Pandas - subtring a column where position number in separate column
1
Dataset
df = pd.DataFrame({'a': [0,3,4], 'b': ['0101010', '0100010', '0111100']})
Basically trying to create a column where it takes the substring of length 1 of column b starting at position number in column a
Attempt
position = df['a']
df['c'] = df['b'].str[position]
Desired Output
a b c
0 0101010 0
3 0100010 0
4 0111100 1
pandas dataframe
asked Nov 20 '18 at 12:22
AK91
777
add a comment |
1
Dataset
df = pd.DataFrame({'a': [0,3,4], 'b': ['0101010', '0100010', '0111100']})
Basically trying to create a column where it takes the substring of length 1 of column b starting at position number in column a
Attempt
position = df['a']
df['c'] = df['b'].str[position]
Desired Output
a b c
0 0101010 0
3 0100010 0
4 0111100 1
pandas dataframe
asked Nov 20 '18 at 12:22
AK91
777
add a comment |
1
1
1
Dataset
df = pd.DataFrame({'a': [0,3,4], 'b': ['0101010', '0100010', '0111100']})
Basically trying to create a column where it takes the substring of length 1 of column b starting at position number in column a
Attempt
position = df['a']
df['c'] = df['b'].str[position]
Desired Output
a b c
0 0101010 0
3 0100010 0
4 0111100 1
pandas dataframe
asked Nov 20 '18 at 12:22
AK91
777
Dataset
df = pd.DataFrame({'a': [0,3,4], 'b': ['0101010', '0100010', '0111100']})
Basically trying to create a column where it takes the substring of length 1 of column b starting at position number in column a
Attempt
position = df['a']
df['c'] = df['b'].str[position]
Desired Output
a b c
0 0101010 0
3 0100010 0
4 0111100 1
pandas dataframe
pandas dataframe
asked Nov 20 '18 at 12:22
AK91
777
asked Nov 20 '18 at 12:22
AK91
777
asked Nov 20 '18 at 12:22
AK91
777
asked Nov 20 '18 at 12:22
AK91
777
asked Nov 20 '18 at 12:22
AK91
777
777
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
Use list comprehension with zip:
df['c'] = [b[a] for a, b in zip(df.a, df.b)]
Or apply:
df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
print (df)
a b c
0 0 0101010 0
1 3 0100010 0
2 4 0111100 1
Performance is different:
#[3000 rows x 2 columns]
df = pd.concat([df] * 1000, ignore_index=True)
In [236]: %timeit df['c'] = [b[a] for a, b in zip(df.a, df.b)]
557 µs ± 25.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [237]: %timeit df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
57.3 ms ± 358 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 20 '18 at 12:24
jezrael
321k22263341
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
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
3
Use list comprehension with zip:
df['c'] = [b[a] for a, b in zip(df.a, df.b)]
Or apply:
df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
print (df)
a b c
0 0 0101010 0
1 3 0100010 0
2 4 0111100 1
Performance is different:
#[3000 rows x 2 columns]
df = pd.concat([df] * 1000, ignore_index=True)
In [236]: %timeit df['c'] = [b[a] for a, b in zip(df.a, df.b)]
557 µs ± 25.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [237]: %timeit df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
57.3 ms ± 358 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 20 '18 at 12:24
jezrael
321k22263341
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
add a comment |
3
Use list comprehension with zip:
df['c'] = [b[a] for a, b in zip(df.a, df.b)]
Or apply:
df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
print (df)
a b c
0 0 0101010 0
1 3 0100010 0
2 4 0111100 1
Performance is different:
#[3000 rows x 2 columns]
df = pd.concat([df] * 1000, ignore_index=True)
In [236]: %timeit df['c'] = [b[a] for a, b in zip(df.a, df.b)]
557 µs ± 25.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [237]: %timeit df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
57.3 ms ± 358 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 20 '18 at 12:24
jezrael
321k22263341
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
add a comment |
3
3
3
Use list comprehension with zip:
df['c'] = [b[a] for a, b in zip(df.a, df.b)]
Or apply:
df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
print (df)
a b c
0 0 0101010 0
1 3 0100010 0
2 4 0111100 1
Performance is different:
#[3000 rows x 2 columns]
df = pd.concat([df] * 1000, ignore_index=True)
In [236]: %timeit df['c'] = [b[a] for a, b in zip(df.a, df.b)]
557 µs ± 25.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [237]: %timeit df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
57.3 ms ± 358 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 20 '18 at 12:24
jezrael
321k22263341
Use list comprehension with zip:
df['c'] = [b[a] for a, b in zip(df.a, df.b)]
Or apply:
df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
print (df)
a b c
0 0 0101010 0
1 3 0100010 0
2 4 0111100 1
Performance is different:
#[3000 rows x 2 columns]
df = pd.concat([df] * 1000, ignore_index=True)
In [236]: %timeit df['c'] = [b[a] for a, b in zip(df.a, df.b)]
557 µs ± 25.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [237]: %timeit df['c'] = df.apply(lambda x: x['b'][x['a']], axis=1)
57.3 ms ± 358 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
answered Nov 20 '18 at 12:24
jezrael
321k22263341
answered Nov 20 '18 at 12:24
jezrael
321k22263341
answered Nov 20 '18 at 12:24
jezrael
321k22263341
answered Nov 20 '18 at 12:24
jezrael
321k22263341
321k22263341
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
add a comment |
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
1
1
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
@jezreal - legend, thanks
– AK91
Nov 20 '18 at 12:51
add a comment |
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