使用 dplyr 跨共享相似名称的列计算按行汇总统计信息，例如平均值、最大值、最小值

Question

使用 R 分析数据时，我需要计算具有相似名称（例如，相同前缀）的列的逐行汇总统计信息（平均值、最小值、最大值、标准差）。 虽然我可以使用诸如 matrixStats 之类的 package 来实现它，但我想知道是否有更优雅的方法可以使用 dplyr 来实现。 下面附上生成示例数据集的代码和我用于计算逐行汇总统计的解决方案。 谢谢！

######SAMPEL CODE#####
library("tidyverse")
library("matrixStats")

# create a sample dataset
sample_data <- data.frame(matrix(nrow = 10, ncol = 6))
names(sample_data) <- c("ID", "score_1", "score_2", "score_3", "score_4", "score_5")
sample_data <- sample_data %>%
  mutate(ID = seq(1:10),
         score_1 = round(runif(10, 1, 5)),
         score_2 = round(runif(10, 1, 5)),
         score_3 = round(runif(10, 1, 5)),
         score_4 = round(runif(10, 1, 5)),
         score_5 = round(runif(10, 1, 5)))

score_columns <- grep("score_", names(sample_data))

sample_data <- sample_data %>% 
  mutate(mean_score = rowMeans(select(., starts_with("score_")), na.rm = TRUE),
         max_score = rowMaxs(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE),
         sd_score = rowSds(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE)) 

Answer 1

您可以使用 dplyr逐行操作结合c_across function来“逐行”计算这些统计信息，例如

library(tidyverse)
library(matrixStats)
#> 
#> Attaching package: 'matrixStats'
#> The following object is masked from 'package:dplyr':
#> 
#>     count

# create a sample dataset
sample_data <- data.frame(matrix(nrow = 10, ncol = 6))
names(sample_data) <- c("ID", "score_1", "score_2", "score_3", "score_4", "score_5")
sample_data <- sample_data %>%
  mutate(ID = seq(1:10),
         score_1 = round(runif(10, 1, 5)),
         score_2 = round(runif(10, 1, 5)),
         score_3 = round(runif(10, 1, 5)),
         score_4 = round(runif(10, 1, 5)),
         score_5 = round(runif(10, 1, 5)))

score_columns <- grep("score_", names(sample_data))

output <- sample_data %>% 
  mutate(mean_score = rowMeans(select(., starts_with("score_")), na.rm = TRUE),
         max_score = rowMaxs(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE),
         sd_score = rowSds(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE)) 
output
#>    ID score_1 score_2 score_3 score_4 score_5 mean_score max_score  sd_score
#> 1   1       1       5       4       3       2        3.0         5 1.5811388
#> 2   2       3       5       1       5       2        3.2         5 1.7888544
#> 3   3       2       5       3       2       4        3.2         5 1.3038405
#> 4   4       3       4       3       3       5        3.6         5 0.8944272
#> 5   5       2       3       2       2       3        2.4         3 0.5477226
#> 6   6       4       2       4       3       2        3.0         4 1.0000000
#> 7   7       2       2       1       3       1        1.8         3 0.8366600
#> 8   8       3       4       1       3       4        3.0         4 1.2247449
#> 9   9       3       2       3       4       2        2.8         4 0.8366600
#> 10 10       5       2       3       3       4        3.4         5 1.1401754

output2 <- sample_data %>%
  rowwise() %>%
  mutate(mean_score = mean(c_across(starts_with("score_"))),
         max_score = max(c_across(!!score_columns)),
         sd_score = sd(c_across(!!score_columns)))
output2
#> # A tibble: 10 × 9
#> # Rowwise: 
#>       ID score_1 score_2 score_3 score_4 score_5 mean_score max_score sd_score
#>    <int>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>      <dbl>     <dbl>    <dbl>
#>  1     1       1       5       4       3       2        3           5    1.58 
#>  2     2       3       5       1       5       2        3.2         5    1.79 
#>  3     3       2       5       3       2       4        3.2         5    1.30 
#>  4     4       3       4       3       3       5        3.6         5    0.894
#>  5     5       2       3       2       2       3        2.4         3    0.548
#>  6     6       4       2       4       3       2        3           4    1    
#>  7     7       2       2       1       3       1        1.8         3    0.837
#>  8     8       3       4       1       3       4        3           4    1.22 
#>  9     9       3       2       3       4       2        2.8         4    0.837
#> 10    10       5       2       3       3       4        3.4         5    1.14

