[python] Pandas: Appending a row to a dataframe and specify its index label

Is there any way to specify the index that I want for a new row, when appending the row to a dataframe?

The original documentation provides the following example:

In [1301]: df = DataFrame(np.random.randn(8, 4), columns=['A','B','C','D'])

In [1302]: df
Out[1302]: 
          A         B         C         D
0 -1.137707 -0.891060 -0.693921  1.613616
1  0.464000  0.227371 -0.496922  0.306389
2 -2.290613 -1.134623 -1.561819 -0.260838
3  0.281957  1.523962 -0.902937  0.068159
4 -0.057873 -0.368204 -1.144073  0.861209
5  0.800193  0.782098 -1.069094 -1.099248
6  0.255269  0.009750  0.661084  0.379319
7 -0.008434  1.952541 -1.056652  0.533946

In [1303]: s = df.xs(3)

In [1304]: df.append(s, ignore_index=True)
Out[1304]: 
          A         B         C         D
0 -1.137707 -0.891060 -0.693921  1.613616
1  0.464000  0.227371 -0.496922  0.306389
2 -2.290613 -1.134623 -1.561819 -0.260838
3  0.281957  1.523962 -0.902937  0.068159
4 -0.057873 -0.368204 -1.144073  0.861209
5  0.800193  0.782098 -1.069094 -1.099248
6  0.255269  0.009750  0.661084  0.379319
7 -0.008434  1.952541 -1.056652  0.533946
8  0.281957  1.523962 -0.902937  0.068159

where the new row gets the index label automatically. Is there any way to control the new label?

There is another solution. The next code is bad (although I think pandas needs this feature):

import pandas as pd

# empty dataframe
a = pd.DataFrame()
a.loc[0] = {'first': 111, 'second': 222}

But the next code runs fine:

import pandas as pd

# empty dataframe
a = pd.DataFrame()
a = a.append(pd.Series({'first': 111, 'second': 222}, name=0))

I shall refer to the same sample of data as posted in the question:

import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randn(8, 4), columns=['A','B','C','D'])
print('The original data frame is: \n{}'.format(df))

Running this code will give you

The original data frame is:

          A         B         C         D
0  0.494824 -0.328480  0.818117  0.100290
1  0.239037  0.954912 -0.186825 -0.651935
2 -1.818285 -0.158856  0.359811 -0.345560
3 -0.070814 -0.394711  0.081697 -1.178845
4 -1.638063  1.498027 -0.609325  0.882594
5 -0.510217  0.500475  1.039466  0.187076
6  1.116529  0.912380  0.869323  0.119459
7 -1.046507  0.507299 -0.373432 -1.024795

Now you wish to append a new row to this data frame, which doesn't need to be copy of any other row in the data frame. @Alon suggested an interesting approach to use df.loc to append a new row with different index. The issue, however, with this approach is if there is already a row present at that index, it will be overwritten by new values. This is typically the case for datasets when row index is not unique, like store ID in transaction datasets. So a more general solution to your question is to create the row, transform the new row data into a pandas series, name it to the index you want to have and then append it to the data frame. Don't forget to overwrite the original data frame with the one with appended row. The reason is df.append returns a view of the dataframe and does not modify its contents. Following is the code:

row = pd.Series({'A':10,'B':20,'C':30,'D':40},name=3)
df = df.append(row)
print('The new data frame is: \n{}'.format(df))

Following would be the new output:

The new data frame is:

           A          B          C          D
0   0.494824  -0.328480   0.818117   0.100290
1   0.239037   0.954912  -0.186825  -0.651935
2  -1.818285  -0.158856   0.359811  -0.345560
3  -0.070814  -0.394711   0.081697  -1.178845
4  -1.638063   1.498027  -0.609325   0.882594
5  -0.510217   0.500475   1.039466   0.187076
6   1.116529   0.912380   0.869323   0.119459
7  -1.046507   0.507299  -0.373432  -1.024795
3  10.000000  20.000000  30.000000  40.000000

df.loc will do the job :

>>> df = pd.DataFrame(np.random.randn(3, 2), columns=['A','B'])
>>> df
          A         B
0 -0.269036  0.534991
1  0.069915 -1.173594
2 -1.177792  0.018381
>>> df.loc[13] = df.loc[1]
>>> df
           A         B
0  -0.269036  0.534991
1   0.069915 -1.173594
2  -1.177792  0.018381
13  0.069915 -1.173594