函数从数据中提取最高的n个分数，并使用‘`apply`’或dplyr ` `rowwise`‘找到它们的平均值。
函数从数据中提取最高的n个分数，并使用‘`apply`’或dplyr ` `rowwise`‘找到它们的平均值。
提问于 2022-01-16 18:47:56
数据文件看起来像这样
df = data.frame(name = c("A","B","C"),
exam1 = c(2,6,4),
exam2 = c(3,5,6),
exam3 = c(5,3,3),
exam4 = c(1,NA,5))我希望提取每个“名称”的前3名，并使用apply()或dplyr 行式()函数找到它们的平均值。
回答 5
Stack Overflow用户
回答已采纳
发布于 2022-01-16 18:49:56
使用apply，使用MARGIN = 1循环数字列上的行，sort，根据decreasing = TRUE/FALSE获取head/tail，并在base R中返回mean
apply(df[-1], 1, FUN = function(x) mean(head(sort(x, decreasing = TRUE), 3)))
[1] 3.333333 4.666667 5.000000或使用dplyr/rowwise
library(dplyr)
df %>%
rowwise %>%
mutate(Mean = mean(head(sort(c_across(where(is.numeric)),
decreasing = TRUE), 3))) %>%
ungroup
# A tibble: 3 × 6
name exam1 exam2 exam3 exam4 Mean
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 A 2 3 5 1 3.33
2 B 6 5 3 NA 4.67
3 C 4 6 3 5 5 Stack Overflow用户
发布于 2022-01-16 18:58:33
下面是一种使用top_n进行旋转和使用的替代方法:这将只返回前3位：
library(dplyr)
library(tidyr)
df %>%
pivot_longer(
-name,
names_to = "exam",
values_to = "value"
) %>%
group_by(name) %>%
top_n(3, value) %>%
mutate(mean = mean(value)) %>%
pivot_wider(
names_from = exam,
values_from = value
) name mean exam1 exam2 exam3 exam4
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 A 3.33 2 3 5 NA
2 B 4.67 6 5 3 NA
3 C 5 4 6 NA 5或：
library(tidyr)
df %>%
pivot_longer(
-name,
names_to = "exam",
values_to = "value"
) %>%
group_by(name) %>%
top_n(3, value) %>%
summarise(mean = mean(value)) name mean
<chr> <dbl>
1 A 3.33
2 B 4.67
3 C 5 Stack Overflow用户
发布于 2022-01-16 19:00:25
使用purrr::pmap_dfr
library(tidyverse)
df = data.frame(name = c("A","B","C"),
exam1 = c(2,6,4),
exam2 = c(3,5,6),
exam3 = c(5,3,3),
exam4 = c(1,NA,5))
df %>%
pmap_dfr(~ list(means = mean(sort(c(..2,..3,..4,..5), decreasing=T)[1:3]))) %>%
bind_cols(df,.)
#> name exam1 exam2 exam3 exam4 means
#> 1 A 2 3 5 1 3.333333
#> 2 B 6 5 3 NA 4.666667
#> 3 C 4 6 3 5 5.000000另一种可能的解决方案，基于tidyr::pivot_longer而不使用rowwise
library(tidyverse)
df = data.frame(name = c("A","B","C"),
exam1 = c(2,6,4),
exam2 = c(3,5,6),
exam3 = c(5,3,3),
exam4 = c(1,NA,5))
df %>%
pivot_longer(cols = 2:5, names_to = "names") %>%
group_by(name) %>%
slice_max(value, n=3) %>%
summarise(mean = mean(value)) %>%
inner_join(df)
#> Joining, by = "name"
#> # A tibble: 3 × 6
#> name mean exam1 exam2 exam3 exam4
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 A 3.33 2 3 5 1
#> 2 B 4.67 6 5 3 NA
#> 3 C 5 4 6 3 5页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/70733133
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