如何在此数据框中传播日期？
How can I spread the dates in this data frame?
我需要扩展数据框的日期值（从长到宽），但我无法实现这一点，因为我需要两个变量。
我想到的一个解决方案可能是创建两个单独的数据框，一个用于每个变量，每小时值列在行中，日期列在列中。
I asked this question differently initially，但后来想到了更好的姿势；因此，我不会删除它，而是发布我修改后的要求，因为原始问题可能会对其他人有所帮助。
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
这是我希望输出的样子（一个用于 offered，一个用于 answered）：
我很确定我可以使用 tidyr::spread() 实现此目的，但无法使其看起来像上图。
我怎样才能做到这一点？
我认为您可以分两部分完成，select 所需的列和 spread 它们为宽格式，然后通过粘贴当前 hour 更改 hour 列] 值与下一个 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 = "-"))
我们可以使用 tidyr 中的 pivot_wider，因为 spread 已被弃用
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
我需要扩展数据框的日期值（从长到宽），但我无法实现这一点，因为我需要两个变量。
我想到的一个解决方案可能是创建两个单独的数据框，一个用于每个变量，每小时值列在行中，日期列在列中。
I asked this question differently initially，但后来想到了更好的姿势；因此，我不会删除它，而是发布我修改后的要求，因为原始问题可能会对其他人有所帮助。
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
这是我希望输出的样子（一个用于 offered，一个用于 answered）：
我很确定我可以使用 tidyr::spread() 实现此目的，但无法使其看起来像上图。
我怎样才能做到这一点？
我认为您可以分两部分完成，select 所需的列和 spread 它们为宽格式，然后通过粘贴当前 hour 更改 hour 列] 值与下一个 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 = "-"))
我们可以使用 tidyr 中的 pivot_wider，因为 spread 已被弃用
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