Python Pandas: Get index of rows which column matches certain value
PythonIndexingPandasPython Problem Overview
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?
Python Solutions
Solution 1 - Python
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
Solution 2 - Python
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']
Solution 3 - Python
If you want to use your dataframe object only once, use:
df['BoolCol'].loc[lambda x: x==True].index
Solution 4 - Python
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!
Solution 5 - Python
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')
Solution 6 - Python
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
Solution 7 - Python
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.