尝试使用dplyr到group_by并应用scale()

Trying to use dplyr to group_by the stud_ID variable in the following data frame, as in this SO question:

尝试在以下数据框架中使用dplyr到group_by的stud_ID变量，如在这个SO问题中:

> str(df)
'data.frame':   4136 obs. of  4 variables:
 $ stud_ID         : chr  "ABB112292" "ABB112292" "ABB112292" "ABB112292" ...
 $ behavioral_scale: num  3.5 4 3.5 3 3.5 2 NA NA 1 2 ...
 $ cognitive_scale : num  3.5 3 3 3 3.5 2 NA NA 1 1 ...
 $ affective_scale : num  2.5 3.5 3 3 2.5 2 NA NA 1 1.5 ...

I tried the following to obtain scale scores by student (rather than scale scores for observations across all students):

我尝试了以下方法来获得学生的分数(而不是所有学生的观察分数):

scaled_data <- 
          df %>%
              group_by(stud_ID) %>%
                  mutate(behavioral_scale_ind = scale(behavioral_scale),
                         cognitive_scale_ind = scale(cognitive_scale),
                         affective_scale_ind = scale(affective_scale))

Here is the result:

这里是结果:

> str(scaled_data)
Classes ‘grouped_df’, ‘tbl_df’, ‘tbl’ and 'data.frame': 4136 obs. of  7 variables:
 $ stud_ID             : chr  "ABB112292" "ABB112292" "ABB112292" "ABB112292" ...
 $ behavioral_scale    : num  3.5 4 3.5 3 3.5 2 NA NA 1 2 ...
 $ cognitive_scale     : num  3.5 3 3 3 3.5 2 NA NA 1 1 ...
 $ affective_scale     : num  2.5 3.5 3 3 2.5 2 NA NA 1 1.5 ...
 $ behavioral_scale_ind: num [1:12, 1] 0.64 1.174 0.64 0.107 0.64 ...
  ..- attr(*, "scaled:center")= num 2.9
  ..- attr(*, "scaled:scale")= num 0.937
 $ cognitive_scale_ind : num [1:12, 1] 1.17 0.64 0.64 0.64 1.17 ...
  ..- attr(*, "scaled:center")= num 2.4
  ..- attr(*, "scaled:scale")= num 0.937
 $ affective_scale_ind : num [1:12, 1] 0 1.28 0.64 0.64 0 ...
  ..- attr(*, "scaled:center")= num 2.5
  ..- attr(*, "scaled:scale")= num 0.782

The three scaled variables (behavioral_scale, cognitive_scale, and affective_scale) have only 12 observations - the same number of observations for the first student, ABB112292.

这三个尺度变量(行为尺度、认知尺度和情感尺度)只有12个观测值——第一个学生ABB112292的观测值相同。

What's going on here? How can I obtain scaled scores by individual?

这是怎么回事?我如何获得个别的评分?

2 个解决方案

#1

The problem seems to be in the base scale() function, which expects a matrix. Try writing your own.

问题似乎出现在base scale()函数中，它需要一个矩阵。尝试编写自己的。

scale_this <- function(x){
  (x - mean(x, na.rm=TRUE)) / sd(x, na.rm=TRUE)
}

Then this works:

那么这个工作原理:

library("dplyr")

# reproducible sample data
set.seed(123)
n = 1000
df <- data.frame(stud_ID = sample(LETTERS, size=n, replace=TRUE),
                 behavioral_scale = runif(n, 0, 10),
                 cognitive_scale = runif(n, 1, 20),
                 affective_scale = runif(n, 0, 1) )
scaled_data <- 
  df %>%
  group_by(stud_ID) %>%
  mutate(behavioral_scale_ind = scale_this(behavioral_scale),
         cognitive_scale_ind = scale_this(cognitive_scale),
         affective_scale_ind = scale_this(affective_scale))

Or, if you're open to a data.table solution:

或者，如果你对数据开放的话。表解决方案:

library("data.table")

setDT(df)

cols_to_scale <- c("behavioral_scale","cognitive_scale","affective_scale")

df[, lapply(.SD, scale_this), .SDcols = cols_to_scale, keyby = factor(stud_ID)]

#2

This was a known problem in dplyr, a fix has been merged to the development version, which you can install via

这在dplyr中是一个已知的问题，补丁已经被合并到开发版本中，您可以通过开发版本进行安装

# install.packages("devtools")
devtools::install_github("hadley/dplyr")

In the stable version, the following should work, too:

在稳定版本中，以下内容也应适用:

scale_this <- function(x) as.vector(scale(x))

#1