在熊猫数据框中移动列

cols = list(df.columns.values) #Make a list of all of the columns in the df
cols.pop(cols.index('b')) #Remove b from list
cols.pop(cols.index('x')) #Remove x from list
df = df[cols+['b','x']] #Create new dataframe with columns in the order you want

You can rearrange columns directly by specifying their order:

df = df[['a', 'y', 'b', 'x']]

In the case of larger dataframes where the column titles are dynamic, you can use a list comprehension to select every column not in your target set and then append the target set to the end.

>>> df[[c for c in df if c not in ['b', 'x']]
+ ['b', 'x']]
a  y  b   x
0  1 -1  2   3
1  2 -2  4   6
2  3 -3  6   9
3  4 -4  8  12

To make it more bullet proof, you can ensure that your target columns are indeed in the dataframe:

cols_at_end = ['b', 'x']
df = df[[c for c in df if c not in cols_at_end]
+ [c for c in cols_at_end if c in df]]

You can use to way below. It's very simple, but similar to the good answer given by Charlie Haley.

df1 = df.pop('b') # remove column b and store it in df1
df2 = df.pop('x') # remove column x and store it in df2
df['b']=df1 # add b series as a 'new' column.
df['x']=df2 # add b series as a 'new' column.

Now you have your dataframe with the columns 'b' and 'x' in the end. You can see this video from OSPY : https://youtu.be/RlbO27N3Xg4

You can also do this as a one-liner:

df.drop(columns=['b', 'x']).assign(b=df['b'], x=df['x'])

You can use pd.Index.difference with np.hstack, then reindex or use label-based indexing. In general, it's a good idea to avoid list comprehensions or other explicit loops with NumPy / Pandas objects.

cols_to_move = ['b', 'x']
new_cols = np.hstack((df.columns.difference(cols_to_move), cols_to_move))


# OPTION 1: reindex
df = df.reindex(columns=new_cols)


# OPTION 2: direct label-based indexing
df = df[new_cols]


# OPTION 3: loc label-based indexing
df = df.loc[:, new_cols]


print(df)


#    a  y  b   x
# 0  1 -1  2   3
# 1  2 -2  4   6
# 2  3 -3  6   9
# 3  4 -4  8  12

This function will reorder your columns without losing data. Any omitted columns remain in the center of the data set:

def reorder_columns(columns, first_cols=[], last_cols=[], drop_cols=[]):
columns = list(set(columns) - set(first_cols))
columns = list(set(columns) - set(drop_cols))
columns = list(set(columns) - set(last_cols))
new_order = first_cols + columns + last_cols
return new_order

Example usage:

my_list = ['first', 'second', 'third', 'fourth', 'fifth', 'sixth']
reorder_columns(my_list, first_cols=['fourth', 'third'], last_cols=['second'], drop_cols=['fifth'])


# Output:
['fourth', 'third', 'first', 'sixth', 'second']

To assign to your dataframe, use:

my_list = df.columns.tolist()
reordered_cols = reorder_columns(my_list, first_cols=['fourth', 'third'], last_cols=['second'], drop_cols=['fifth'])
df = df[reordered_cols]

An alternative, more generic method;

from pandas import DataFrame




def move_columns(df: DataFrame, cols_to_move: list, new_index: int) -> DataFrame:
"""
This method re-arranges the columns in a dataframe to place the desired columns at the desired index.
ex Usage: df = move_columns(df, ['Rev'], 2)
:param df:
:param cols_to_move: The names of the columns to move. They must be a list
:param new_index: The 0-based location to place the columns.
:return: Return a dataframe with the columns re-arranged
"""
other = [c for c in df if c not in cols_to_move]
start = other[0:new_index]
end = other[new_index:]
return df[start + cols_to_move + end]

similar to ROBBAT1's answer above, but hopefully a bit more robust:

df.insert(len(df.columns)-1, 'b', df.pop('b'))
df.insert(len(df.columns)-1, 'x', df.pop('x'))

I use Pokémon database as an example, the columns for my data base are

['Name', '#', 'Type 1', 'Type 2', 'Total', 'HP', 'Attack', 'Defense', 'Sp. Atk', 'Sp. Def', 'Speed', 'Generation', 'Legendary']

Here is the code:

import pandas as pd
df = pd.read_html('https://gist.github.com/armgilles/194bcff35001e7eb53a2a8b441e8b2c6')[0]
cols = df.columns.to_list()
cos_end= ["Name", "Total", "HP", "Defense"]


for i, j in enumerate(cos_end, start=(len(cols)-len(cos_end))):
cols.insert(i, cols.pop(cols.index(j)))
print(cols)
        

df = df.reindex(columns=cols)


print(df)

Simple solution:

old_cols = df.columns.values
new_cols= ['a', 'y', 'b', 'x']
df = df.reindex(columns=new_cols)

For example, to move column "name" to be the first column in df you can use insert:

column_to_move = df.pop("name")


# insert column with insert(location, column_name, column_value)


df.insert(0, "name", column_to_move)

similarly, if you want this column to be e.g. third column from the beginning:

df.insert(2, "name", column_to_move )

This will move any column to the last column :

1. Move any column to the last column of dataframe :

df= df[ [ col for col in df.columns if col != 'col_name_to_moved' ] + ['col_name_to_moved']]

2. Move any column to the first column of dataframe:

df= df[ ['col_name_to_moved'] + [ col for col in df.columns if col != 'col_name_to_moved' ]]

where col_name_to_moved is the column that you want to move.

You can use movecolumn package in Python to move columns:

pip install movecolumn

Then you can write your code as:

import movecolumn as mc
mc.MoveToLast(df,'b')
mc.MoveToLast(df,'x')

Hope that helps.

P.S : The package can be found here. https://pypi.org/project/movecolumn/