我的資料如下:
資料
library(dplyr)
dat_in_one <- structure(list(rn = c("Type_A", "Type_B"
), `[0,25)` = c(5L, 0L), `[25,50)` = c(0L, 0L), `[25,100)` = c(38L,
3L), `[50,100)` = c(0L, 0L), `[100,250)` = c(43L, 5L), `[100,500)` = c(0L,
0L), `[250,500)` = c(27L, 12L), `[500,1000)` = c(44L, 0L), `[1000,1500)` = c(0L,
0L), `[1500,3000)` = c(0L, 0L), `[500,1000000]` = c(0L, 53L),
`[1000,1000000]` = c(20L, 0L), `[3000,1000000]` = c(0L, 0L
), Sum_bin = c(177, 73), strata = list(c(0, 25, 100, 250,
500, 1000, 1e 06), c(0, 25, 100, 250, 500, 1e 06))), row.names = c(NA,
-2L), class = c("data.table", "data.frame"))

library(dplyr)
dat_in_two <- structure(list(rn = c("Type_A", "Type_B"
), `[0,25) East` = c(5L, 0L), `[25,50) South` = c(0L, 0L), `[25,100) West` = c(38L,
3L), `[50,100) North` = c(0L, 0L), `[100,250) East` = c(43L, 5L), `[100,500) South` = c(0L,
0L), `[250,500) South` = c(27L, 12L), `[500,1000) South` = c(44L, 0L), `[1000,1500) South` = c(0L,
0L), `[1500,3000) South` = c(0L, 0L), `[500,1000000] East` = c(0L, 53L),
`[1000,1000000] South` = c(20L, 0L), `[3000,1000000] South` = c(0L, 0L
), Sum_bin = c(177, 73), strata = list(c(0, 25, 100, 250,
500, 1000, 1e 06), c(0, 25, 100, 250, 500, 1e 06))), row.names = c(NA,
-2L), class = c("data.table", "data.frame"))

問題
我想修改下面的這段代碼(它從strata列中提取數字并從中創建兩個新列):
dat_in_one %>%
pivot_longer(-c(rn, strata)) %>%
extract(name, c('lower', 'upper'), '(\\d ),(\\d )', convert = TRUE)
# A tibble: 28 x 5
rn strata lower upper value
<chr> <list> <int> <int> <dbl>
1 Type_A <dbl [7]> 0 25 5
2 Type_A <dbl [7]> 25 50 0
3 Type_A <dbl [7]> 25 100 38
4 Type_A <dbl [7]> 50 100 0
5 Type_A <dbl [7]> 100 250 43
6 Type_A <dbl [7]> 100 500 0
# ... with 22 more rows
它目前忽略East. 我想嘗試調整此代碼,以同時處理和,它在哪里創建第三列。我試圖這樣做,但它根本不起作用。Southdat_in_twodat_in_onedat_in_twodat_in_two
dat_in_two %>%
pivot_longer(-c(rn, strata)) %>%
extract(name, c('lower', 'upper', 'the_rest'), '(\\d ),(\\d ),(\\w )', convert = TRUE)
dat_in_one 的期望結果
# A tibble: 28 x 5
rn strata lower upper value
<chr> <list> <int> <int> <dbl>
1 Type_A <dbl [7]> 0 25 5
2 Type_A <dbl [7]> 25 50 0
3 Type_A <dbl [7]> 25 100 38
4 Type_A <dbl [7]> 50 100 0
5 Type_A <dbl [7]> 100 250 43
6 Type_A <dbl [7]> 100 500 0
# ... with 22 more rows
dat_in_two 的期望結果
# A tibble: 28 x 5
rn strata lower upper rest value
<chr> <list> <int> <int> <char> <dbl>
1 Type_A <dbl [7]> 0 25 East 5
2 Type_A <dbl [7]> 25 50 South 0
3 Type_A <dbl [7]> 25 100 West 38
4 Type_A <dbl [7]> 50 100 North 0
5 Type_A <dbl [7]> 100 250 East 43
6 Type_A <dbl [7]> 100 500 South 0
# ... with 22 more rows
uj5u.com熱心網友回復:
這個怎么樣:
dat_in_one <- structure(list(rn = c("Type_A", "Type_B"
