Remove strings based on condition pandas dataframe column
1
I have the following data:
df = pd.DataFrame({ 'Column_A': [1,2,3,4],
'Column_B': [["X1", "X2", "Y1"],
["X3", "Y2"],
["X4", "X5"],
["X5", "Y3", "Y4"]],})
Column_A Column_B
0 1 [X1, X2, Y1]
1 2 [X3, Y2]
2 3 [X4, X5]
3 4 [X5, Y3, Y4]
I wish to remove all strings starting with Y in the second column. Desired output:
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
python string pandas dataframe
asked Nov 23 '18 at 11:06
prmlmu
816
add a comment |
1
I have the following data:
df = pd.DataFrame({ 'Column_A': [1,2,3,4],
'Column_B': [["X1", "X2", "Y1"],
["X3", "Y2"],
["X4", "X5"],
["X5", "Y3", "Y4"]],})
Column_A Column_B
0 1 [X1, X2, Y1]
1 2 [X3, Y2]
2 3 [X4, X5]
3 4 [X5, Y3, Y4]
I wish to remove all strings starting with Y in the second column. Desired output:
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
python string pandas dataframe
asked Nov 23 '18 at 11:06
prmlmu
816
add a comment |
1
1
1
I have the following data:
df = pd.DataFrame({ 'Column_A': [1,2,3,4],
'Column_B': [["X1", "X2", "Y1"],
["X3", "Y2"],
["X4", "X5"],
["X5", "Y3", "Y4"]],})
Column_A Column_B
0 1 [X1, X2, Y1]
1 2 [X3, Y2]
2 3 [X4, X5]
3 4 [X5, Y3, Y4]
I wish to remove all strings starting with Y in the second column. Desired output:
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
python string pandas dataframe
asked Nov 23 '18 at 11:06
prmlmu
816
I have the following data:
df = pd.DataFrame({ 'Column_A': [1,2,3,4],
'Column_B': [["X1", "X2", "Y1"],
["X3", "Y2"],
["X4", "X5"],
["X5", "Y3", "Y4"]],})
Column_A Column_B
0 1 [X1, X2, Y1]
1 2 [X3, Y2]
2 3 [X4, X5]
3 4 [X5, Y3, Y4]
I wish to remove all strings starting with Y in the second column. Desired output:
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
python string pandas dataframe
python string pandas dataframe
asked Nov 23 '18 at 11:06
prmlmu
816
asked Nov 23 '18 at 11:06
prmlmu
816
asked Nov 23 '18 at 11:06
prmlmu
816
asked Nov 23 '18 at 11:06
prmlmu
816
asked Nov 23 '18 at 11:06
prmlmu
816
816
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
Use nested list comprehension with filtering with startswith:
df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
apply alternative:
df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
Or use filter:
df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
print (df)
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
Performance:
Depends of number of rows, number of values in lists and number of matched values:
#[40000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [142]: %timeit df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
23.7 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [143]: %timeit df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
36.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [144]: %timeit df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
30.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 23 '18 at 11:09
jezrael
320k22259338
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Use nested list comprehension with filtering with startswith:
df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
apply alternative:
df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
Or use filter:
df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
print (df)
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
Performance:
Depends of number of rows, number of values in lists and number of matched values:
#[40000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [142]: %timeit df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
23.7 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [143]: %timeit df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
36.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [144]: %timeit df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
30.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 23 '18 at 11:09
jezrael
320k22259338
add a comment |
2
Use nested list comprehension with filtering with startswith:
df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
apply alternative:
df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
Or use filter:
df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
print (df)
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
Performance:
Depends of number of rows, number of values in lists and number of matched values:
#[40000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [142]: %timeit df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
23.7 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [143]: %timeit df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
36.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [144]: %timeit df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
30.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 23 '18 at 11:09
jezrael
320k22259338
add a comment |
2
2
2
Use nested list comprehension with filtering with startswith:
df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
apply alternative:
df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
Or use filter:
df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
print (df)
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
Performance:
Depends of number of rows, number of values in lists and number of matched values:
#[40000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [142]: %timeit df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
23.7 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [143]: %timeit df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
36.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [144]: %timeit df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
30.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 23 '18 at 11:09
jezrael
320k22259338
Use nested list comprehension with filtering with startswith:
df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
apply alternative:
df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
Or use filter:
df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
print (df)
Column_A Column_B
0 1 [X1, X2]
1 2 [X3]
2 3 [X4, X5]
3 4 [X5]
Performance:
Depends of number of rows, number of values in lists and number of matched values:
#[40000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [142]: %timeit df['Column_B'] = [[y for y in x if not y.startswith('Y')] for x in df['Column_B']]
23.7 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [143]: %timeit df['Column_B'] = [list(filter(lambda y: not y.startswith('Y'), x)) for x in df['Column_B']]
36.5 ms ± 204 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [144]: %timeit df['Column_B'] = df['Column_B'].apply(lambda x: [y for y in x if not y.startswith('Y')])
30.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 23 '18 at 11:09
jezrael
320k22259338
edited Nov 23 '18 at 11:15
answered Nov 23 '18 at 11:09
jezrael
320k22259338
answered Nov 23 '18 at 11:09
jezrael
320k22259338
answered Nov 23 '18 at 11:09
jezrael
320k22259338
320k22259338
add a comment |
add a comment |
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