## The Question :

*321 people think this question is useful*

Given a DataFrame with a column “BoolCol”, we want to find the indexes of the DataFrame in which the values for “BoolCol” == True

I currently have the iterating way to do it, which works perfectly:

for i in range(100,3000):
if df.iloc[i]['BoolCol']== True:
print i,df.iloc[i]['BoolCol']

But this is not the correct panda’s way to do it.
After some research, I am currently using this code:

df[df['BoolCol'] == True].index.tolist()

This one gives me a list of indexes, but they dont match, when I check them by doing:

df.iloc[i]['BoolCol']

The result is actually False!!

Which would be the correct Pandas way to do this?

*The Question Comments :*

## The Answer 1

*504 people think this answer is useful*

`df.iloc[i]`

returns the `ith`

row of `df`

. `i`

does not refer to the index label, `i`

is a 0-based index.

In contrast, **the attribute **`index`

returns actual index labels, not numeric row-indices:

df.index[df['BoolCol'] == True].tolist()

or equivalently,

df.index[df['BoolCol']].tolist()

You can see the difference quite clearly by playing with a DataFrame with
a non-default index that does not equal to the row’s numerical position:

df = pd.DataFrame({'BoolCol': [True, False, False, True, True]},
index=[10,20,30,40,50])
In [53]: df
Out[53]:
BoolCol
10 True
20 False
30 False
40 True
50 True
[5 rows x 1 columns]
In [54]: df.index[df['BoolCol']].tolist()
Out[54]: [10, 40, 50]

**If you want to use the index**,

In [56]: idx = df.index[df['BoolCol']]
In [57]: idx
Out[57]: Int64Index([10, 40, 50], dtype='int64')

**then you can select the rows using **`loc`

instead of `iloc`

:

In [58]: df.loc[idx]
Out[58]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]

Note that `loc`

can also accept boolean arrays:

In [55]: df.loc[df['BoolCol']]
Out[55]:
BoolCol
10 True
40 True
50 True
[3 rows x 1 columns]

**If you have a boolean array, **`mask`

, and need ordinal index values, you can compute them using `np.flatnonzero`

:

In [110]: np.flatnonzero(df['BoolCol'])
Out[112]: array([0, 3, 4])

Use `df.iloc`

to select rows by ordinal index:

In [113]: df.iloc[np.flatnonzero(df['BoolCol'])]
Out[113]:
BoolCol
10 True
40 True
50 True

## The Answer 2

*35 people think this answer is useful*

Can be done using numpy where() function:

import pandas as pd
import numpy as np
In [716]: df = pd.DataFrame({"gene_name": ['SLC45A1', 'NECAP2', 'CLIC4', 'ADC', 'AGBL4'] , "BoolCol": [False, True, False, True, True] },
index=list("abcde"))
In [717]: df
Out[717]:
BoolCol gene_name
a False SLC45A1
b True NECAP2
c False CLIC4
d True ADC
e True AGBL4
In [718]: np.where(df["BoolCol"] == True)
Out[718]: (array([1, 3, 4]),)
In [719]: select_indices = list(np.where(df["BoolCol"] == True)[0])
In [720]: df.iloc[select_indices]
Out[720]:
BoolCol gene_name
b True NECAP2
d True ADC
e True AGBL4

Though you don’t always need index for a match, but incase if you need:

In [796]: df.iloc[select_indices].index
Out[796]: Index([u'b', u'd', u'e'], dtype='object')
In [797]: df.iloc[select_indices].index.tolist()
Out[797]: ['b', 'd', 'e']

## The Answer 3

*9 people think this answer is useful*

If you want to use your dataframe object only once, use:

df['BoolCol'].loc[lambda x: x==True].index

## The Answer 4

*3 people think this answer is useful*

Simple way is to reset the index of the DataFrame prior to filtering:

df_reset = df.reset_index()
df_reset[df_reset['BoolCol']].index.tolist()

Bit hacky, but it’s quick!

## The Answer 5

*1 people think this answer is useful*

First you may check `query`

when the target column is type `bool`

(PS: about how to use it please check link )

df.query('BoolCol')
Out[123]:
BoolCol
10 True
40 True
50 True

After we filter the original df by the Boolean column we can pick the index .

df=df.query('BoolCol')
df.index
Out[125]: Int64Index([10, 40, 50], dtype='int64')

Also pandas have `nonzero`

, we just select the *position* of `True`

row and using it slice the `DataFrame`

or `index`

df.index[df.BoolCol.nonzero()[0]]
Out[128]: Int64Index([10, 40, 50], dtype='int64')

## The Answer 6

*1 people think this answer is useful*

I extended this question that is how to gets the `row`

, `column`

and `value`

of all matches value?

here is solution:

import pandas as pd
import numpy as np
def search_coordinate(df_data: pd.DataFrame, search_set: set) -> list:
nda_values = df_data.values
tuple_index = np.where(np.isin(nda_values, [e for e in search_set]))
return [(row, col, nda_values[row][col]) for row, col in zip(tuple_index[0], tuple_index[1])]
if __name__ == '__main__':
test_datas = [['cat', 'dog', ''],
['goldfish', '', 'kitten'],
['Puppy', 'hamster', 'mouse']
]
df_data = pd.DataFrame(test_datas)
print(df_data)
result_list = search_coordinate(df_data, {'dog', 'Puppy'})
print(f"\n\n{'row':<4} {'col':<4} {'name':>10}")
[print(f"{row:<4} {col:<4} {name:>10}") for row, col, name in result_list]

Output:

0 1 2
0 cat dog
1 goldfish kitten
2 Puppy hamster mouse
row col name
0 1 dog
2 0 Puppy