Getting the integer index of a Pandas DataFrame row fulfilling a condition?

Question

I have the following DataFrame:

   a  b  c b 2  1  2  3 5  4  5  6 

As you can see, column b is used as an index. I want to get the ordinal number of the row fulfilling ('b' == 5), which in this case would be 1.

The column being tested can be either an index column (as with b in this case) or a regular column, e.g. I may want to find the index of the row fulfilling ('c' == 6).

Answer 1

Use Index.get_loc instead.

Reusing @unutbu's set up code, you'll achieve the same results.

>>> import pandas as pd >>> import numpy as np   >>> df = pd.DataFrame(np.arange(1,7).reshape(2,3),                   columns = list('abc'),                   index=pd.Series([2,5], name='b')) >>> df    a  b  c b 2  1  2  3 5  4  5  6 >>> df.index.get_loc(5) 1 

Answer 2

You could use np.where like this:

import pandas as pd import numpy as np df = pd.DataFrame(np.arange(1,7).reshape(2,3),                   columns = list('abc'),                    index=pd.Series([2,5], name='b')) print(df) #    a  b  c # b          # 2  1  2  3 # 5  4  5  6 print(np.where(df.index==5)[0]) # [1] print(np.where(df['c']==6)[0]) # [1] 

The value returned is an array since there could be more than one row with a particular index or value in a column.