更新值和融化列的功能

Question

我有一个 data.table，它目前是宽格式且非常大（超过 20,000 行）。数据当前的格式设置为大约有 20-30 列的值为 0 或 1，我需要将它们组合成一列。我可以使用melt 或do.call(paste())，但我不知道如何确定0 和1 来自哪一列。我当前的工作正常的过程是单独更新每一列，以便任何 1 成为与该列名称相同的字符串，然后使用 do.call(paste()) 将所有这些列合并为一个。我觉得必须有一种更优雅的方式来完成这个过程，但我想不出任何东西。有没有比我目前的方法更好的方法（当前方法见下文）？

非常缩小的data.table：

dput(head(photos01a))
structure(list(photo_name = c("BENT-5023-2-150927-CHECK1 (1).JPG", 
"BENT-5023-2-150927-CHECK1 (10).JPG", "BENT-5023-2-150927-CHECK1 (100).JPG", 
"BENT-5023-2-150927-CHECK1 (101).JPG", "BENT-5023-2-150927-CHECK1 (102).JPG", 
"BENT-5023-2-150927-CHECK1 (103).JPG"), BAAS = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), BIRD = c(NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_), CADO = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_), 
CAFA = c(NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_), CALA = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_
), CALL = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_), CEEL = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_), Crew = c("CREW", NA, NA, NA, NA, NA)), row.names = c(NA, 
-6L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000012d1fe71ef0>)

当前方法：

# Adjust the species columns so they can be combined into one column
  photos01a[BAAS == 1, BAAS := "BAAS"]
  photos01a[BIRD == 1, BIRD := "BIRD"] 
  photos01a[CADO == 1, CADO := "CADO"] 
  photos01a[CAFA == 1, CAFA := "CAFA"]
  photos01a[CALA == 1, CALA := "CALA"]
  photos01a[CALL == 1, CALL := "CALL"] 
  photos01a[CEEL == 1, CEEL := "CEEL"] 
  photos01a[Crew == 1, Crew := "CREW"] 

# Create a list of all the species columns
  species_cols <- c("BAAS", "BIRD", "CADO", "CAFA", "CALA", "CALL", "CEEL", "Crew")
  
# Merge the species columns into one column, any photos that have more then one species tagged will have the 4 letter codes pushed together (i.e. TASP and TADO become TASPTADO)
photos01a[, "org_species" := .(col_test = do.call(paste, c(replace(.SD, is.na(.SD), ""), sep = ""))), .SDcols = species_cols]
# Separate out species for photos with multiple tags
photos01a[, "species_1" := substr(org_species, 1, 4)]
photos01a[nchar(org_species) > 4, "species_2" := substr(org_species, 5, 8)]
photos01a[nchar(org_species) > 8, "species_3" := substr(org_species, 9, nchar(org_species))]
# Bring those seperate species back into one column and get rid of unneeded columns
photos01 <- photos01a %>% 
    melt(measure.vars = c("species_1", "species_2", "species_3"),# Pivot table so there 
         value.name = "species",                                 # is only one species column, 
         value.factor = TRUE,                                    # and photos with 2 tags are 
         variable.name = "tag_order",                            # duplicated with one tag in each
         variable.factor = TRUE,
         na.rm = TRUE)

Answer 1

我相信您提供了一个中间对象作为数据，因此您的示例实际上不可重现。

我已对数据进行了编辑，以使其更具说明性。

数据

                            photo_name BAAS BIRD CADO CAFA CALA CALL CEEL Crew
1:   BENT-5023-2-150927-CHECK1 (1).JPG    1   NA   NA   NA   NA   NA   NA    1
2:  BENT-5023-2-150927-CHECK1 (10).JPG   NA    1   NA   NA   NA   NA   NA   NA
3: BENT-5023-2-150927-CHECK1 (100).JPG   NA   NA   NA   NA   NA   NA   NA   NA
4: BENT-5023-2-150927-CHECK1 (101).JPG   NA   NA   NA   NA   NA   NA   NA   NA
5: BENT-5023-2-150927-CHECK1 (102).JPG   NA   NA   NA   NA   NA   NA   NA   NA
6: BENT-5023-2-150927-CHECK1 (103).JPG   NA   NA   NA   NA   NA   NA   NA   NA

我有一个 dplyr 的解决方案，它基本上依赖于replace 和tidyr::unite：

library(dplyr)
library(tidyr)

df %>% mutate(across(BAAS:Crew, ~replace(.x, .x==1, deparse(substitute(.x))))) %>%
        unite(BAAS:Crew, col=new, na.rm=TRUE, sep="")

                            photo_name      new
1:   BENT-5023-2-150927-CHECK1 (1).JPG BAASCrew
2:  BENT-5023-2-150927-CHECK1 (10).JPG     BIRD
3: BENT-5023-2-150927-CHECK1 (100).JPG         
4: BENT-5023-2-150927-CHECK1 (101).JPG         
5: BENT-5023-2-150927-CHECK1 (102).JPG         
6: BENT-5023-2-150927-CHECK1 (103).JPG         

数据输入：

structure(list(photo_name = c("BENT-5023-2-150927-CHECK1 (1).JPG", 
"BENT-5023-2-150927-CHECK1 (10).JPG", "BENT-5023-2-150927-CHECK1 (100).JPG", 
"BENT-5023-2-150927-CHECK1 (101).JPG", "BENT-5023-2-150927-CHECK1 (102).JPG", 
"BENT-5023-2-150927-CHECK1 (103).JPG"), BAAS = c(1, NA, NA, NA, 
NA, NA), BIRD = c(NA, 1, NA, NA, NA, NA), CADO = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), CAFA = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), CALA = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), CALL = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), CEEL = c(NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), Crew = c(1, 
NA, NA, NA, NA, NA)), row.names = c(NA, -6L), class = c("data.table", 
"data.frame"))