我如何添加一个颜色列到下面的数据框架,使颜色='绿色'如果设置== 'Z',和颜色='红色'否则?

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

当前回答

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')

其他回答

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

下面的方法比这里计时的方法慢,但是我们可以基于多个列的内容计算额外的列,并且可以为额外的列计算两个以上的值。

使用“Set”列的简单示例:

def set_color(row):
    if row["Set"] == "Z":
        return "red"
    else:
        return "green"

df = df.assign(color=df.apply(set_color, axis=1))

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

考虑到更多颜色和更多列的例子:

def set_color(row):
    if row["Set"] == "Z":
        return "red"
    elif row["Type"] == "C":
        return "blue"
    else:
        return "green"

df = df.assign(color=df.apply(set_color, axis=1))

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

编辑(21/06/2019):使用plydata

也可以使用plydata来做这类事情(不过,这似乎比使用assign和apply还要慢)。

from plydata import define, if_else

简单的if_else:

df = define(df, color=if_else('Set=="Z"', '"red"', '"green"'))

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

嵌套if_else:

df = define(df, color=if_else(
    'Set=="Z"',
    '"red"',
    if_else('Type=="C"', '"green"', '"blue"')))

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

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')

一个使用np.select的更简洁的方法:

a = np.array([['A','Z'],['B','Z'],['B','X'],['C','Y']])
df = pd.DataFrame(a,columns=['Type','Set'])

conditions = [
    df['Set'] == 'Z'
]

outputs = [
    'Green'
    ]
             # conditions Z is Green, Red Otherwise.
res = np.select(conditions, outputs, 'Red')
res 
array(['Green', 'Green', 'Red', 'Red'], dtype='<U5')
df.insert(2, 'new_column',res)    

df
    Type    Set new_column
0   A   Z   Green
1   B   Z   Green
2   B   X   Red
3   C   Y   Red

df.to_numpy()    
    
array([['A', 'Z', 'Green'],
       ['B', 'Z', 'Green'],
       ['B', 'X', 'Red'],
       ['C', 'Y', 'Red']], dtype=object)

%%timeit conditions = [df['Set'] == 'Z'] 
outputs = ['Green'] 
np.select(conditions, outputs, 'Red')

134 µs ± 9.71 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

df2 = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
%%timeit conditions = [df2['Set'] == 'Z'] 
outputs = ['Green'] 
np.select(conditions, outputs, 'Red')

188 ms ± 26.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

如果你在处理海量数据,记忆方法是最好的:

# First create a dictionary of manually stored values
color_dict = {'Z':'red'}

# Second, build a dictionary of "other" values
color_dict_other = {x:'green' for x in df['Set'].unique() if x not in color_dict.keys()}

# Next, merge the two
color_dict.update(color_dict_other)

# Finally, map it to your column
df['color'] = df['Set'].map(color_dict)

当您有许多重复的值时,这种方法将是最快的。我的一般经验法则是记住data_size > 10**4 & n_distinct < data_size/4

在一种情况下,记忆10,000行,不同值不超过2,500。