如何使用具有 NA 值的 scale() 函数计算 z-score
how to calculate z-score using scale() function with NA values
我有一个包含 98790 obs 的数据框。 143 个变量。它包含数字和 NA。我想为每一行执行 z-score。我尝试了以下方法：
>df
sample1 sample2 sample3 sample4 sample5 sampl6 sample7 sample8
1: 6.96123 3.021311 NA NA 7.464205 7.902878 -1.194076 7.771018
2: NA NA NA NA NA NA NA NA
3: NA NA NA NA NA NA 2.784635 NA
4: NA NA 8.342075 NA 8.464205 NA 6.462707 7.118941
5: NA 7.243703 10.149430 NA NA 8.317915 NA NA
并且：
>res <- t(scale(t(df)))
上述函数会忽略所有NA并计算z-score吗？如果不是，我如何在不考虑 NAs 的情况下计算 z 分数？
您可能希望在 transposing/scaling/re-transposing 之前转换为矩阵（数据框 -> 矩阵 -> 转置 -> 缩放 -> 转置 -> 数据框）
否则，似乎工作正常。这是一个包含一些 NA 值的示例：
set.seed(101)
m <- matrix(rnorm(25),5,5)
m[sample(1:25,size=8)] <- NA
m
## [,1] [,2] [,3] [,4] [,5]
## [1,] -0.3260365 NA 0.5264481 -0.1933380 NA
## [2,] 0.5524619 NA -0.7948444 -0.8497547 0.7085221
## [3,] -0.6749438 NA 1.4277555 0.0584655 -0.2679805
## [4,] 0.2143595 NA -1.4668197 -0.8176704 -1.4639218
## [5,] NA NA -0.2366834 NA 0.7444358
scale(m)
## [,1] [,2] [,3] [,4] [,5]
## [1,] -0.4885685 NA 0.5628440 0.5661203 NA
## [2,] 1.1159619 NA -0.6077977 -0.8785073 0.7475404
## [3,] -1.1258292 NA 1.3613864 1.1202838 -0.1904198
## [4,] 0.4984359 NA -1.2031558 -0.8078967 -1.3391573
## [5,] NA NA -0.1132769 NA 0.7820366
## attr(,"scaled:center")
## [1] -0.05853976 NaN -0.10882877 -0.45057439 -0.06973609
## attr(,"scaled:scale")
## [1] 0.5475112 0.0000000 1.1286908 0.4543848 1.0410918
文档 (?scale) 也非常明确地说明了 NA 值的处理方式：
... centering is done by subtracting the
column means (omitting ‘NA’s) of ‘x’ from their corresponding
columns ...
... the root-mean-square for a (possibly centered) column is defined
as sqrt(sum(x^2)/(n-1)), where x is a vector of the non-missing
values and n is the number of non-missing values ...
（强调）
我有一个包含 98790 obs 的数据框。 143 个变量。它包含数字和 NA。我想为每一行执行 z-score。我尝试了以下方法：
>df
sample1 sample2 sample3 sample4 sample5 sampl6 sample7 sample8
1: 6.96123 3.021311 NA NA 7.464205 7.902878 -1.194076 7.771018
2: NA NA NA NA NA NA NA NA
3: NA NA NA NA NA NA 2.784635 NA
4: NA NA 8.342075 NA 8.464205 NA 6.462707 7.118941
5: NA 7.243703 10.149430 NA NA 8.317915 NA NA
并且：
>res <- t(scale(t(df)))
上述函数会忽略所有NA并计算z-score吗？如果不是，我如何在不考虑 NAs 的情况下计算 z 分数？
您可能希望在 transposing/scaling/re-transposing 之前转换为矩阵（数据框 -> 矩阵 -> 转置 -> 缩放 -> 转置 -> 数据框）
否则，似乎工作正常。这是一个包含一些 NA 值的示例：
set.seed(101)
m <- matrix(rnorm(25),5,5)
m[sample(1:25,size=8)] <- NA
m
## [,1] [,2] [,3] [,4] [,5]
## [1,] -0.3260365 NA 0.5264481 -0.1933380 NA
## [2,] 0.5524619 NA -0.7948444 -0.8497547 0.7085221
## [3,] -0.6749438 NA 1.4277555 0.0584655 -0.2679805
## [4,] 0.2143595 NA -1.4668197 -0.8176704 -1.4639218
## [5,] NA NA -0.2366834 NA 0.7444358
scale(m)
## [,1] [,2] [,3] [,4] [,5]
## [1,] -0.4885685 NA 0.5628440 0.5661203 NA
## [2,] 1.1159619 NA -0.6077977 -0.8785073 0.7475404
## [3,] -1.1258292 NA 1.3613864 1.1202838 -0.1904198
## [4,] 0.4984359 NA -1.2031558 -0.8078967 -1.3391573
## [5,] NA NA -0.1132769 NA 0.7820366
## attr(,"scaled:center")
## [1] -0.05853976 NaN -0.10882877 -0.45057439 -0.06973609
## attr(,"scaled:scale")
## [1] 0.5475112 0.0000000 1.1286908 0.4543848 1.0410918
文档 (?scale) 也非常明确地说明了 NA 值的处理方式：
... centering is done by subtracting the column means (omitting ‘NA’s) of ‘x’ from their corresponding columns ...
... the root-mean-square for a (possibly centered) column is defined as sqrt(sum(x^2)/(n-1)), where x is a vector of the non-missing values and n is the number of non-missing values ...
（强调）