), `[0,25)` = c(5L, 0L), `[25,50)` = c(0L, 0L), `[25,100)` = c(38L,
3L), `[50,100)` = c(0L, 0L), `[100,250)` = c(43L, 5L), `[100,500)` = c(0L,
0L), `[250,500)` = c(27L, 12L), `[500,1000)` = c(44L, 0L), `[1000,1500)` = c(0L,
0L), `[1500,3000)` = c(0L, 0L), `[500,1000000]` = c(0L, 53L),
`[1000,1000000]` = c(20L, 0L), `[3000,1000000]` = c(0L, 0L
), Sum_bin = c(177, 73), strata = list(c(0, 25, 100, 250,
500, 1000, 1e 06), c(0, 25, 100, 250, 500, 1e 06))), row.names = c(NA,
-2L), class = c("data.table", "data.frame"))
dat_in_two <- structure(list(rn = c("Type_A", "Type_B"
), `[0,25) East` = c(5L, 0L), `[25,50) South` = c(0L, 0L), `[25,100) West` = c(38L,
3L), `[50,100) North` = c(0L, 0L), `[100,250) East` = c(43L, 5L), `[100,500) South` = c(0L,
0L), `[250,500) South` = c(27L, 12L), `[500,1000) South` = c(44L, 0L), `[1000,1500) South` = c(0L,
0L), `[1500,3000) South` = c(0L, 0L), `[500,1000000] East` = c(0L, 53L),
`[1000,1000000] South` = c(20L, 0L), `[3000,1000000] South` = c(0L, 0L
), Sum_bin = c(177, 73), strata = list(c(0, 25, 100, 250,
500, 1000, 1e 06), c(0, 25, 100, 250, 500, 1e 06))), row.names = c(NA,
-2L), class = c("data.table", "data.frame"))
library(dplyr)
library(tidyr)
is_all_na <- function(x)all(is.na(x))
dat_in_two %>%
pivot_longer(-c(rn, strata)) %>%
extract(name, c('lower', 'upper', 'rest'), '(\\d ),(\\d )[\\]\\)]\\s*(\\w*)', convert = TRUE) %>%
select(-where(is_all_na))
#> # A tibble: 28 × 6
#> rn strata lower upper rest value
#> <chr> <list> <int> <int> <chr> <dbl>
#> 1 Type_A <dbl [7]> 0 25 East 5
#> 2 Type_A <dbl [7]> 25 50 South 0
#> 3 Type_A <dbl [7]> 25 100 West 38
#> 4 Type_A <dbl [7]> 50 100 North 0
#> 5 Type_A <dbl [7]> 100 250 East 43
#> 6 Type_A <dbl [7]> 100 500 South 0
#> 7 Type_A <dbl [7]> 250 500 South 27
#> 8 Type_A <dbl [7]> 500 1000 South 44
#> 9 Type_A <dbl [7]> 1000 1500 South 0
#> 10 Type_A <dbl [7]> 1500 3000 South 0
#> # … with 18 more rows
dat_in_one %>%
pivot_longer(-c(rn, strata)) %>%
extract(name, c('lower', 'upper', 'rest'), '(\\d ),(\\d )[\\]\\)]\\s*(\\w*)', convert = TRUE) %>%
select(-where(is_all_na))
#> # A tibble: 28 × 5
#> rn strata lower upper value
#> <chr> <list> <int> <int> <dbl>
#> 1 Type_A <dbl [7]> 0 25 5
#> 2 Type_A <dbl [7]> 25 50 0
#> 3 Type_A <dbl [7]> 25 100 38
#> 4 Type_A <dbl [7]> 50 100 0
#> 5 Type_A <dbl [7]> 100 250 43
#> 6 Type_A <dbl [7]> 100 500 0
#> 7 Type_A <dbl [7]> 250 500 27
#> 8 Type_A <dbl [7]> 500 1000 44
#> 9 Type_A <dbl [7]> 1000 1500 0
#> 10 Type_A <dbl [7]> 1500 3000 0
#> # … with 18 more rows
由reprex 包于 2022-05-04 創建(v2.0.1)
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