Here is my df:
Net Upper Lower Mid Zsore
Answer option
More than once a da
I prefer this solution:
col = df.pop("Mid")
df.insert(0, col.name, col)
It's simpler to read and faster than other suggested answers.
def move_column_inplace(df, col, pos):
col = df.pop(col)
df.insert(pos, col.name, col)
Performance assessment:
For this test, the currently last column is moved to the front in each repetition. In-place methods generally perform better. While citynorman's solution can be made in-place, Ed Chum's method based on .loc
and sachinnm's method based on reindex
cannot.
While other methods are generic, citynorman's solution is limited to pos=0
. I didn't observe any performance difference between df.loc[cols]
and df[cols]
, which is why I didn't include some other suggestions.
I tested with python 3.6.8 and pandas 0.24.2 on a MacBook Pro (Mid 2015).
import numpy as np
import pandas as pd
n_cols = 11
df = pd.DataFrame(np.random.randn(200000, n_cols),
columns=range(n_cols))
def move_column_inplace(df, col, pos):
col = df.pop(col)
df.insert(pos, col.name, col)
def move_to_front_normanius_inplace(df, col):
move_column_inplace(df, col, 0)
return df
def move_to_front_chum(df, col):
cols = list(df)
cols.insert(0, cols.pop(cols.index(col)))
return df.loc[:, cols]
def move_to_front_chum_inplace(df, col):
col = df[col]
df.drop(col.name, axis=1, inplace=True)
df.insert(0, col.name, col)
return df
def move_to_front_elpastor(df, col):
cols = [col] + [ c for c in df.columns if c!=col ]
return df[cols] # or df.loc[cols]
def move_to_front_sachinmm(df, col):
cols = df.columns.tolist()
cols.insert(0, cols.pop(cols.index(col)))
df = df.reindex(columns=cols, copy=False)
return df
def move_to_front_citynorman_inplace(df, col):
# This approach exploits that reset_index() moves the index
# at the first position of the data frame.
df.set_index(col, inplace=True)
df.reset_index(inplace=True)
return df
def test(method, df):
col = np.random.randint(0, n_cols)
method(df, col)
col = np.random.randint(0, n_cols)
ret_mine = move_to_front_normanius_inplace(df.copy(), col)
ret_chum1 = move_to_front_chum(df.copy(), col)
ret_chum2 = move_to_front_chum_inplace(df.copy(), col)
ret_elpas = move_to_front_elpastor(df.copy(), col)
ret_sach = move_to_front_sachinmm(df.copy(), col)
ret_city = move_to_front_citynorman_inplace(df.copy(), col)
# Assert equivalence of solutions.
assert(ret_mine.equals(ret_chum1))
assert(ret_mine.equals(ret_chum2))
assert(ret_mine.equals(ret_elpas))
assert(ret_mine.equals(ret_sach))
assert(ret_mine.equals(ret_city))
Results:
# For n_cols = 11:
%timeit test(move_to_front_normanius_inplace, df)
# 1.05 ms ± 42.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.68 ms ± 46.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit test(move_to_front_sachinmm, df)
# 3.24 ms ± 96.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 3.84 ms ± 114 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_elpastor, df)
# 3.85 ms ± 58.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 9.67 ms ± 101 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# For n_cols = 31:
%timeit test(move_to_front_normanius_inplace, df)
# 1.26 ms ± 31.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_citynorman_inplace, df)
# 1.95 ms ± 260 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_sachinmm, df)
# 10.7 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum, df)
# 11.5 ms ± 869 µs per loop (mean ± std. dev. of 7 runs, 100 loops each
%timeit test(move_to_front_elpastor, df)
# 11.4 ms ± 598 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit test(move_to_front_chum_inplace, df)
# 31.4 ms ± 1.89 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Don't forget the two (()) 'brackets' around columns names.Otherwise, it'll give you an error.
# here you can add below line and it should work
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df
Mid Upper Lower Net Zsore
Answer option
More than once a day 2 0.22% -0.12% 0% 65
Once a day 3 0.32% -0.19% 0% 45
Several times a week 4 2.45% 1.10% 2% 78
Once a week 6 1.63% -0.40% 1% 65
We can use ix
to reorder by passing a list:
In [27]:
# get a list of columns
cols = list(df)
# move the column to head of list using index, pop and insert
cols.insert(0, cols.pop(cols.index('Mid')))
cols
Out[27]:
['Mid', 'Net', 'Upper', 'Lower', 'Zsore']
In [28]:
# use ix to reorder
df = df.ix[:, cols]
df
Out[28]:
Mid Net Upper Lower Zsore
Answer_option
More_than_once_a_day 2 0% 0.22% -0.12% 65
Once_a_day 3 0% 0.32% -0.19% 45
Several_times_a_week 4 2% 2.45% 1.10% 78
Once_a_week 6 1% 1.63% -0.40% 65
Another method is to take a reference to the column and reinsert it at the front:
In [39]:
mid = df['Mid']
df.drop(labels=['Mid'], axis=1,inplace = True)
df.insert(0, 'Mid', mid)
df
Out[39]:
Mid Net Upper Lower Zsore
Answer_option
More_than_once_a_day 2 0% 0.22% -0.12% 65
Once_a_day 3 0% 0.32% -0.19% 45
Several_times_a_week 4 2% 2.45% 1.10% 78
Once_a_week 6 1% 1.63% -0.40% 65
You can also use loc
to achieve the same result as ix
will be deprecated in a future version of pandas from 0.20.0 onwards:
df = df.loc[:, cols]