我发现自己一直在编写这段代码来为组方法产生标准错误(然后用于绘制置信区间)。
但是,编写我自己的函数在一行代码中执行此操作会很好。我已经阅读了dplyr中关于非标准评估的nse小插图以及this blog post。我得到它有些,但我太过于自我了解这个问题。任何人都可以帮忙吗?谢谢。
var1<-sample(c('red', 'green'), size=10, replace=T)
var2<-rnorm(10, mean=5, sd=1)
df<-data.frame(var1, var2)
df %>%
group_by(var1) %>%
summarize(avg=mean(var2), n=n(), sd=sd(var2), se=sd/sqrt(n))
答案 0 :(得分:3)
您可以使用函数enquo
在函数调用中明确命名变量:
my_fun <- function(x, cat_var, num_var){
cat_var <- enquo(cat_var)
num_var <- enquo(num_var)
x %>%
group_by(!!cat_var) %>%
summarize(avg = mean(!!num_var), n = n(),
sd = sd(!!num_var), se = sd/sqrt(n))
}
给你:
> my_fun(df, var1, var2)
# A tibble: 2 x 5
var1 avg n sd se
<fctr> <dbl> <int> <dbl> <dbl>
1 green 4.873617 7 0.7515280 0.2840509
2 red 5.337151 3 0.1383129 0.0798550
并且匹配示例的输出:
> df %>%
+ group_by(var1) %>%
+ summarize(avg=mean(var2), n=n(), sd=sd(var2), se=sd/sqrt(n))
# A tibble: 2 x 5
var1 avg n sd se
<fctr> <dbl> <int> <dbl> <dbl>
1 green 4.873617 7 0.7515280 0.2840509
2 red 5.337151 3 0.1383129 0.0798550
修改强>
OP要求从函数中删除group_by
语句,以便为group_by添加多个变量的能力。有两种方法可以解决这个IMO问题。首先,您可以简单地删除group_by
语句并将分组数据框管道输入到函数中。那种方法看起来像这样:
my_fun <- function(x, num_var){
num_var <- enquo(num_var)
x %>%
summarize(avg = mean(!!num_var), n = n(),
sd = sd(!!num_var), se = sd/sqrt(n))
}
df %>%
group_by(var1) %>%
my_fun(var2)
另一种方法是使用...
和quos
来允许函数捕获group_by
语句的多个参数。这看起来像这样:
#first, build the new dataframe
var1<-sample(c('red', 'green'), size=10, replace=T)
var2<-rnorm(10, mean=5, sd=1)
var3 <- sample(c("A", "B"), size = 10, replace = TRUE)
df<-data.frame(var1, var2, var3)
# using the first version `my_fun`, it would look like this
df %>%
group_by(var1, var3) %>%
my_fun(var2)
# A tibble: 4 x 6
# Groups: var1 [?]
var1 var3 avg n sd se
<fctr> <fctr> <dbl> <int> <dbl> <dbl>
1 green A 5.248095 1 NaN NaN
2 green B 5.589881 2 0.7252621 0.5128378
3 red A 5.364265 2 0.5748759 0.4064986
4 red B 4.908226 5 1.1437186 0.5114865
# Now doing it with a new function `my_fun2`
my_fun2 <- function(x, num_var, ...){
group_var <- quos(...)
num_var <- enquo(num_var)
x %>%
group_by(!!!group_var) %>%
summarize(avg = mean(!!num_var), n = n(),
sd = sd(!!num_var), se = sd/sqrt(n))
}
df %>%
my_fun2(var2, var1, var3)
# A tibble: 4 x 6
# Groups: var1 [?]
var1 var3 avg n sd se
<fctr> <fctr> <dbl> <int> <dbl> <dbl>
1 green A 5.248095 1 NaN NaN
2 green B 5.589881 2 0.7252621 0.5128378
3 red A 5.364265 2 0.5748759 0.4064986
4 red B 4.908226 5 1.1437186 0.5114865