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?
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
, 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
Source: Stackoverflow.com