Question

任何人都可以在幕后解释数学吗？为什么Python和R会给我不同的结果？我应该将哪一个用于实际业务场景？

原始数据

id  cost    sales   item
1   300      50     pen
2   3        88     wf
3   1        70     gher
4   5        80     dger
5   2        999    ww

Python代码：

import pandas as pd
from sklearn.preprocessing import StandardScaler
df = pd.read_csv('Scale.csv')
df[['cost', 'sales']] = StandardScaler().fit_transform(df[['cost', 'sales']])
df

Python规范化结果

    id     cost        sales    item
0   1   1.999876    -0.559003   pen
1   2   -0.497867   -0.456582   wf
2   3   -0.514686   -0.505097   gher
3   4   -0.481047   -0.478144   dger
4   5   -0.506276   1.998826    ww

和R代码

library(readr)
library(dplyr)
df <- read_csv("C:/Users/Ho/Desktop/Scale.csv")
df <- df %>% mutate_each_(funs(scale(.) %>% as.vector), 
                             vars=c("cost","sales"))

R标准化结果

   id   cost     sales     item 

1   1   1.7887437   -0.4999873  pen
2   2   -0.4453054  -0.4083792  wf
3   3   -0.4603495  -0.4517725  gher
4   4   -0.4302613  -0.4276651  dger
5   5   -0.4528275  1.7878041   ww

谢谢@Wen

Answer 1

我没有在Python中使用这些函数，但数据似乎暗示不同之处在于Python中的函数使用了＆＃39; n＆＃39;在计算方差以标准化时，R使用＆＃39; n-1＆＃39;。我们可以通过乘法在两者之间进行转换，以下显示在乘以sqrt（5/4）后，R中的数据与Python值匹配。

> tab <- read.table(textConnection("1   1   1.7887437   -0.4999873  pen
+ 2   2   -0.4453054  -0.4083792  wf
+ 3   3   -0.4603495  -0.4517725  gher
+ 4   4   -0.4302613  -0.4276651  dger
+ 5   5   -0.4528275  1.7878041   ww"))
> tab
  V1 V2                   V3                   V4   V5
1  1  1  1.78874369999999994 -0.49998730000000002  pen
2  2  2 -0.44530540000000002 -0.40837920000000000   wf
3  3  3 -0.46034950000000002 -0.45177250000000002 gher
4  4  4 -0.43026130000000001 -0.42766510000000002 dger
5  5  5 -0.45282749999999999  1.78780410000000001   ww
> # To transform as if we used n in the denominator instead of
> # n-1 we just multiply by sqrt(n/(n-1))
> tab$V3 * sqrt(5/4)
[1]  1.99987625376224520 -0.49786657257386746 -0.51468638770401975
[4] -0.48104675744371517 -0.50627653604064304
> tab$V4 * sqrt(5/4)
[1] -0.55900279534329034 -0.45658182589849106 -0.50509701018251196
[4] -0.47814411760212272  1.99882574902641608

为什么Python和R有两种不同的标准化结果

1 个答案: