Question

我有一个数据框需要分成两个表来满足Codd的第三范式。在一个简单的例子中，原始数据框看起来像这样：

library(lubridate)
> (df <- data.frame(hh_id = 1:2,
                   income = c(55000, 94000),
                   bday_01 = ymd(c(20150309, 19890211)),
                   bday_02 = ymd(c(19850911, 20000815)),
                   gender_01 = factor(c("M", "F")),
                   gender_02 = factor(c("F", "F"))))

    hh_id income    bday_01    bday_02 gender_01 gender_02
  1     1  55000 2015-03-09 1985-09-11         M         F
  2     2  94000 1989-02-11 2000-08-15         F         F

当我使用收集功能时，它会警告属性不相同，并且会丢失性别因素和bday的润滑（或实际示例中的其他属性）。是否有一个很好的tidyr解决方案，以避免丢失每列的数据类型？

library(tidyr)
> (person <- df %>% 
      select(hh_id, bday_01:gender_02) %>% 
      gather(key, value, -hh_id) %>%
      separate(key, c("key", "per_num"), sep = "_") %>%
      spread(key, value))

     hh_id per_num       bday gender
   1     1      01 1425859200      M
   2     1      02  495244800      F
   3     2      01  603158400      F
   4     2      02  966297600      F

   Warning message:
   attributes are not identical across measure variables; they will be dropped

> lapply(person, class)

  $hh_id
  [1] "integer"

  $per_num
  [1] "character"

  $bday
  [1] "character"

  $gender
  [1] "character"

我可以想象一种方法，通过分别收集具有相同数据类型的每组变量然后加入所有表来实现它，但必须有一个我更缺失的更优雅的解决方案。

Answer 1

您可以将日期转换为字符，然后将其转换回日期结束日期：

(person <- df %>% 
      select(hh_id, bday_01:gender_02) %>% 
      mutate_each(funs(as.character), contains('bday')) %>%
      gather(key, value, -hh_id) %>%
      separate(key, c("key", "per_num"), sep = "_") %>%
      spread(key, value) %>%
      mutate(bday=ymd(bday)))

  hh_id per_num       bday gender
1     1      01 2015-03-09      M
2     1      02 1985-09-11      F
3     2      01 1989-02-11      F
4     2      02 2000-08-15      F

或者，如果您使用Date代替POSIXct，则可以执行以下操作：

(person <- df %>% 
      select(hh_id, bday_01:gender_02) %>% 
      gather(per_num1, gender, contains('gender'), convert=TRUE) %>%
      gather(per_num2, bday, contains('bday'), convert=TRUE) %>%
      mutate(bday=as.Date(bday)) %>%
      mutate_each(funs(str_extract(., '\\d+')), per_num1, per_num2) %>%
      filter(per_num1 == per_num2) %>%
      rename(per_num=per_num1) %>%
      select(-per_num2))

修改

您看到的警告：

Warning: attributes are not identical across measure variables; they will be dropped

来自收集性别列，这些列是因素并具有不同的级别向量（请参阅str(df)）。如果您要将性别列转换为字符，或者您要将其级别与类似的内容同步，

df <- mutate(df, gender_02 = factor(gender_02, levels=levels(gender_01)))

然后您将看到执行
时警告消失
person <- df %>% select(hh_id, bday_01:gender_02) %>% gather(key, value, contains('gender'))

Answer 2

您似乎不喜欢我的base solutions。让我再次诱惑你

(df <- data.frame(hh_id = 1:2,
                  income = c(55000, 94000),
                  bday_01 = ymd(c(20150309, 19890211)),
                  bday_02 = ymd(c(19850911, 20000815)),
                  gender_01 = factor(c("M", "F")),
                  gender_02 = factor(c("F", "F"))))


reshape(df, idvar = 'hh_id', varying = list(3:4, 5:6), direction = 'long',
        v.names = c('bday','gender'), timevar = 'per_num')

#     hh_id income    per_num       bday gender
# 1.1     1  55000          1 2015-03-09      M
# 2.1     2  94000          1 1989-02-11      F
# 1.2     1  55000          2 1985-09-11      F
# 2.2     2  94000          2 2000-08-15      F

Answer 3

使用 tidyr 1.0.0 可以按以下步骤进行操作：

suppressPackageStartupMessages({
  library(tidyr)
  library(lubridate)
})
df <- data.frame(hh_id = 1:2,
                 income = c(55000, 94000),
                 bday_01 = ymd(c(20150309, 19890211)),
                 bday_02 = ymd(c(19850911, 20000815)),
                 gender_01 = factor(c("M", "F")),
                 gender_02 = factor(c("F", "F")))

pivot_longer(df, -(1:2), names_to = c(".value","per_num"),names_sep = "_" )
#> # A tibble: 4 x 5
#>   hh_id income per_num bday       gender
#>   <int>  <dbl> <chr>   <date>     <fct> 
#> 1     1  55000 01      2015-03-09 M     
#> 2     1  55000 02      1985-09-11 F     
#> 3     2  94000 01      1989-02-11 F     
#> 4     2  94000 02      2000-08-15 F

^{由reprex package（v0.3.0）于2019-09-14创建}

使用来自tidyr的聚集时保留属性（属性不相同）

3 个答案: