我试图弄清楚如何同时添加多个列的熊猫与熊猫。我希望在一个步骤中做到这一点,而不是重复多个步骤。

import pandas as pd

df = {'col_1': [0, 1, 2, 3],
        'col_2': [4, 5, 6, 7]}
df = pd.DataFrame(df)

df[[ 'column_new_1', 'column_new_2','column_new_3']] = [np.nan, 'dogs',3]  # I thought this would work here...

当前回答

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。

其他回答

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

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)

在编写Pandas时,我的目标是编写可以链接的高效可读代码。我不会在这里解释为什么我这么喜欢链接,我在我的书《Effective Pandas》中对此进行了阐述。

我经常希望以简洁的方式添加新列,这也允许我进行链接。我的一般规则是使用.assign方法更新或创建列。

为了回答你的问题,我将使用以下代码:

(df
 .assign(column_new_1=np.nan,
         column_new_2='dogs',
         column_new_3=3
        )
)

再深入一点。我经常有一个数据框架,其中有我想要添加到我的数据框架的新列。让我们假设它看起来像…一个你想要的三列的数据框架:

df2 = pd.DataFrame({'column_new_1': np.nan,
                    'column_new_2': 'dogs',
                    'column_new_3': 3},
                   index=df.index
                  )

在这种情况下,我将编写以下代码:

(df
 .assign(**df2)
)
import pandas as pd
df = pd.DataFrame({
 'col_1': [0, 1, 2, 3], 
 'col_2': [4, 5, 6, 7]
 })
df['col_3'],  df['col_4'] =  [df.col_1]*2

>> df
col_1   col_2   col_3   col_4
0      4       0       0
1      5       1       1
2      6       2       2
3      7       3       3

你可以使用元组解包:

df = pd.DataFrame({'col1': [1, 2], 'col2': [3, 4]})

df['col3'], df['col4'] = 'a', 10

结果:

   col1  col2 col3  col4
0     1     3    a    10
1     2     4    a    10