^{由代表 package (v2.0.1) 于 2022 年 1 月 31 日创建}

由于这是一个特定于 dplyr 的解决方案，您可以在其他解决方案将失败或需要笨拙的解决方法的情况下合并进一步的数据操作（例如filter()到 select 特定行），例如

library(tidyverse)
library(matrixStats)
#> 
#> Attaching package: 'matrixStats'
#> The following object is masked from 'package:dplyr':
#> 
#>     count

# create a sample dataset
sample_data <- data.frame(matrix(nrow = 10, ncol = 6))
names(sample_data) <- c("ID", "score_1", "score_2", "score_3", "score_4", "score_5")
sample_data <- sample_data %>%
  mutate(ID = seq(1:10),
         score_1 = round(runif(10, 1, 5)),
         score_2 = round(runif(10, 1, 5)),
         score_3 = round(runif(10, 1, 5)),
         score_4 = round(runif(10, 1, 5)),
         score_5 = round(runif(10, 1, 5)))

score_columns <- grep("score_", names(sample_data))

output <- sample_data %>%
  filter(ID < 6) %>%
  mutate(mean_score = rowMeans(select(., starts_with("score_")), na.rm = TRUE),
         max_score = rowMaxs(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE),
         sd_score = rowSds(as.matrix(sample_data[,c(score_columns)]), na.rm = TRUE)) 
#> Error: Problem with `mutate()` column `max_score`.
#> ℹ `max_score = rowMaxs(as.matrix(sample_data[, c(score_columns)]), na.rm = TRUE)`.
#> ℹ `max_score` must be size 5 or 1, not 10.


output2 <- sample_data %>%
  filter(ID < 6) %>%
  rowwise() %>%
  mutate(mean_score = mean(c_across(starts_with("score_"))),
         max_score = max(c_across(!!score_columns)),
         sd_score = sd(c_across(!!score_columns)))
output2
#> # A tibble: 5 × 9
#> # Rowwise: 
#>      ID score_1 score_2 score_3 score_4 score_5 mean_score max_score sd_score
#>   <int>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>      <dbl>     <dbl>    <dbl>
#> 1     1       5       2       5       5       1        3.6         5     1.95
#> 2     2       1       2       4       1       2        2           4     1.22
#> 3     3       1       2       2       4       5        2.8         5     1.64
#> 4     4       5       4       3       3       5        4           5     1   
#> 5     5       4       2       2       5       4        3.4         5     1.34

^{由代表 package (v2.0.1) 于 2022 年 1 月 31 日创建}

Answer 2

我用来回答您的问题的方式略有不同：

library("tidyverse")
library("matrixStats")

# create a sample dataset
sample_data <- data.frame(matrix(nrow = 10, ncol = 6))
names(sample_data) <- c("ID", "score_1", "score_2", "score_3", "score_4", "score_5")
sample_data <- sample_data %>%
  mutate(ID = seq(1:10),
         score_1 = round(runif(10, 1, 5)),
         score_2 = round(runif(10, 1, 5)),
         score_3 = round(runif(10, 1, 5)),
         score_4 = round(runif(10, 1, 5)),
         score_5 = round(runif(10, 1, 5)))

score_columns <- colnames(sample_data)[grep("score_", names(sample_data))]

sample_data<- sample_data %>% 
  rowwise(ID) %>% 
  mutate(mean_score = mean(c_across(score_columns[1]:length(score_columns))),
         max_score = max(c_across(score_columns[1]:length(score_columns))),
         sd_score = sd(c_across(score_columns[1]:length(score_columns))))

如果您想要更轻松地使用基函数，请尝试以下操作：

score_columns <- grep("score_", names(sample_data))
sample_data['mean_score'] <- apply(sample_data[,score_columns], 1, mean)
sample_data['max_score'] <- apply(sample_data[,score_columns], 1, max)
sample_data['sd_score'] <- apply(sample_data[,score_columns], 1, sd)