R：使用具有意外行为的ifelse的dplyr管道条件超前/滞后

I'm trying to use a conditional lead/lag function in a dplyr pipe using ifelse but getting an error. However, using the same approach outside the pipe it seems to work. What am I missing?

我正在尝试使用ifelse在dplyr管道中使用条件超前/滞后函数但是会出错。但是，在管道外使用相同的方法似乎可行。我错过了什么？

require(dplyr)

Data:

数据：

test <- data.frame(a = c("b","b","b","b","b","b",
                         "m","m","m","m","m","m",
                         "s","s","s","s","s","s"), 
                   b = replicate(1,n=18), 
                   stringsAsFactors=F)

dplyr pipe:

dplyr管道：

test %>%
  mutate(delta = ifelse(a == "s", b + lag(b, n = 2*6),
                        ifelse(a == "m", b + lag(b, n = 1*6), 0)))

# Error: could not convert second argument to an integer. type=LANGSXP, length = 3

Without the pipe it works:

没有管道它工作：

test$delta <- ifelse(test$a == "s", test$b + lag(test$b, n = 2*6),
                     ifelse(test$a == "m", test$b + lag(test$b, n = 1*6), 0))

I found some indication that there was an issue with dplyr lead/lag in combination with grouped data frames. But I am not grouping here.

我发现有一些迹象表明dplyr超前/滞后与分组数据帧有关。但我不是在这里分组。

Version info: R 3.1.1 and dplyr_0.4.1.

版本信息：R 3.1.1和dplyr_0.4.1。

2 个解决方案

#1

dplyr fails to parse the expression. One solution is to define the function first:

dplyr无法解析表达式。一种解决方案是首先定义函数：

foo <- function(a, b)
    ifelse(a=="s",b+lag(b,n=2*6), ifelse(a=="m",b+lag(b,n=1*6),0))
test %>% mutate(delta = foo(a,b))

#2

This:

这个：

test %>%
    mutate(delta = ifelse(a=="s",b+lag(b,n=12),
                          ifelse(a=="m",b+lag(b,n=6),0)))

works. This means that you cannot pass expressions in lag arguments.

作品。这意味着您无法在滞后参数中传递表达式。

#1