Find integer row-index from pandas index
<p>The following code find index where df['A'] == 1</p>

<pre><code>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()
</code></pre>

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

<pre><code>   A
8  0
4  0
6  0
3  1
7  1
1  1
5  2
0  2
2  2
9  2
[3, 7, 1]
</code></pre>

<p>Here is one way:</p>

<pre><code>df.reset_index().index[df.A == 1].tolist()
</code></pre>

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

<hr>

<p><strong>Edit</strong> Credits to @Max for the <code>index[df.A == 1]</code> idea.</p>

python pandas dataframe row indices