Python: Pandas Dataframe how to multiply entire column with a scalar
Asked Answered
Q

13

116

How do I multiply each element of a given column of my dataframe with a scalar? (I have tried looking on SO, but cannot seem to find the right solution)

Doing something like:

df['quantity'] *= -1 # trying to multiply each row's quantity column with -1

gives me a warning:

A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

Note: If possible, I do not want to be iterating over the dataframe and do something like this...as I think any standard math operation on an entire column should be possible w/o having to write a loop:

for idx, row in df.iterrows():
    df.loc[idx, 'quantity'] *= -1

EDIT:

I am running 0.16.2 of Pandas

full trace:

 SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self.obj[item] = s
Quinze answered 17/11, 2015 at 22:17 Comment(6)
Check type of that column using dtype. I can't replicate that error, it's also good to give full traceback.Botheration
I've edited to add full trace...also its not an error, its a warning (for clarity)Quinze
I think it's being caused by something other than that line, or maybe that line is causing the warning to rise that was generated from earlier. What you're getting is related to slicing the dataframe.Botheration
Curious, did you ever figure this out? I'm dealing with the same problem.Atalante
At some point before this piece of code you have filtered df to reduce the number of rows or something. Perhaps you did df = BigDF.query("X == 1") or df = BigDF[BigDF.X == 1] or somesuch and that means df is actually just a view on BigDF. The warning is telling you that it is forcing it to make a copy, since otherwise it would cause a change in BigDF.Melainemelamed
So I had the same problem, my solution was make a copy of the df df2 = df.copy(). Then continue your code as before using the the copy df2['quantity'] *= -1. If I am doing something wrong please don't slaughter me, I am a beginner however this solution removed the warning for me. Please correct me if I am giving the incorrect solution.Winser
Q
57

Here's the answer after a bit of research:

df.loc[:,'quantity'] *= -1 #seems to prevent SettingWithCopyWarning 
Quinze answered 18/11, 2015 at 12:50 Comment(7)
This throws a SettingWithCopyWarning in pandas 0.18.0.Santos
Seems outrageous how many gotchas there are in Pandas, and how much easier this is in R: require(data.table); df[,quantity]*-1. No need to remember colons, .ix,.loc, iloc, quoting field names, nor updating copies when you meant to update the original.Leifeste
The real problem of why you are getting the error is not that there is anything with your code: you can use iloc, loc, or apply. The real problem that you have is due to how you created the df DataFrame. Most likely you created your df as a slice of another DataFrame without using .copy(). The correct way to create your df as a slice of another DataFrame is df = original_df.loc[some slicing].copy().Absa
@Absa is correct, almost all the answers here fail when operating on a slice of a dataframe.Hedberg
@Absa your explanation is very clear and really the error does not occur after using copy(). But, even if I modify the df which was made without copy(), the original_df is not modified. Why is that?Fayum
@starriet, Pandas does not change the original data, rather it throws the warning to let you know it didn't change the data: i.e. you changed a copy, rather than the original. This is to catch chained assignment, i.e. df["a"]["b"] = 1, which has unpredictable results due to python's interpreter. The pandas documentation has a discussion on the details and reasons for this.Linkous
@MattWalck Ah, thanks! So you mean, whether or not using copy(), Pandas tries to copy anyway, right? But if not using copy() explicitly, it throws an error to tell the user "Hey user, I think you thought the copying wouldn't happen, but in fact it does. Didn't you make a mistake?". Did I get you correctly?Fayum
N
88

try using apply function.

df['quantity'] = df['quantity'].apply(lambda x: x*-1)
Nonprofit answered 18/11, 2015 at 10:32 Comment(3)
this is pretty graceful when compared to looping, though I still get the SettingWithCopyWarningQuinze
Series.apply is a loop and should not be used for simple multiplication. The unnecessary lambda only makes it worse.Knuckleduster
@Knuckleduster What alternative do you propose?Mv
M
85

