Question

我有一个像这样的data.frame：

df <- structure(list(sample = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                          2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), sub_sample = structure(c(1L, 
                                                                                        1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
                                                                                        3L), .Label = c("A", "B", "C"), class = "factor"), value = c(111L, 
                                                                                                                                                     233L, NA, NA, NA, 56L, 48L, 23L, 48L, 567L, 98L, 75L, 7578L, 
                                                                                                                                                     NA, 56L, 48L, NA, NA)), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                 -18L))

有一些缺失值（NA），我想计算每个组中非NA的百分比。我现在的操作方式是这样的：

total_nr <- df %>% 
  group_by(sample, sub_sample) %>%
  tally()

nr_wo_NA <- df %>% 
  group_by(sample, sub_sample) %>%
  na.omit() %>% 
  tally()

nr_wo_NA$n <- (nr_wo_NA$n / total_nr$n) * 100

这给了我我想要的东西：

# A tibble: 6 x 3
# Groups:   sample [2]
  sample sub_sample     n
   <int> <fct>      <dbl>
1      1 A           66.7
2      1 B           33.3
3      1 C          100  
4      2 A          100  
5      2 B           66.7
6      2 C           33.3

但是有没有一种方法可以在不创建两个单独的data.frames的情况下做到这一点？

Answer 1

您可以这样做：

df %>%
 group_by(sample, sub_sample) %>%
 summarise(value_non_na = sum(!is.na(value))/n()*100)

  sample sub_sample value_non_na
   <int> <fct>             <dbl>
1      1 A                  66.7
2      1 B                  33.3
3      1 C                 100  
4      2 A                 100  
5      2 B                  66.7
6      2 C                  33.3

Answer 2

将mean与is.na进行比较后，我们可以取其逻辑值

library(dplyr)
df %>% group_by(sample, sub_sample)%>% summarise(value = mean(!is.na(value)) * 100)

#  sample sub_sample value
#   <int> <fct>      <dbl>
#1      1 A           66.7
#2      1 B           33.3
#3      1 C          100  
#4      2 A          100  
#5      2 B           66.7
#6      2 C           33.3

我们可以对基数R使用相同的逻辑

aggregate(value~sample+sub_sample, df, function(x) mean(!is.na(x)), na.action = na.pass)

和data.table

library(data.table)
setDT(df)[, mean(!is.na(value)), .(sample, sub_sample)]

计算子组的非NA值的百分比

2 个答案: