
    logy=False,
    title=f'Remapping values in a column (len(di)={m}, {int(n*100)}% replaced)',
    equality_check=pd.Series.equals);

# fourth plot
m, n = 100, 0.01
di = {k: (k+1)/2 for k in range(m)}
plot(
    setup=lambda N: pd.DataFrame({'col1': np.resize(np.arange((n-1)*m, n*m), N)}),
    kernels=kernels, labels=labels,
    n_range=[2**k for k in range(6, 21)], 
    xlabel='Length of dataframe',
    title=f'Remapping values in a column (len(di)={m}, {int(n*100)}% replaced)',
    equality_check=pd.Series.equals);

Ar

Answer 7

You can update your mapping dictionary with missing pairs from the dataframe. For example:

df = pd.DataFrame({'col1': ['a', 'b', 'c', 'd', np.nan]})
map_ = {'a': 'A', 'b': 'B', 'd': np.nan}

# Get mapping from df
uniques = df['col1'].unique()
map_new = dict(zip(uniques, uniques))
# {'a': 'a', 'b': 'b', 'c': 'c', 'd': 'd', nan: nan}

# Update mapping
map_new.update(map_)
# {'a': 'A', 'b': 'B', 'c': 'c', 'd': nan, nan: nan}

df['col2'] = df['col1'].map(dct_map_new)

Result:

  col1 col2
0    a    A
1    b    B
2    c    c
3    d  NaN
4  NaN  NaN

Transfigure

Answer 8

Adding to this question if you ever have more than one columns to remap in a data dataframe:

def remap(data,dict_labels):
    """
    This function take in a dictionnary of labels : dict_labels 
    and replace the values (previously labelencode) into the string.

    ex: dict_labels = {{'col1':{1:'A',2:'B'}}

    """
    for field,values in dict_labels.items():
        print("I am remapping %s"%field)
        data.replace({field:values},inplace=True)
    print("DONE")

    return data

Hope it can be useful to someone.

Cheers

Unhurried

This functionality is already provided by DataFrame.replace(), although I don't know when it was added. – Garneau

Answer 9

Or do apply:

df['col1'].apply(lambda x: {1: "A", 2: "B"}.get(x,x))

Demo:

>>> df['col1']=df['col1'].apply(lambda x: {1: "A", 2: "B"}.get(x,x))
>>> df
  col1 col2
0    w    a
1    1    2
2    2  NaN
>>> 

Casteel

Answer 10

A nice complete solution that keeps a map of your class labels:

labels = features['col1'].unique()
labels_dict = dict(zip(labels, range(len(labels))))
features = features.replace({"col1": labels_dict})

This way, you can at any point refer to the original class label from labels_dict.

Diocesan

Answer 11

As an extension to what have been proposed by Nico Coallier (apply to multiple columns) and U10-Forward(using apply style of methods), and summarising it into a one-liner I propose:

df.loc[:,['col1','col2']].transform(lambda x: x.map(lambda x: {1: "A", 2: "B"}.get(x,x))

The .transform() processes each column as a series. Contrary to .apply()which passes the columns aggregated in a DataFrame.

Consequently you can apply the Series method map().

Finally, and I discovered this behaviour thanks to U10, you can use the whole Series in the .get() expression. Unless I have misunderstood its behaviour and it processes sequentially the series instead of bitwisely.
The .get(x,x)accounts for the values you did not mention in your mapping dictionary which would be considered as Nan otherwise by the .map() method

Colchicine

The .transform() processes each column as a series. Contrary to .apply() which passes the columns aggregated in a DataFrame. I just tried, apply() works fine. There's no need to use loc either, this seems overly complex. df[["col1", "col2"]].apply(lambda col: col.map(lambda elem: my_dict.get(elem, elem))) should work just fine. The .get(x,x)accounts for the values you did not mention in your mapping dictionary which would be considered as Nan otherwise by the .map() method You could also use fillna() afterwards. – Garneau

Finally, and I discovered this behaviour thanks to U10, you can use the whole Series in the .get() expression. Unless I have misunderstood its behaviour and it processes sequentially the series instead of bitwisely. I can't reproduce this, can you elaborate? The identically named variables are likely playing some role here. – Garneau

Answer 12

A more native pandas approach is to apply a replace function as below:

def multiple_replace(dict, text):
  # Create a regular expression  from the dictionary keys
  regex = re.compile("(%s)" % "|".join(map(re.escape, dict.keys())))

  # For each match, look-up corresponding value in dictionary
  return regex.sub(lambda mo: dict[mo.string[mo.start():mo.end()]], text) 

Once you defined the function, you can apply it to your dataframe.

di = {1: "A", 2: "B"}
df['col1'] = df.apply(lambda row: multiple_replace(di, row['col1']), axis=1)

Annul

A more native pandas approach is to apply a replace function as below How is that more "native" (idiomatic?) than the much simpler methods provided by Pandas? – Garneau