Adding new column to existing DataFrame in Python

Question

I have the following indexed DataFrame with named columns and rows not- continuous numbers:

          a         b         c         d
2  0.671399  0.101208 -0.181532  0.241273
3  0.446172 -0.243316  0.051767  1.577318
5  0.614758  0.075793 -0.451460 -0.012493

I would like to add a new column, 'e', to the existing data frame and do not want to change anything in the data frame (i.e., the new column always has the same length as the DataFrame).

0   -0.335485
1   -1.166658
2   -0.385571
dtype: float64

I tried different versions of join, append, merge, but I did not get the result I wanted, only errors at most. How can I add column e to the above example?

Answer 1

I was looking for a general way of adding a column of numpy.nans to a dataframe without getting the dumb SettingWithCopyWarning.

From the following:

• the answers here
• this question about passing a variable as a keyword argument
• this method for generating a numpy array of NaNs in-line

I came up with this:

col = 'column_name'
df = df.assign(**{col:numpy.full(len(df), numpy.nan)})

Answer 2

This is the simple way of adding a new column: df['e'] = e

Answer 3

To add a new column, 'e', to the existing data frame

 df1.loc[:,'e'] = Series(np.random.randn(sLength))

Answer 4

It seems that in recent Pandas versions the way to go is to use df.assign:

df1 = df1.assign(e=np.random.randn(sLength))

It doesn't produce SettingWithCopyWarning.

Answer 5

1. First create a python's list_of_e that has relevant data.
2. Use this: df['e'] = list_of_e

Answer 6

For the sake of completeness - yet another solution using DataFrame.eval() method:

Data:

In [44]: e
Out[44]:
0    1.225506
1   -1.033944
2   -0.498953
3   -0.373332
4    0.615030
5   -0.622436
dtype: float64

In [45]: df1
Out[45]:
          a         b         c         d
0 -0.634222 -0.103264  0.745069  0.801288
4  0.782387 -0.090279  0.757662 -0.602408
5 -0.117456  2.124496  1.057301  0.765466
7  0.767532  0.104304 -0.586850  1.051297
8 -0.103272  0.958334  1.163092  1.182315
9 -0.616254  0.296678 -0.112027  0.679112

Solution:

In [46]: df1.eval("e = @e.values", inplace=True)

In [47]: df1
Out[47]:
          a         b         c         d         e
0 -0.634222 -0.103264  0.745069  0.801288  1.225506
4  0.782387 -0.090279  0.757662 -0.602408 -1.033944
5 -0.117456  2.124496  1.057301  0.765466 -0.498953
7  0.767532  0.104304 -0.586850  1.051297 -0.373332
8 -0.103272  0.958334  1.163092  1.182315  0.615030
9 -0.616254  0.296678 -0.112027  0.679112 -0.622436