为什么dplyr的mutate()会改变时间格式？

Question

我readr用来读取包含时间格式的日期列的数据.我可以使用col_types选项正确读取它readr.

library(dplyr)
library(readr)

sample <- "time,id
2015-03-05 02:28:11,1674
2015-03-03 13:10:59,36749
2015-03-05 07:55:48,NA
2015-03-05 06:13:19,NA
"

mydf <- read_csv(sample, col_types="Ti")
mydf
                 time    id
1 2015-03-05 02:28:11  1674
2 2015-03-03 13:10:59 36749
3 2015-03-05 07:55:48    NA
4 2015-03-05 06:13:19    NA

这很好.但是,如果我想操作此列dplyr,则时间列将丢失其格式.

mydf %>% mutate(time = ifelse(is.na(id), NA, time))
        time    id
1 1425522491  1674
2 1425388259 36749
3         NA    NA
4         NA    NA

为什么会这样？

我知道我可以通过将它转换为字符来解决这个问题,但是如果不来回转换它会更方便.

mydf %>% mutate(time = as.character(time)) %>% 
    mutate(time = ifelse(is.na(id), NA, time))

Answer 1

这实际上ifelse()是造成这个问题,而不是dplyr::mutate().属性剥离问题的一个例子如下所示help(ifelse)-

## ifelse() strips attributes
## This is important when working with Dates and factors
x <- seq(as.Date("2000-02-29"), as.Date("2004-10-04"), by = "1 month")
## has many "yyyy-mm-29", but a few "yyyy-03-01" in the non-leap years
y <- ifelse(as.POSIXlt(x)$mday == 29, x, NA)
head(y) # not what you expected ... ==> need restore the class attribute:
class(y) <- class(x)

所以你有它.如果你想使用它,这是一些额外的工作ifelse().这里有两种可能的方法,可以让您达到理想的结果ifelse().第一个是非常简单和使用is.na<-.

## mark 'time' as NA if 'id' is NA
is.na(mydf$time) <- is.na(mydf$id)

## resulting in
mydf
#                  time    id
# 1 2015-03-05 02:28:11  1674
# 2 2015-03-03 13:10:59 36749
# 3                <NA>    NA
# 4                <NA>    NA

如果您不想选择该路线,并希望继续使用该dplyr方法,则可以使用replace()而不是ifelse().

mydf %>% mutate(time = replace(time, is.na(id), NA))
#                  time    id
# 1 2015-03-05 02:28:11  1674
# 2 2015-03-03 13:10:59 36749
# 3                <NA>    NA
# 4                <NA>    NA

数据:

mydf <- structure(list(time = structure(c(1425551291, 1425417059, 1425570948, 
1425564799), class = c("POSIXct", "POSIXt"), tzone = ""), id = c(1674L, 
36749L, NA, NA)), .Names = c("time", "id"), class = "data.frame", row.names = c(NA, 
-4L))