考虑数据框:
data = data.frame(a=c(1,1,1,2,2,3),
b=c("apples", "oranges", "apples", "apples", "apples", "grapefruit"),
c=c(12, 22, 22, 45, 67, 28),
d=c("Monday", "Monday", "Monday", "Tuesday", "Wednesday", "Tuesday"),
out = c(12, 14, 16, 18, 20, 22),
rate = c(0.01, 0.02, 0.03, 0.04, 0.07, 0.06))
我正在尝试group_by并进行总结,但是,我不断收到错误消息
Error in new_quosures(NextMethod()) :
could not find function "new_quosures"
我正在使用的代码如下:
model.data.dim.names <- c("a", "b", "c")
data2 <- data %>% group_by_(.dots = model.data.dim.names) %>% summarise(
mean_adj1 = (mean(out, na.rm=FALSE)),
mean_adj2 = (mean(out)/mean(rate))
)
请注意,这是伪数据,并且在Windows操作系统中使用伪数据重现错误。此外,我正在Windows OS上工作。此外,我尝试了以下方法:
能否请您帮助我了解错误的根本原因或可以使用的替代方法?
Answer:
1) tidyr library screws up with it. Removing tidyr helps
2) use most updated dplyr library and group_by/ group_by_at/group_by(!!!syms(model.data.dim.names) works
答案 0 :(得分:1)
不推荐使用group-by_
函数,当前的tidyeval方法是将字符向量转换为符号,然后取消引号将它们拼接为group_by
:
library(dplyr)
data %>%
group_by(!!!syms(model.data.dim.names)) %>%
summarise(
mean_adj1 = mean(out, na.rm=FALSE),
mean_adj2 = mean(out) / mean(rate)
)
## A tibble: 6 x 5
## Groups: a, b [4]
# a b c mean_adj1 mean_adj2
# <dbl> <fct> <dbl> <dbl> <dbl>
#1 1 apples 12 12 1200
#2 1 apples 22 16 533.
#3 1 oranges 22 14 700
#4 2 apples 45 18 450
#5 2 apples 67 20 286.
#6 3 grapefruit 28 22 367.
答案 1 :(得分:1)
我们可以使用group_by_at
中的dplyr
,它可以将字符串作为输入
library(dplyr)
data %>%
group_by_at(model.data.dim.names) %>%
summarise(
mean_adj1 = mean(out, na.rm=FALSE),
mean_adj2 = mean(out) / mean(rate)
)
# A tibble: 6 x 5
# Groups: a, b [4]
# a b c mean_adj1 mean_adj2
# <dbl> <fct> <dbl> <dbl> <dbl>
#1 1 apples 12 12 1200
#2 1 apples 22 16 533.
#3 1 oranges 22 14 700
#4 2 apples 45 18 450
#5 2 apples 67 20 286.
#6 3 grapefruit 28 22 367.
答案 2 :(得分:1)
我在2周前的代码中也遇到了同样的错误。应用dplyr::group_by()
时发生。我的dplyr软件包版本为0.7.6,并将其更新为0.8.0.1。这样就解决了问题。