使用.apply()方法的一行代码如下:

df['color'] = df['Set'].apply(lambda set_: 'green' if set_=='Z' else 'red')

之后，df数据帧是这样的:

>>> print(df)
  Type Set  color
0    A   Z  green
1    B   Z  green
2    B   X    red
3    C   Y    red

如果你只有两种选择:

df['color'] = np.where(df['Set']=='Z', 'green', 'red')

例如,

import pandas as pd
import numpy as np

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
df['color'] = np.where(df['Set']=='Z', 'green', 'red')
print(df)

收益率

  Set Type  color
0   Z    A  green
1   Z    B  green
2   X    B    red
3   Y    C    red

如果你有两个以上的条件，那么使用np.select。例如，如果你想要颜色

黄色时(df['设置']= = ' Z ') & (df(“类型”)= =“一”) 否则蓝色当(df['设置']= = ' Z ') & (df(“类型”)= = ' B ') 否则为紫色，当(df['Type'] == 'B') 否则黑,

然后使用

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
conditions = [
    (df['Set'] == 'Z') & (df['Type'] == 'A'),
    (df['Set'] == 'Z') & (df['Type'] == 'B'),
    (df['Type'] == 'B')]
choices = ['yellow', 'blue', 'purple']
df['color'] = np.select(conditions, choices, default='black')
print(df)

的收益率

  Set Type   color
0   Z    A  yellow
1   Z    B    blue
2   X    B  purple
3   Y    C   black

你可以使用pandas方法:

df['color'] = 'green'
df['color'] = df['color'].where(df['Set']=='Z', other='red')
# Replace values where the condition is False

or

df['color'] = 'red'
df['color'] = df['color'].mask(df['Set']=='Z', other='green')
# Replace values where the condition is True

或者，你也可以使用lambda函数的transform方法:

df['color'] = df['Set'].transform(lambda x: 'green' if x == 'Z' else 'red')

输出:

  Type Set  color
1    A   Z  green
2    B   Z  green
3    B   X    red
4    C   Y    red

@chai的性能比较:

import pandas as pd
import numpy as np
df = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
 
%timeit df['color1'] = 'red'; df['color1'].where(df['Set']=='Z','green')
%timeit df['color2'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color3'] = np.where(df['Set']=='Z', 'red', 'green')
%timeit df['color4'] = df.Set.map(lambda x: 'red' if x == 'Z' else 'green')

397 ms ± 101 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
976 ms ± 241 ms per loop
673 ms ± 139 ms per loop
796 ms ± 182 ms per loop

另一种实现这一目标的方法是

df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')

您可以简单地使用强大的.loc方法，并根据需要使用一个或多个条件(使用pandas=1.0.5进行测试)。

代码总结:

df=pd.DataFrame(dict(Type='A B B C'.split(), Set='Z Z X Y'.split()))
df['Color'] = "red"
df.loc[(df['Set']=="Z"), 'Color'] = "green"

#practice!
df.loc[(df['Set']=="Z")&(df['Type']=="B")|(df['Type']=="C"), 'Color'] = "purple"

解释:

df=pd.DataFrame(dict(Type='A B B C'.split(), Set='Z Z X Y'.split()))

# df so far: 
  Type Set  
0    A   Z 
1    B   Z 
2    B   X 
3    C   Y

添加“color”列，并将所有值设置为“red”

df['Color'] = "red"

应用你的单一条件:

df.loc[(df['Set']=="Z"), 'Color'] = "green"


# df: 
  Type Set  Color
0    A   Z  green
1    B   Z  green
2    B   X    red
3    C   Y    red

或者多重条件:

df.loc[(df['Set']=="Z")&(df['Type']=="B")|(df['Type']=="C"), 'Color'] = "purple"

你可以在这里阅读Pandas逻辑运算符和条件选择: Pandas中用于布尔索引的逻辑运算符

列表推导式是有条件地创建另一列的另一种方法。如果您在列中使用对象dtype，就像您的示例一样，列表推导式通常优于大多数其他方法。

示例列表理解:

df['color'] = ['red' if x == 'Z' else 'green' for x in df['Set']]

%时间它测试：

import pandas as pd
import numpy as np

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
%timeit df['color'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color'] = np.where(df['Set']=='Z', 'green', 'red')
%timeit df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')

1000 loops, best of 3: 239 µs per loop
1000 loops, best of 3: 523 µs per loop
1000 loops, best of 3: 263 µs per loop