Copy all values in a column to a new column in a pandas dataframe

This is a very basic question, I just can not seem to find an answer.

I have a dataframe like this, called df:

  A     B     C
 a.1   b.1   c.1
 a.2   b.2   c.2
 a.3   b.3   c.3

Then I extract all the rows from df, where column 'B' has a value of 'b.2'. I assign these results to df_2.

df_2 = df[df['B'] == 'b.2']

df_2 becomes:

  A     B     C
 a.2   b.2   c.2

Then, I copy all the values in column 'B' to a new column named 'D'. Causing df_2 to become:

  A     B     C     D
 a.2   b.2   c.2   b.2

When I preform an assignment like this:

df_2['D'] = df_2['B']

I get the following 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 > > See the the caveats in the documentation: > http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

---

I have also tried using .loc when creating df_2 like this:

df_2 = df.loc[df['B'] == 'b.2']

However, I still get the warning.

Any help is greatly appreciated.

You can simply assign the B to the new column , Like -

df['D'] = df['B']

---

Example/Demo -

In [1]: import pandas as pd

In [2]: df = pd.DataFrame([['a.1','b.1','c.1'],['a.2','b.2','c.2'],['a.3','b.3','c.3']],columns=['A','B','C'])

In [3]: df
Out[3]:
     A    B    C
0  a.1  b.1  c.1
1  a.2  b.2  c.2
2  a.3  b.3  c.3

In [4]: df['D'] = df['B']                  #<---What you want.

In [5]: df
Out[5]:
     A    B    C    D
0  a.1  b.1  c.1  b.1
1  a.2  b.2  c.2  b.2
2  a.3  b.3  c.3  b.3

In [6]: df.loc[0,'D'] = 'd.1'

In [7]: df
Out[7]:
     A    B    C    D
0  a.1  b.1  c.1  d.1
1  a.2  b.2  c.2  b.2
2  a.3  b.3  c.3  b.3

The problem is in the line before the one that throws the warning. When you create df_2 that's where you're creating a copy of a slice of a dataframe. Instead, when you create df_2, use .copy() and you won't get that warning later on.

df_2 = df[df['B'] == 'b.2'].copy()

I think the correct access method is using the index:

df_2.loc[:,'D'] = df_2['B']

How about:

df['D'] = df['B'].values

Here is your dataframe:

import pandas as pd
df = pd.DataFrame({
    'A': ['a.1', 'a.2', 'a.3'],
    'B': ['b.1', 'b.2', 'b.3'],
    'C': ['c.1', 'c.2', 'c.3']})

Your answer is in the paragraph "Setting with enlargement" in the section on "Indexing and selecting data" in the documentation on Pandas.

It says:

> A DataFrame can be enlarged on either axis via .loc.

So what you need to do is simply one of these two:

df.loc[:, 'D'] = df.loc[:, 'B']
df.loc[:, 'D'] = df['B']

You can use the method assign. It returns a new DataFrame so you can use it in chains with other methods.

df.assign(D=df.B)

Output:

     A    B    C    D
0  a.1  b.1  c.1  b.1
1  a.2  b.2  c.2  b.2
2  a.3  b.3  c.3  b.3

The question was asked a while ago, but my response could help others.

I had a similar situation. When you sliced a dataframe into df_2, you need to reset index,

df_2 = df_2.reset_index(drop = True)  

Now you can run the command without warning

df_2['D'] = df_2['B']

Following up on these solutions, here is some helpful code illustrating :

#
# Copying columns in pandas without slice warning
#
import numpy as np
df = pd.DataFrame(np.random.randn(10, 3), columns=list('ABC'))

#
# copies column B into new column D
df.loc[:,'D'] = df['B']
print df

#
# creates new column 'E' with values -99
# 
# But copy command replaces those where 'B'>0 while others become NaN (not copied)
df['E'] = -99
print df
df['E'] = df[df['B']>0]['B'].copy()
print df

#
# creates new column 'F' with values -99
# 
# Copy command only overwrites values which meet criteria 'B'>0
df['F']=-99
df.loc[df['B']>0,'F'] = df[df['B']>0]['B'].copy()
print df

eval lets you assign B to the new columns D right away:

In [8]: df.eval('D=B', inplace=True)

In [9]: df
Out[9]: 
     A    B    C    D
0  a.1  b.1  c.1  b.1
1  a.2  b.2  c.2  b.2
2  a.3  b.3  c.3  b.3

Since inplace=True you don't need to assign it back to df.