我有以下索引DataFrame命名列和行不连续的数字:

          a         b         c         d
2  0.671399  0.101208 -0.181532  0.241273
3  0.446172 -0.243316  0.051767  1.577318
5  0.614758  0.075793 -0.451460 -0.012493

我想添加一个新列,'e',到现有的数据帧,并不想改变数据帧中的任何东西(即,新列始终具有与DataFrame相同的长度)。

0   -0.335485
1   -1.166658
2   -0.385571
dtype: float64

如何将列e添加到上面的例子中?


当前回答

在最近的Pandas版本中,似乎应该使用df.assign:

df1 = df1.assign(e=np.random)

它不会产生SettingWithCopyWarning。

其他回答

如果你得到SettingWithCopyWarning,一个简单的解决方法是复制你想要添加列的数据帧。

df = df.copy()
df['col_name'] = values

如果数据帧和Series对象具有相同的索引,则为pandas。Concat也在这里工作:

import pandas as pd
df
#          a            b           c           d
#0  0.671399     0.101208   -0.181532    0.241273
#1  0.446172    -0.243316    0.051767    1.577318
#2  0.614758     0.075793   -0.451460   -0.012493

e = pd.Series([-0.335485, -1.166658, -0.385571])    
e
#0   -0.335485
#1   -1.166658
#2   -0.385571
#dtype: float64

# here we need to give the series object a name which converts to the new  column name 
# in the result
df = pd.concat([df, e.rename("e")], axis=1)
df

#          a            b           c           d           e
#0  0.671399     0.101208   -0.181532    0.241273   -0.335485
#1  0.446172    -0.243316    0.051767    1.577318   -1.166658
#2  0.614758     0.075793   -0.451460   -0.012493   -0.385571

以防它们没有相同的索引:

e.index = df.index
df = pd.concat([df, e.rename("e")], axis=1)

如果你要添加的列是一个序列变量,那么只需:

df["new_columns_name"]=series_variable_name #this will do it for you

即使要替换现有列,这种方法也很有效。只需输入与要替换的列相同的new_columns_name。它只会用新的系列数据覆盖现有的列数据。

我得到了可怕的SettingWithCopyWarning,它没有通过使用iloc语法修复。我的DataFrame是由read_sql从ODBC源创建的。根据上面low - tech的建议,以下方法对我来说是有效的:

df.insert(len(df.columns), 'e', pd.Series(np.random.randn(sLength),  index=df.index))

This worked fine to insert the column at the end. I don't know if it is the most efficient, but I don't like warning messages. I think there is a better solution, but I can't find it, and I think it depends on some aspect of the index. Note. That this only works once and will give an error message if trying to overwrite and existing column. Note As above and from 0.16.0 assign is the best solution. See documentation http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.assign.html#pandas.DataFrame.assign Works well for data flow type where you don't overwrite your intermediate values.

向pandas数据框架插入新列的4种方法

using simple assignment, insert(), assign() and Concat() methods.

import pandas as pd

df = pd.DataFrame({
    'col_a':[True, False, False], 
    'col_b': [1, 2, 3],
})
print(df)
    col_a  col_b
0   True     1
1  False     2
2  False     3

使用简单赋值

ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
print(ser)
0    a
1    b
2    c
dtype: object

df['col_c'] = pd.Series(['a', 'b', 'c'], index=[1, 2, 3])
print(df)
     col_a  col_b col_c
0   True     1  NaN
1  False     2    a
2  False     3    b

使用分配()

e = pd.Series([1.0, 3.0, 2.0], index=[0, 2, 1])
ser = pd.Series(['a', 'b', 'c'], index=[0, 1, 2])
df.assign(colC=s.values, colB=e.values)
     col_a  col_b col_c
0   True   1.0    a
1  False   3.0    b
2  False   2.0    c

使用insert ()

df.insert(len(df.columns), 'col_c', ser.values)
print(df)
    col_a  col_b col_c
0   True     1    a
1  False     2    b
2  False     3    c

使用concat ()

ser = pd.Series(['a', 'b', 'c'], index=[10, 20, 30])
df = pd.concat([df, ser.rename('colC')], axis=1)
print(df)
     col_a  col_b col_c
0    True   1.0  NaN
1   False   2.0  NaN
2   False   3.0  NaN
10    NaN   NaN    a
20    NaN   NaN    b
30    NaN   NaN    c