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?

7 Answers

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

9

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']

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

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

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!

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')

I extended this question that is how to gets the row, columnand 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

For known index candidate that we interested, a faster way by not checking the whole column can be done like this:

np.array(index_slice)[np.where(df.loc[index_slice]['column_name'] >= threshold)[0]]

Full comparison:

import pandas as pd
import numpy as np
index_slice = list(range(50,150)) # know index location for our inteterest
data = np.zeros(10000)
data[(index_slice)] = np.random.random(len(index_slice))
df = pd.DataFrame( {'column_name': data},
)
threshold = 0.5

%%timeit
np.array(index_slice)[np.where(df.loc[index_slice]['column_name'] >= threshold)[0]]
# 600 µs ± 1.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%%timeit
[i for i in index_slice if i in df.index[df['column_name'] >= threshold].tolist()]
# 22.5 ms ± 29.1 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

The way it works is like this:

# generate Boolean satisfy condition only in sliced column
df.loc[index_slice]['column_name'] >= threshold
# convert Boolean to index, but start from 0 and increment by 1
np.where(...)[0]
# list of index to be sliced
np.array(index_slice)[...]

Note: It needs to be noted that np.array(index_slice) can't be substituted by df.index due to np.where(...)[0] indexing start from 0 and increment by 1, but you can make something like df.index[index_slice]. And I think this is not worth the hassle if you just do it one time with small number of rows.