R语言 tidyverse系列学习笔记（系列5）dplyr 数据分析之across

news2024/11/24 20:41:04

成绩单 score

install.packages("dplyr")
library(dplyr)

install.packages("tibble")
library(tibble)

install.packages("stringr")
library(stringr)

score = tibble(ID=c("1222-1","2001-0","3321-1","4898-0","2782-0","1002-8","4211-0","1023-1","3325-1"),
               gender=c("female","male","male","male","female","female","male","female","male") ,
               chinese_mid_score=round(runif(9,80,90),digits = 0),
               chinese_final_score=round(runif(9,80,90),digits = 0),
               english_score=c(round(runif(8,80,90),digits = 0),NA),
               match_score=round(runif(9,80,90),digits = 0),
               musci_score=round(runif(9,80,90),digits = 0)
              )

view(score)

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在之前的系列3中，介绍了成绩单的计算，链接: https://blog.csdn.net/qq_43792777/article/details/131049634?spm=1001.2014.3001.5502

本系列继续以成绩单为例，对系列3进行改进。
之前介绍了rowwise+summary组合来处理NA，但是当数据量特别大时，会出现bug

新的函数，`rowSums`

score1 = score |> 
          mutate(rowSums(select(score,musci_score,english_score,match_score),na.rm = TRUE))
          
view(score1)

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select有个缺点就是当需要计算的数据列过多时，需要一一列举很麻烦，用contains会很简单

计算每个学生各科分数之和

score_sum = score |> 
  mutate(rowSums(select(score, contains("_score")),na.rm = TRUE))

view(score_sum)

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新的问题来了，我的数据命名没有规律该怎么办呢？
可以根据数据类型来计算

查看数据类型的方式：`typeof`

typeof(score$ID)
typeof(score$english_score)

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`across`应用范围比`select`+`contains`更广泛

score_sum1 = score |> 
  mutate(total_score = rowSums(across(where(is.numeric)),na.rm = TRUE))

view(score_sum1)

double属于numeric,可以用 where(is.numeric) 也可以用 where(is.double)

考试题目太难了，给学生提升一下成绩的 task，给每科都除以0.95

score_curve = score |> 
  mutate(across(.cols = where(is.numeric),  .fns = ~.x/0.95, .names = "curve_{col}"))

view(score_curve)

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原成绩 >= 85分为 pass, otherwise 则为 fail

score_pass = score |> 
  mutate(across(.cols = where(is.numeric),  .fns = ~if_else(.x>=85,"pass","fail"), .names = "pass_{col}"))

view(score_pass)

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`.cols` 还可以是 `everything()`

score_pass_everything = score |> 
  mutate(across(.cols = everything(),  .fns = ~if_else(.x>=85,"pass","fail"), .names = "pass_{col}"))

view(score_pass_everything)

看一下结果，ID 列和 gender 列也输出为 pass or fail
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当然，本例子中将 ID 列和 gender 列强制转换成 numeric 是不恰当的，只需要了解该方法即可

修改后的成绩 >= 85分为 pass, otherwise 则为 fail

score_curve_pass = score |> 
  mutate(across(.cols = where(is.numeric), 
                .fns = c(curve = ~.x/0.95, pass = ~if_else(.>=85,"pass","fail")), 
                .names = "{fn}_{col}"))

view(score_curve_pass)

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`summarise`+`across`

计算均值、方差、5%、95%

score_summary = score |> 
  summarise(across(.cols = where(is.numeric),
                   .fns = c(
                     mean = ~mean(.x, na.rm = TRUE),
                     var = ~var(.x, na.rm = TRUE),
                     q5 = ~quantile(.x, 0.05, na.rm = TRUE),
                     q95 = ~quantile(.x, 0.95, na.rm = TRUE)
                   ),
                   .names = "{fn}:{col}"
                   ))
view(score_summary)