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
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
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
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
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)
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')
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
pyjanitor中的case_when函数是pd.Series.mask的包装器,并为多种条件提供了可链接/方便的形式:
对于单一条件:
df.case_when(
df.col1 == "Z", # condition
"green", # value if True
"red", # value if False
column_name = "color"
)
Type Set color
1 A Z green
2 B Z green
3 B X red
4 C Y red
适用于多种情况:
df.case_when(
df.Set.eq('Z') & df.Type.eq('A'), 'yellow', # condition, result
df.Set.eq('Z') & df.Type.eq('B'), 'blue', # condition, result
df.Type.eq('B'), 'purple', # condition, result
'black', # default if none of the conditions evaluate to True
column_name = 'color'
)
Type Set color
1 A Z yellow
2 B Z blue
3 B X purple
4 C Y black
更多的例子可以在这里找到