Tidyr聚集()与NA

时间:2015-03-10 01:11:02

标签: r lubridate tidyr

我正在使用tidyrlubridate将宽表转换为长表。以下工作正常。

> (df <- data.frame(hh_id = 1:2,
                   bday_01 = ymd(20150309),
                   bday_02 = ymd(19850911),
                   bday_03 = ymd(19801231)))

  hh_id    bday_01    bday_02    bday_03
1     1 2015-03-09 1985-09-11 1980-12-31
2     2 2015-03-09 1985-09-11 1980-12-31

> gather(df, person_num, bday, starts_with("bday_0"))

  hh_id  person_num        bday
1     1     bday_01  2015-03-09
2     2     bday_01  2015-03-09
3     1     bday_02  1985-09-11
4     2     bday_02  1985-09-11
5     1     bday_03  1980-12-31
6     2     bday_03  1980-12-31

但是,当混音中有NA时,日期会转换为字符串。

> (df <- data.frame(hh_id = 1:2,
                   bday_01 = ymd(20150309),
                   bday_02 = ymd(19850911),
                   bday_03 = NA))

  hh_id    bday_01    bday_02    bday_03
1     1 2015-03-09 1985-09-11         NA
2     2 2015-03-09 1985-09-11         NA

> gather(df, person_num, bday, starts_with("bday_0"))

  hh_id person_num       bday
1     1    bday_01 1425859200
2     2    bday_01 1425859200
3     1    bday_02  495244800
4     2    bday_02  495244800
5     1    bday_03         NA
6     2    bday_03         NA
Warning message:
attributes are not identical across measure variables; they will be dropped 

请注意,当常规字符串与NA混合使用时仍会出现警告。

> (df <- data.frame(hh_id = 1:2,
                    bday_01 = '20150309',
                    bday_02 = '19850911',
                    bday_03 = NA))

  hh_id  bday_01  bday_02 bday_03
1     1 20150309 19850911      NA
2     2 20150309 19850911      NA

> gather(df, person_num, bday, starts_with("bday_0"))

  hh_id person_num     bday
1     1    bday_01 20150309
2     2    bday_01 20150309
3     1    bday_02 19850911
4     2    bday_02 19850911
5     1    bday_03     <NA>
6     2    bday_03     <NA>
Warning message:
attributes are not identical across measure variables; they will be dropped 

是否可以在避免警告和保留格式的同时使用带有NA的tidyr?

1 个答案:

答案 0 :(得分:2)

数据未转换为字符串,它将回退到1970-01-01以来秒的整数表示形式,这是Date中原始df值表示的内容:< / p>

x <- df$bday_01
x
#[1] "2015-03-09 UTC" "2015-03-09 UTC"
attributes(x) <- NULL
x
#[1] 1425859200 1425859200

警告消息为您提供了解决方法的提示:

  测量变量的

属性不相同;他们会   丢弃

所以,试试:

attributes(df$bday_03) <- attributes(df$bday_02)
gather(df, person_num, bday, starts_with("bday_0"))

#  hh_id person_num       bday
#1     1    bday_01 2015-03-09
#2     2    bday_01 2015-03-09
#3     1    bday_02 1985-09-11
#4     2    bday_02 1985-09-11
#5     1    bday_03       <NA>
#6     2    bday_03       <NA>