pandas 将数据帧的特定列乘以常数标量值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35094256/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Multiplying specific columns of dataframe by constant scalar value
提问by Korean_Of_the_Mountain
How do I multiply only specific columns of a dataframe by a constant value?
如何仅将数据帧的特定列乘以常数值？
df0 = pd.DataFrame({'A' : 1.,
'B' : 1,
'C' : 1,
'D' : np.array([1] * 4,dtype='int32')})
mult_by_two = df0.iloc[:,2:].mul(2)
print mult_by_two
I get this:
我明白了：
C D
0 2 2
1 2 2
2 2 2
3 2 2
but what I want is this:
但我想要的是：
A B C D
0 1 1 2 2
1 1 1 2 2
2 1 1 2 2
3 1 1 2 2
回答by Mike Müller
You can assign the result to df0:
您可以将结果分配给df0：
>>> df0.iloc[:,2:] = df0.iloc[:,2:].mul(2)
>>> df0
A B C D
0 1 1 2 2
1 1 1 2 2
2 1 1 2 2
3 1 1 2 2
If you want a copy, make before the assigment:
如果您想要副本，请在分配之前制作：
df1 = df0.copy()
df1.iloc[:,2:] = df1.iloc[:,2:].mul(2)
回答by Anton Protopopov
If you need to multiply on scalar you don't need to call mulmethod you could use usual *operator:
如果您需要乘以标量，则不需要调用mul方法，您可以使用通常的*运算符：
In [24]: df0.iloc[:,2:] * 2
Out[24]:
C D
0 2 2
1 2 2
2 2 2
3 2 2
For your question you could use pd.concatwith your first columns and columns you are multiplying:
对于您的问题，您可以pd.concat将第一列和要相乘的列一起使用：
In [25]: pd.concat([df0.iloc[:,:2], df0.iloc[:,2:] * 2], axis=1)
Out[25]:
A B C D
0 1 1 2 2
1 1 1 2 2
2 1 1 2 2
3 1 1 2 2