Note: for those using pandas 0.20.3 and above, and are looking for an answer, all these options will work:

df = pd.DataFrame(np.ones((5,6)),columns=['one','two','three',
                                       'four','five','six'])
df.one *=5
df.two = df.two*5
df.three = df.three.multiply(5)
df['four'] = df['four']*5
df.loc[:, 'five'] *=5
df.iloc[:, 5] = df.iloc[:, 5]*5

which results in

   one  two  three  four  five  six
0  5.0  5.0    5.0   5.0   5.0  5.0
1  5.0  5.0    5.0   5.0   5.0  5.0
2  5.0  5.0    5.0   5.0   5.0  5.0
3  5.0  5.0    5.0   5.0   5.0  5.0
4  5.0  5.0    5.0   5.0   5.0  5.0
Morris answered 26/9, 2017 at 23:41 Comment(2)
i tried this and my allocation which is running 1.2 sec now running in 0.05 secDemonetize
Also note, df.loc[:, 'five'] *=5 would NOT work it will put it in a new row (unlike df['four'] = df['four']*5 df.loc[:, 'five'] *=5Khasi
Q
57

Here's the answer after a bit of research:

df.loc[:,'quantity'] *= -1 #seems to prevent SettingWithCopyWarning 
Quinze answered 18/11, 2015 at 12:50 Comment(7)
This throws a SettingWithCopyWarning in pandas 0.18.0.Santos
Seems outrageous how many gotchas there are in Pandas, and how much easier this is in R: require(data.table); df[,quantity]*-1. No need to remember colons, .ix,.loc, iloc, quoting field names, nor updating copies when you meant to update the original.Leifeste
The real problem of why you are getting the error is not that there is anything with your code: you can use iloc, loc, or apply. The real problem that you have is due to how you created the df DataFrame. Most likely you created your df as a slice of another DataFrame without using .copy(). The correct way to create your df as a slice of another DataFrame is df = original_df.loc[some slicing].copy().Absa
@Absa is correct, almost all the answers here fail when operating on a slice of a dataframe.Hedberg
@Absa your explanation is very clear and really the error does not occur after using copy(). But, even if I modify the df which was made without copy(), the original_df is not modified. Why is that?Fayum
@starriet, Pandas does not change the original data, rather it throws the warning to let you know it didn't change the data: i.e. you changed a copy, rather than the original. This is to catch chained assignment, i.e. df["a"]["b"] = 1, which has unpredictable results due to python's interpreter. The pandas documentation has a discussion on the details and reasons for this.Linkous
@MattWalck Ah, thanks! So you mean, whether or not using copy(), Pandas tries to copy anyway, right? But if not using copy() explicitly, it throws an error to tell the user "Hey user, I think you thought the copying wouldn't happen, but in fact it does. Didn't you make a mistake?". Did I get you correctly?Fayum
S
27

More recent pandas versions have the pd.DataFrame.multiply function.

df['quantity'] = df['quantity'].multiply(-1)
Saturated answered 14/1, 2019 at 23:46 Comment(0)
A
22

The real problem of why you are getting the error is not that there is anything wrong with your code: you can use either iloc, loc, or apply, or *=, another of them could have worked.

The real problem that you have is due to how you created the df DataFrame. Most likely you created your df as a slice of another DataFrame without using .copy(). The correct way to create your df as a slice of another DataFrame is df = original_df.loc[some slicing].copy().

The problem is already stated in the error message you got " SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead"
You will get the same message in the most current version of pandas too.

Whenever you receive this kind of error message, you should always check how you created your DataFrame. Chances are you forgot the .copy()

Absa answered 13/11, 2019 at 2:23 Comment(2)
This should now be the accepted answer. Adding .copy() to the previous slicing operation is the key to prevent the mentioned warning.Prisilla
I wasted my time with most other answers. This answer helps understand the problem clearly and provides a simple solution to the problem with .copy(). Thank you !Forras
H
5

Try df['quantity'] = df['quantity'] * -1.

