如何按行计算字符串数字并计算百分比（仅考虑空列）

Question

对于以下数据框：

df <- structure(list(test = c("A", "B", "C"), `2019` = c("true", "", 
"false"), `2020` = c("false", "true", "true"), `2021` = c("true", 
"false", "true"), `2022` = c("", "false", "false")), class = "data.frame", row.names = c(NA, 
-3L))

输出：

  test  2019  2020  2021  2022
1    A  true false  true      
2    B        true false false
3    C false  true  true false

我需要计算从 2019 列到 2022 的 true 的数量，然后从 not NA 计算 true 的百分比或非空值:

预期的结果可能是这样的：

  test  2019  2020  2021  2022 true_pct
1    A  true false  true           0.67 # 2/3
2    B        true false false     0.33 # 1/3
3    C false  true  true false     0.50 # 2/4

下面的代码 return 的答案是错误的，因为它需要所有年份的列，因为数据中有空字符串，但我需要过滤掩码，例如 !is.na(df[,2:5]) | df[,2:5] != '':

df$count <- rowSums(df[-1] == "true")
df$not_na <- rowSums(!is.na(df[,2:5]))
# df$not_na <- rowSums(!complete.cases(df[,2:5]))
df$true_pct <- df$count/df$not_na

输出：

  test  2019  2020  2021  2022 count not_na true_pct
1    A  true false  true           2      4     0.50
2    B        true false false     1      4     0.25
3    C false  true  true false     2      4     0.50

请分享更好的解决方案，谢谢。

更新： 对于以下数据集，代码似乎没有 return 正确的结果：

df <- structure(list(test = c("A", "B", "C"), `2018` = c("true", NA, NA
), `2019` = c("true", "", "false"), `2020` = c("false", "true", 
                                               "true"), `2021` = c("true", "false", "true"), `2022` = c("", 
                                                                                                        "false", "false")), class = "data.frame", row.names = c(NA, -3L
                                                                                                        ))

tmp <- df[-1]
df$true_pct <- rowSums(tmp == "true")/rowSums(tmp != "" & !is.na(tmp))
df

或：

df$count <- rowSums(df[-1] == "true")
df$not_na <- rowSums(sapply(df[, 2:6], function(x){x %in% c("true", "false")}))
df$true_pct <- df$count/df$not_na
df

生成相同的结果：

  test 2018  2019  2020  2021  2022 true_pct
1    A true  true false  true           0.75
2    B <NA>        true false false       NA
3    C <NA> false  true  true false       NA

Answer 1

你可以使用-

#To select only required columns
tmp <- df[-1]
#Total true values divided by total values which are not `NA` or empty.
df$true_pct <- rowSums(tmp == "true")/rowSums(tmp != "" & !is.na(tmp))
df

#  test  2019  2020  2021  2022  true_pct
#1    A  true false  true       0.6666667
#2    B        true false false 0.3333333
#3    C false  true  true false 0.5000000

Answer 2

rowSums 和 na.rm = TRUE 怎么样：

df$true_pct <- rowSums(df[-1] == "true", na.rm = TRUE)/rowSums(df[-1] != "" & !is.na(df[-1]), na.rm = TRUE)

或此 dplyr 解决方案 apply:

library(dplyr)
library(stringr)
    df %>%
      mutate(true_pct = apply(.[-1], 1, paste, collapse = " "),
             true_pct = str_count(true_pct, "true")/str_count(true_pct, "true|false"))
  test 2018  2019  2020  2021  2022  true_pct
1    A true  true false  true       0.7500000
2    B <NA>        true false false 0.3333333
3    C <NA> false  true  true false 0.5000000

数据：

df <- structure(list(test = c("A", "B", "C"), `2018` = c("true", NA, NA
), `2019` = c("true", "", "false"), `2020` = c("false", "true", 
                                               "true"), `2021` = c("true", "false", "true"), `2022` = c("", 
                                                                                                        "false", "false")), class = "data.frame", row.names = c(NA, -3L
                                                                                                        ))

如何按行计算字符串数字并计算百分比（仅考虑空列）

How to count string numbers row-wisely and count percentage (only not empty columns taking into account)

r

pipe

dataframe

dplyr

tidyverse