我在group_by中使用summarise和dplyr命令来为我的数据生成一些摘要。我想为整体数据集获得相同的摘要，并将其合并为一个tibble。
是否有一种简便的方法？我下面的解决方案感觉像是有效执行此操作所需代码量的四倍！
谢谢。
# reprex
library(tidyverse)
tidy_data <- tibble::tribble(
~drug, ~gender, ~condition, ~value,
"control", "f", "work", 0.06,
"treatment", "m", "work", 0.42,
"treatment", "f", "work", 0.22,
"control", "m", "work", 0.38,
"treatment", "m", "work", 0.57,
"treatment", "f", "work", 0.24,
"control", "f", "work", 0.61,
"control", "f", "play", 0.27,
"treatment", "m", "play", 0.3,
"treatment", "f", "play", 0.09,
"control", "m", "play", 0.84,
"control", "m", "play", 0.65,
"treatment", "m", "play", 0.98,
"treatment", "f", "play", 0.38
)
tidy_summaries <- tidy_data %>%
# Group by the required variables
group_by(drug, gender, condition) %>%
summarise(mean = mean(value),
median = median(value),
min = min(value),
max = max(value)) %>%
# Bind rows will bind this output to the following one
bind_rows(
# Now for the overall version
tidy_data %>%
# Generate the overall summary values
mutate(mean = mean(value),
median = median(value),
min = min(value),
max = max(value)) %>%
# We need to know what the structure of the 'grouped_by' tibble first
# as the overall output format needs to match that
select(drug, gender, condition, mean:max) %>% # Keep columns of interest
# The same information will be appended to all rows, so we just need to retain one
filter(row_number() == 1) %>%
# Change the values in drug, gender, condition to "overall"
mutate_at(vars(drug:condition),
list(~ifelse(is.character(.), "overall", .)))
)
这是我想要的输出，但是没有我希望的那么简单。
tidy_summaries
#> # A tibble: 9 x 7
#> # Groups: drug, gender [5]
#> drug gender condition mean median min max
#> <chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 control f play 0.27 0.27 0.27 0.27
#> 2 control f work 0.335 0.335 0.06 0.61
#> 3 control m play 0.745 0.745 0.65 0.84
#> 4 control m work 0.38 0.38 0.38 0.38
#> 5 treatment f play 0.235 0.235 0.09 0.38
#> 6 treatment f work 0.23 0.23 0.22 0.24
#> 7 treatment m play 0.64 0.64 0.3 0.98
#> 8 treatment m work 0.495 0.495 0.42 0.570
#> 9 overall overall overall 0.429 0.38 0.06 0.98
答案 0 :(得分：0)
尝试
tidy_data %>%
group_by(drug, gender, condition) %>%
summarise(mean = mean(value), median = median(value), min = min(value), max = max(value)) %>%
bind_rows(.,
tidy_data %>%
summarise(drug = "Overall", gender = "Overall", condition = "Overall", mean = mean(value), median = median(value), min = min(value), max = max(value))
)
这给出了：
# A tibble: 9 x 7
# Groups: drug, gender [5]
drug gender condition mean median min max
<chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 control f play 0.27 0.27 0.27 0.27
2 control f work 0.335 0.335 0.06 0.61
3 control m play 0.745 0.745 0.65 0.84
4 control m work 0.38 0.38 0.38 0.38
5 treatment f play 0.235 0.235 0.09 0.38
6 treatment f work 0.23 0.23 0.22 0.24
7 treatment m play 0.64 0.64 0.3 0.98
8 treatment m work 0.495 0.495 0.42 0.570
9 Overall Overall Overall 0.429 0.38 0.06 0.98
代码首先通过分组对其进行汇总，然后根据原始数据创建最终的摘要行，并将其绑定在最底部。
答案 1 :(得分：0)
有趣的问题。我的回答基本上与@sumshyftw相同，但是使用了mutate_if和summarise_at。
代码
library(hablar)
funs <- list(mean = ~mean(.),
median = ~median(.),
min = ~min(.),
max = ~max(.))
tidy_data %>%
group_by(drug, gender, condition) %>%
summarise_at(vars(value), funs) %>%
ungroup() %>%
bind_rows(., tidy_data %>% summarise_at(vars(value), funs)) %>%
mutate_if(is.character, ~if_na(., "Overall"))
结果
drug gender condition mean median min max
<chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 control f play 0.27 0.27 0.27 0.27
2 control f work 0.335 0.335 0.06 0.61
3 control m play 0.745 0.745 0.65 0.84
4 control m work 0.38 0.38 0.38 0.38
5 treatment f play 0.235 0.235 0.09 0.38
6 treatment f work 0.23 0.23 0.22 0.24
7 treatment m play 0.64 0.64 0.3 0.98
8 treatment m work 0.495 0.495 0.42 0.570
9 Overall Overall Overall 0.429 0.38 0.06 0.98