今天我在使用.
和%>%
时遇到了一些我不太了解的事情。现在,我不确定我是否能理解其中任何一个。
set.seed(1)
df <- setDT(data.frame(id = sample(1:5, 10, replace = T), value = runif(10)))
df[, .(Mean = mean(value)), by = .(id)] %>% .$Mean %>% sum()
[1] 3.529399
df[, .(Mean = mean(value)), by = .(id)] %>% {sum(.$Mean)}
[1] 3.529399
sum(df[, .(Mean = mean(value)), by = .(id)]$Mean)
[1] 3.529399
df[, .(Mean = mean(value)), by = .(id)] %>% sum(.$Mean)
[1] 22.0588
有人可以向我解释管道操作员实际上如何使用.
。我过去常常考虑去获取%>%
左侧的内容。
我尝试将sum
替换为print
,以查看实际发生的情况
# As Expected
df[, .(Mean = mean(value)), by = .(id)] %>% .$Mean %>% print()
[1] 0.5111589 0.7698414 0.7475319 0.9919061 0.5089610
df[, .(Mean = mean(value)), by = .(id)] %>% print(.$Mean) %>% sum()
[1] 3.529399
# Surprised
df[, .(Mean = mean(value)), by = .(id)] %>% print(.$Mean)
id Mean
1: 1 0.5111589
---
5: 3 0.5089610
# Same
df[, .(Mean = mean(value)), by = .(id)] %>% sum(print(.$Mean))
[1] 22.0588
# Utterly Confused
df[, .(Mean = mean(value)), by = .(id)] %>% print(.$Mean) %>% sum()
[1] 18.5294 #Not even the same as above??
编辑:似乎与 data.table 或它的分组方式无关,与 data.frame 相同:>
x <- data.frame(x1 = 1:3, x2 = 4:6)
sum(x$x1)
# [1] 6
sum(x$x2)
# [1] 15
x %>% .$x1 %>% sum
# [1] 6
x %>% .$x2 %>% sum
# [1] 15
# Why?
x %>% sum(.$x1)
# [1] 27
x %>% sum(.$x2)
# [1] 36
答案 0 :(得分:1)
更新后的简短示例会有所帮助。
我们知道使用管道时,第一个参数来自LHS
(除非我们用{}
“停止”它),所以发生的事情是:
x %>% sum(.$x1)
#[1] 27
等同于
sum(x, x$x1)
#[1] 27
数据帧的总和与列x1
相加。
就原始示例而言,我们可以验证相同的行为
library(data.table)
temp <- df[, .(Mean = mean(value)), by = .(id)]
sum(temp, temp$Mean)
#[1] 22.0588