Question

我需要散布数据框的日期值（从长到宽），但是由于需要两个变量，因此难以实现。

我想到的一种解决方案可能是创建两个单独的数据框，每个变量一个，时值列在行中，日期列在列中。

I asked this question differently initially，但此后就想到了一种更好的摆放方式；因此，我不会删除它，而是发布我的修订要求，因为原始问题可能会帮助其他人。

我的数据框：

df <- structure(list(date = structure(c(17563, 17563, 17563, 17563, 
17563, 17563, 17563, 17563, 17563, 17563, 17563, 17563, 17563, 
17563, 17563, 17563, 17563, 17563, 17563, 17563, 17563, 17563, 
17563, 17563, 17564, 17564, 17564, 17564, 17564, 17564, 17564, 
17564, 17564, 17564, 17564, 17564, 17564, 17564, 17564, 17564, 
17564, 17564, 17564, 17564, 17564, 17564, 17564, 17564, 17565, 
17565, 17565, 17565, 17565, 17565, 17565, 17565, 17565, 17565, 
17565, 17565, 17565, 17565, 17565, 17565, 17565, 17565, 17565, 
17565, 17565, 17565, 17565, 17565, 17566, 17566, 17566, 17566, 
17566, 17566, 17566, 17566, 17566, 17566, 17566, 17566, 17566, 
17566, 17566, 17566, 17566, 17566, 17566, 17566, 17566, 17566, 
17566, 17566), class = "Date"), hour = c("00", "01", "02", "03", 
"04", "05", "06", "07", "08", "09", "10", "11", "12", "13", "14", 
"15", "16", "17", "18", "19", "20", "21", "22", "23", "00", "01", 
"02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12", 
"13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23", 
"00", "01", "02", "03", "04", "05", "06", "07", "08", "09", "10", 
"11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", 
"22", "23", "00", "01", "02", "03", "04", "05", "06", "07", "08", 
"09", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", 
"20", "21", "22", "23"), offered = c(30L, 28L, 15L, 21L, 11L, 
14L, 18L, 35L, 42L, 36L, 37L, 38L, 54L, 45L, 37L, 52L, 40L, 66L, 
84L, 69L, 75L, 51L, 39L, 38L, 25L, 21L, 18L, 20L, 7L, 14L, 14L, 
28L, 37L, 50L, 46L, 31L, 45L, 45L, 39L, 31L, 48L, 69L, 91L, 117L, 
74L, 66L, 60L, 37L, 20L, 31L, 15L, 26L, 18L, 12L, 21L, 42L, 107L, 
118L, 138L, 137L, 93L, 109L, 102L, 91L, 102L, 76L, 76L, 70L, 
68L, 74L, 55L, 54L, 28L, 19L, 23L, 12L, 16L, 12L, 18L, 39L, 96L, 
119L, 111L, 95L, 65L, 81L, 67L, 76L, 64L, 64L, 68L, 71L, 54L, 
65L, 51L, 41L), answered = c(30L, 28L, 15L, 21L, 11L, 14L, 18L, 
35L, 42L, 36L, 37L, 38L, 54L, 45L, 37L, 51L, 40L, 66L, 83L, 68L, 
74L, 51L, 39L, 38L, 25L, 21L, 18L, 20L, 7L, 14L, 14L, 28L, 37L, 
49L, 46L, 31L, 43L, 45L, 39L, 31L, 47L, 65L, 81L, 83L, 61L, 65L, 
58L, 37L, 20L, 31L, 15L, 25L, 17L, 12L, 21L, 42L, 106L, 115L, 
134L, 127L, 93L, 107L, 97L, 88L, 94L, 74L, 74L, 66L, 65L, 69L, 
52L, 51L, 28L, 19L, 23L, 12L, 16L, 12L, 17L, 39L, 91L, 115L, 
104L, 95L, 65L, 79L, 67L, 73L, 64L, 64L, 68L, 70L, 53L, 64L, 
48L, 38L)), row.names = c(NA, -96L), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), groups = structure(list(date = structure(c(17563, 
17564, 17565, 17566), class = "Date"), .rows = list(1:24, 25:48, 
    49:72, 73:96)), row.names = c(NA, -4L), class = c("tbl_df", 
"tbl", "data.frame"), .drop = TRUE))

看起来像这样：

> head(df)
# A tibble: 6 x 4
# Groups:   date [1]
  date       hour  offered answered
  <date>     <chr>   <int>    <int>
1 2018-02-01 00         30       30
2 2018-02-01 01         28       28
3 2018-02-01 02         15       15
4 2018-02-01 03         21       21
5 2018-02-01 04         11       11
6 2018-02-01 05         14       14

这就是我希望输出看起来的样子（一个用于offered，一个用于answered）：

我非常确定我可以使用tidyr::spread()来实现这一点，但是还无法使其看起来像上面的图像。

我该如何实现？

Answer 1

我认为您可以分为两部分，select必需的列，然后spread将其更改为宽格式，然后通过粘贴当前的hour值来更改hour列与下一个hour值。

对于offered

library(tidyverse)

df %>%
  select(date, hour, offered) %>%
  spread(date, offered) %>%
  mutate(hour = paste(hour, lead(hour, default = first(hour)), sep = "-")) 


# A tibble: 24 x 5
#   hour  `2018-02-01` `2018-02-02` `2018-02-03` `2018-02-04`
#   <chr>        <int>        <int>        <int>        <int>
# 1 00-01           30           25           20           28
# 2 01-02           28           21           31           19
# 3 02-03           15           18           15           23
# 4 03-04           21           20           26           12
# 5 04-05           11            7           18           16
# 6 05-06           14           14           12           12
# 7 06-07           18           14           21           18
# 8 07-08           35           28           42           39
# 9 08-09           42           37          107           96
#10 09-10           36           50          118          119
# … with 14 more rows

和answered

df %>%
  select(date, hour, answered) %>%
  spread(date, answered) %>%
  mutate(hour = paste(hour, lead(hour, default = first(hour)), sep = "-"))

Answer 2

随着pivot_wider被弃用，我们可以使用tidyr中的spread

library(dplyr)
library(tidyr)
library(stringr)
df %>% 
  select(-answered) %>%
  pivot_wider(names_from = date, values_from = offered) %>% 
  mutate(hour = str_c(hour, lead(hour, default = first(hour)), sep="_"))
# A tibble: 24 x 5
#   hour  `2018-02-01` `2018-02-02` `2018-02-03` `2018-02-04`
#   <chr>        <int>        <int>        <int>        <int>
# 1 00_01           30           25           20           28
# 2 01_02           28           21           31           19
# 3 02_03           15           18           15           23
# 4 03_04           21           20           26           12
# 5 04_05           11            7           18           16
# 6 05_06           14           14           12           12
# 7 06_07           18           14           21           18
# 8 07_08           35           28           42           39
# 9 08_09           42           37          107           96
#10 09_10           36           50          118          119
# … with 14 more rows

如何在此数据框中散布日期？

2 个答案: