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 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.
