Python Pandas: Get index of rows which column matches certain value

Question

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?

Answer 1

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

Answer 2

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']

Answer 3

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

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

Answer 4

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!

Answer 5

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')

Answer 6

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

Answer 7

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.