R:使用具有意外行为的ifelse的dplyr管道条件超前/滞后

Question

我正在尝试使用ifelse在dplyr管道中使用条件lead/ lag函数但是收到错误.但是,在管道外使用相同的方法似乎有效.我错过了什么？

require(dplyr)

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

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))

我发现有一些迹象表明dplyr lead/ lag与分组数据帧相结合存在问题.但我不是在这里分组.

版本信息:R 3.1.1和dplyr_0.4.1.

Answer 1

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))