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如果在dplyr管道中有日期

lubridate dplyr date r

-2

Thomas Speidel · 技术社区 · 6 年前

假设我有这些数据:

df <- structure(list(end = structure(c(2932896, 2932896, 17434, 2932896, 
2932896, 2932896), class = "Date"), start = structure(c(15397, 
16847, 14249, 13801, 12101, 13360), class = "Date")), class = "data.frame", row.names = c(NA, 
-6L))

> df
         end      start
1 9999-12-31 2012-02-27
2 9999-12-31 2016-02-16
3 2017-09-25 2009-01-05
4 9999-12-31 2007-10-15
5 9999-12-31 2003-02-18
6 9999-12-31 2006-07-31

我想创建第三个变量, dur ,以某个语句为条件:

library(dplyr)
library(lubridate)

df %>%
  mutate(dur = if_else(end == "9999-12-31",
                       as.duration(today() - max("2012-01-01", start)),
                       as.duration(max(start, "2012-01-01") - end)
                       )
         )

Error in mutate_impl(.data, dots) : 
  Evaluation error: non-numeric argument to binary operator.

我知道一些建议 DT ifelse 有约会,但我想呆在房间里。

更新1

这里是专栏表示预期输出:

| end           | start         | dur                       | code                                              |
|------------   |------------   |-------------------------- |-------------------------------------------------- |
| 9999-12-31    | 2012-02-27    | 207100800s (~6.56 years)  | as.duration(today()-ymd("2012-02-27"))            |
| 9999-12-31    | 2016-02-16    | 81820800s (~2.59 years)   | as.duration(today()-ymd("2016-02-16"))            |
| 2017-09-25    | 2009-01-05    | 180921600s (~5.73 years)  | as.duration(ymd("2017-09-25")-ymd("2012-01-01"))  |
| 9999-12-31    | 2007-10-15    | 212025600s (~6.72 years)  | as.duration(today()-ymd("2012-01-01"))            |
| 9999-12-31    | 2003-02-18    | 212025600s (~6.72 years)  | as.duration(today()-ymd("2012-01-01"))            |
| 9999-12-31    | 2006-07-31    | 212025600s (~6.72 years)  | as.duration(today()-ymd("2012-01-01"))            |

更新2

df %>%
  mutate(dur = if_else(end == ymd("9999-12-31"),
                       as.duration(today() - max(ymd("2012-01-01"), start)),
                       as.duration(max(start, ymd("2012-01-01")) - end)
                       )
         )

生产:

         end      start                     dur
1 9999-12-31 2012-02-27 81820800s (~2.59 years)
2 9999-12-31 2016-02-16 81820800s (~2.59 years)
3 2017-09-25 2009-01-05 50716800s (~1.61 years)
4 9999-12-31 2007-10-15 81820800s (~2.59 years)
5 9999-12-31 2003-02-18 81820800s (~2.59 years)
6 9999-12-31 2006-07-31 81820800s (~2.59 years)

更新3(已解决!)

感谢 @jdobres pmax 而不是 max . 我想不出原因,但文件上说: pmax和pmin还将使用适当的比较方法来处理分类的S3或S4对象,is.na和rep(如果需要循环使用参数) . 我怀疑这与此有关。

df %>%
  mutate(dur = if_else(end == ymd("9999-12-31"),
                       as.duration(today() - pmax(ymd("2012-01-01"), start)),
                       as.duration(pmax(start, ymd("2012-01-01")) - end)
                       )
         )

         end      start                      dur
1 9999-12-31 2012-02-27 207100800s (~6.56 years)
2 9999-12-31 2016-02-16  81820800s (~2.59 years)
3 2017-09-25 2009-01-05 180921600s (~5.73 years)
4 9999-12-31 2007-10-15 212025600s (~6.72 years)
5 9999-12-31 2003-02-18 212025600s (~6.72 years)
6 9999-12-31 2006-07-31 212025600s (~6.72 years)

2 回复 | 直到 4 年前

jdobres 6 年前

ymd 或 as.Date pmax 以获得两个持续时间中的较大值。提供具有相同元素数的向量间的并行最大值。例如:

 pmax(1:10, rep(5, 10))

 [1]  5  5  5  5  5  6  7  8  9 10

df %>%
  mutate(dur = if_else(end == ymd("9999-12-31"),
                       as.duration(today() - pmax(ymd("2012-01-01"), start)),
                       as.duration(pmax(start, ymd("2012-01-01")) - end)
                       )
                    )

Paul 6 年前

我试着做一个变异,就像你做的,在使用 lubridate::ymd() 在你输入的字符日期上,但是 max() 不是逐行比较,而是抓住了所有起始值的最大值——也许有人能解释为什么?

最后我和阿普去了。

library(dplyr)
library(lubridate)

df %>%
  mutate(dur =
           apply(tbl_df(df), 1, function(x){
             print(x)

             ifelse(
               x["end"] == ymd("9999-12-31"),

               interval(today(), max(ymd("2012-01-01"), ymd(x["start"]))) %>%
                 as.duration() %>%
                 as.numeric("years"),

               interval(max(x["start"], ymd("2012-01-01")), ymd(x["end"])) %>%
                 as.duration() %>%
                 as.numeric("years")
             )

           }))

#          end      start       dur
# 1 9999-12-31 2012-02-27 -6.562628
# 2 9999-12-31 2016-02-16 -2.592745
# 3 2017-09-25 2009-01-05  8.720055
# 4 9999-12-31 2007-10-15 -6.718686
# 5 9999-12-31 2003-02-18 -6.718686
# 6 9999-12-31 2006-07-31 -6.718686

ymd() 再次在函数中。我以前就注意到了,我不知道为什么。