Hiles answered 17/11, 2015 at 22:23 Comment(1)
this is no different than df['quantity'] *= -1 (and yes I get the same warning)Quinze
B
5

A bit old, but I was still getting the same SettingWithCopyWarning. Here was my solution:

df.loc[:, 'quantity'] = df['quantity'] * -1
Belen answered 18/5, 2016 at 17:40 Comment(0)
L
4

A little late to the game, but for future searchers, this also should work:

df.quantity = df.quantity  * -1
Lacour answered 30/4, 2019 at 11:33 Comment(1)
Or df.quantity *= -1Stanstance
T
2

I got this warning using Pandas 0.22. You can avoid this by being very explicit using the assign method:

df = df.assign(quantity = df.quantity.mul(-1))
Toney answered 24/7, 2018 at 14:15 Comment(2)
this is the only mentioned solution that is working and doesn't throw the warningMientao
It is not working properly for me. Besides, the answer with apply function is way simple and better. So, why do you think we should use this solution?Wildeyed
S
1

You can use the index of the column you want to apply the multiplication for

df.loc[:,6] *= -1

This will multiply the column with index 6 with -1.

Speedboat answered 16/5, 2019 at 3:45 Comment(0)
B
0

Also it's possible to use numerical indeces with .iloc.

df.iloc[:,0]  *= -1
Balough answered 25/5, 2020 at 14:28 Comment(0)
C
0

Update 2022-08-10

Python: 3.10.5 - pandas: 1.4.3

As Mentioned in Previous comments, one the applicable approaches is using lambda. But, Be Careful with data types when using lambda approach.

Suppose you have a pandas Data Frame like this:

# Create List of lists
products = [[1010, 'Nokia', '200', 1800], [2020, 'Apple', '150', 3000], [3030, 'Samsung', '180', 2000]]

# Create the pandas DataFrame
df = pd.DataFrame(products, columns=['ProductId', 'ProductName', 'Quantity', 'Price'])

# print DataFrame
print(df)

   ProductId ProductName Quantity  Price
0       1010       Nokia      200   1800
1       2020       Apple      150   3000
2       3030     Samsung      180   2000

So, if you want to triple the value of Quantity for all rows in Products and use the following Statement:

# This statement considers the values of Quantity as string and updates the DataFrame
df['Quantity'] = df['Quantity'].apply(lambda x:x*3)

# print DataFrame
print(df)

The Result will be:

   ProductId ProductName   Quantity  Price
0       1010       Nokia  200200200   1800
1       2020       Apple  150150150   3000
2       3030     Samsung  180180180   2000

The above statement considers the values of Quantity as string.

So, in order to do the multiplication in the right way, the following statement with a convert could generate correct output:

# This statement considers the values of Quantity as integer and updates the DataFrame
df['Quantity'] = df['Quantity'].apply(lambda x:int(x)*3)

# print DataFrame
print(df)

Therefore the output will be like this:

   ProductId ProductName  Quantity  Price
0       1010       Nokia       600   1800
1       2020       Apple       450   3000
2       3030     Samsung       540   2000

I Hope this could help :)

Catling answered 10/8, 2022 at 12:38 Comment(0)
O
0

Here I'm providing code with output:

Code:

import pandas as pd

data = {
    'Name': ['Alice', 'Bob', 'Charlie', 'David', 'Emily'],
    'Age': [25, 30, 22, 35, 28],
    'Salary': [5000, 6000, 4000, 7000, 5500]
}

df = pd.DataFrame(data)

i_want_to_multiply = 2
df['Salary'] = df['Salary'] * i_want_to_multiply

print("Updated DataFrame:")
print(df)

Output:

Updated DataFrame:
Name  Age  Salary
0    Alice   25   10000
1      Bob   30   12000
2  Charlie   22    8000
3    David   35   14000
4    Emily   28   11000
Ozoniferous answered 2/1 at 7:46 Comment(0)

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