Set value to an entire column of a pandas dataframe


I’m trying to set the entire column of a dataframe to a specific value.

In  [1]: df
Out [1]: 
     issueid   industry
0        001        xxx
1        002        xxx
2        003        xxx
3        004        xxx
4        005        xxx

From what I’ve seen, loc is the best practice when replacing values in a dataframe (or isn’t it?):

In  [2]: df.loc[:,'industry'] = 'yyy'

However, I still received this much talked-about warning message:

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

If I do

In  [3]: df['industry'] = 'yyy'

I got the same warning message.

Any ideas? Working with Python 3.5.2 and pandas 0.18.1.

Asked By: Jingwei Yu



You can do :

df['industry'] = 'yyy'
Answered By: HH1

Assuming your Data frame is like ‘Data’ you have to consider if your data is a string or an integer. Both are treated differently. So in this case you need be specific about that.

import pandas as pd

data = [('001','xxx'), ('002','xxx'), ('003','xxx'), ('004','xxx'), ('005','xxx')]

df = pd.DataFrame(data,columns=['issueid', 'industry'])

print("Old DataFrame")

df.loc[:,'industry'] = str('yyy')

print("New DataFrame")

Now if want to put numbers instead of letters you must create and array

list_of_ones = [1,1,1,1,1]
df.loc[:,'industry'] = list_of_ones

Or if you are using Numpy

import numpy as np
n = len(df)
df.loc[:,'industry'] = np.ones(n)

Python can do unexpected things when new objects are defined from existing ones. You stated in a comment above that your dataframe is defined along the lines of df = df_all.loc[df_all['issueid']==specific_id,:]. In this case, df is really just a stand-in for the rows stored in the df_all object: a new object is NOT created in memory.

To avoid these issues altogether, I often have to remind myself to use the copy module, which explicitly forces objects to be copied in memory so that methods called on the new objects are not applied to the source object. I had the same problem as you, and avoided it using the deepcopy function.

In your case, this should get rid of the warning message:

from copy import deepcopy
df = deepcopy(df_all.loc[df_all['issueid']==specific_id,:])
df['industry'] = 'yyy'

EDIT: Also see David M.’s excellent comment below!

df = df_all.loc[df_all['issueid']==specific_id,:].copy()
df['industry'] = 'yyy'
Answered By: Alex P. Miller

This provides you with the possibility of adding conditions on the rows and then change all the cells of a specific column corresponding to those rows:

df.loc[(df['issueid'] == '001'), 'industry'] = str('yyy')
Answered By: Azim

You can use the assign function:

df = df.assign(industry='yyy')
Answered By: Mina HE
df.loc[:,'industry'] = 'yyy'

This does the magic. You are to add ‘.loc’ with ‘:’ for all rows. Hope it helps

Answered By: Nwoye CID

Seems to me that:

df1 = df[df['col1']==some_value] will not create a new DataFrame, basically, changes in df1 will be reflected in the parent df. This leads to the warning.
Whereas, df1 = df[df['col1]]==some_value].copy() will create a new DataFrame, and changes in df1 will not be reflected in df. The copy method is recommended if you don’t want to make changes to your original df.

Answered By: hukai916

I had a similar issue before even with this approach df.loc[:,'industry'] = 'yyy', but once I refreshed the notebook, it ran well.

You may want to try refreshing the cells after you have df.loc[:,'industry'] = 'yyy'.

Answered By: John Mutuma

if you just create new but empty data frame, you cannot directly sign a value to a whole column. This will show as NaN because the system wouldn’t know how many rows the data frame will have!You need to either define the size or have some existing columns.

df = pd.DataFrame()
df["A"] = 1
df["B"] = 2
df["C"] = 3
Answered By: Tao

Only use them instead:

df.iloc[:]['industry'] = 'yyy'

remember: this only works with exist columns in dataframe

this for people who didn’t work .loc

Answered By: Erwing Forero

For anyone else coming for this answer and doesn’t want to use copy –

df['industry'] = df['industry'].apply(lambda x: '')
Answered By: Abhinav Saini
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