Find integer row-index from pandas index

Question

The following code find index where df['A'] == 1

import pandas as pd
import numpy as np
import random

index = range(10)
random.shuffle(index)
df = pd.DataFrame(np.zeros((10,1)).astype(int), columns = ['A'], index = index)

df.A.iloc[3:6] = 1
df.A.iloc[6:] = 2

print df

print df.loc[df['A'] == 1].index.tolist()

It returns pandas index correctly. How do I get the integer index ([3,4,5]) instead using pandas API?

   A
8  0
4  0
6  0
3  1
7  1
1  1
5  2
0  2
2  2
9  2
[3, 7, 1]

Answer 1

what about?

In [12]: df.index[df.A == 1]
Out[12]: Int64Index([3, 7, 1], dtype='int64')

or (depending on your goals):

In [15]: df.reset_index().index[df.A == 1]
Out[15]: Int64Index([3, 4, 5], dtype='int64')

Demo:

In [11]: df
Out[11]:
   A
8  0
4  0
6  0
3  1
7  1
1  1
5  2
0  2
2  2
9  2

In [12]: df.index[df.A == 1]
Out[12]: Int64Index([3, 7, 1], dtype='int64')

In [15]: df.reset_index().index[df.A == 1]
Out[15]: Int64Index([3, 4, 5], dtype='int64')

Answer 2

Here is one way:

df.reset_index().index[df.A == 1].tolist()

This re-indexes the data frame with [0, 1, 2, ...], then extracts the integer index values based on the boolean mask df.A == 1.

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Edit Credits to @Max for the index[df.A == 1] idea.

Answer 3

No need for numpy, you're right. Just pure python with a listcomp:

Just find the indexes where the values are 1

print([i for i,x in enumerate(df['A'].values) if x == 1])