你可以实例化一个字典的值，如果你想为每一列不同的值&你不介意在行之前创建一个字典。

>>> import pandas as pd
>>> import numpy as np
>>> df = pd.DataFrame({
  'col_1': [0, 1, 2, 3], 
  'col_2': [4, 5, 6, 7]
})
>>> df
   col_1  col_2
0      0      4
1      1      5
2      2      6
3      3      7
>>> cols = {
  'column_new_1':np.nan,
  'column_new_2':'dogs',
  'column_new_3': 3
}
>>> df[list(cols)] = pd.DataFrame(data={k:[v]*len(df) for k,v in cols.items()})
>>> df
   col_1  col_2  column_new_1 column_new_2  column_new_3
0      0      4           NaN         dogs             3
1      1      5           NaN         dogs             3
2      2      6           NaN         dogs             3
3      3      7           NaN         dogs             3

不一定比公认的答案更好，但这是另一种尚未列出的方法。

concat的用法:

In [128]: df
Out[128]: 
   col_1  col_2
0      0      4
1      1      5
2      2      6
3      3      7

In [129]: pd.concat([df, pd.DataFrame(columns = [ 'column_new_1', 'column_new_2','column_new_3'])])
Out[129]: 
   col_1  col_2 column_new_1 column_new_2 column_new_3
0    0.0    4.0          NaN          NaN          NaN
1    1.0    5.0          NaN          NaN          NaN
2    2.0    6.0          NaN          NaN          NaN
3    3.0    7.0          NaN          NaN          NaN

不太确定你想用它做什么。南,“狗”,3)。也许现在把它们设置为默认值?

In [142]: df1 = pd.concat([df, pd.DataFrame(columns = [ 'column_new_1', 'column_new_2','column_new_3'])])
In [143]: df1[[ 'column_new_1', 'column_new_2','column_new_3']] = [np.nan, 'dogs', 3]

In [144]: df1
Out[144]: 
   col_1  col_2  column_new_1 column_new_2  column_new_3
0    0.0    4.0           NaN         dogs             3
1    1.0    5.0           NaN         dogs             3
2    2.0    6.0           NaN         dogs             3
3    3.0    7.0           NaN         dogs             3

如果用相同的值添加很多缺失的列(a, b, c，....)，这里是0，我这样做:

    new_cols = ["a", "b", "c" ] 
    df[new_cols] = pd.DataFrame([[0] * len(new_cols)], index=df.index)

这是基于公认答案的第二种变体。

使用.assign()映射字典:

这是在处理多个列时用值分配新列的最具可读性和最动态的方式。

import pandas as pd
import numpy as np

new_cols = ["column_new_1", "column_new_2", "column_new_3"]
new_vals = [np.nan, "dogs", 3]
# Map new columns as keys and new values as values
col_val_mapping = dict(zip(new_cols, new_vals))
# Unpack new column/new value pairs and assign them to the data frame
df = df.assign(**col_val_mapping)

如果你只是想把新列的值初始化为空，因为你不知道这些值会是什么，或者你有很多新列。

import pandas as pd
import numpy as np

new_cols = ["column_new_1", "column_new_2", "column_new_3"]
new_vals = [None for item in new_cols]
# Map new columns as keys and new values as values
col_val_mapping = dict(zip(new_cols, new_vals))
# Unpack new column/new value pairs and assign them to the data frame
df = df.assign(**col_val_mapping)

使用列表理解，pd。DataFrame和pd.concat

pd.concat(
    [
        df,
        pd.DataFrame(
            [[np.nan, 'dogs', 3] for _ in range(df.shape[0])],
            df.index, ['column_new_1', 'column_new_2','column_new_3']
        )
    ], axis=1)

如果您只是想添加空的新列，重新索引将完成这项工作

df
   col_1  col_2
0      0      4
1      1      5
2      2      6
3      3      7

df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1)
   col_1  col_2  column_new_1  column_new_2  column_new_3
0      0      4           NaN           NaN           NaN
1      1      5           NaN           NaN           NaN
2      2      6           NaN           NaN           NaN
3      3      7           NaN           NaN           NaN

完整的代码示例

import numpy as np
import pandas as pd

df = {'col_1': [0, 1, 2, 3],
        'col_2': [4, 5, 6, 7]}
df = pd.DataFrame(df)
print('df',df, sep='\n')
print()
df=df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1)
print('''df.reindex(list(df)+['column_new_1', 'column_new_2','column_new_3'], axis=1)''',df, sep='\n')

否则就用赋值来赋0