英文:
pivot columns with irregular names into rows
问题
df2 <- structure(list(A = c("A_1_01", "A_1_01", "A_1_01"), B = c("A", "A", "A"),C = c("1", "1", "1"), D = c("inside", "eating", "sleeping"),`1` = c("1", "1", "0"), `2` = c("1", "0", "0"), `3` = c("0", "0", "1"),`4` = c("0", "1", "1"), `1_Location` = c("I", "I", "I"), `2_Location` = c("I", "I", "I"),`3_Location` = c("O", "O", "O"), `4_Location` = c("O", "O", "O")),class = "data.frame", row.names = c(NA, -3L))df3 <- structure(list(H = c("1", "2", "3", "4", "1", "2", "3", "4", "1", "2", "3", "4"),A = c("A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01", "A_1_01"),B = c("A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A"),C = c("1", "1", "1", "1", "1", "1", "1", "1", "1", "1", "1", "1"),D = c("inside", "inside", "inside", "inside", "eating", "eating", "eating", "eating", "sleeping", "sleeping", "sleeping", "sleeping"),Value = c(1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1),Location = c("I", "I", "O", "O", "I", "I", "O", "O", "I", "I", "O", "O")),class = "data.frame", row.names = c(NA, -12L))
英文:
Using R how do I pivot the columns into rows to get the required structure in df3 from df2, while extracting new information from the previous dataframe:
df2<- structure(list(A=c("A_1_01", "A_1_01", "A_1_01"), B=c("A", "A", "A"), C=c("1", "1", "1"), D=c("inside", "eating", "sleeping"), "1"=c("1","1","0"), "2"=c("1","0","0"), "3"=c("0","0","1"), "4"=c("0","1","1"), "1_Location"=c("I","I", "I"), "2_Location"=c("I","I", "I"), "3_Location"=c("O","O", "O"), "4_Location"=c("O","O", "O")), class= "data.frame", row.names = c(NA,-3L))
df3<- structure(list(H=c("1","2","3","4","1","2","3","4","1","2","3","4"),A=c("A_1_01", "A_1_01", "A_1_01","A_1_01", "A_1_01","A_1_01","A_1_01", "A_1_01", "A_1_01","A_1_01","A_1_01", "A_1_01"),B=c("A", "A", "A","A", "A", "A","A", "A", "A","A", "A", "A"),C=c("1", "1", "1","1", "1", "1","1", "1", "1","1", "1", "1"),D=c("inside","inside","inside","inside","eating","eating","eating","eating","sleeping","sleeping","sleeping","sleeping"),Value=c(1,1,0,0,1,0,0,1,0,0,1,1),Location=c("I","I","O","O","I","I","O","O","I","I","O","O")),class= "data.frame", row.names = c(NA,-12L))
Thank you
答案1
得分: 3
尝试在仅包含数字的列名称后添加后缀“_Value”后，使用pivot_longer函数进行操作：
library(stringr)library(tidyr)library(dplyr)df2 %>%rename_with(~ str_c(.x, "_Value"), matches("^\\d+$")) %>%pivot_longer(cols = contains("_"), names_to = c("H", ".value"),names_pattern = "(\\d+)_(.*)")
-output
# A tibble: 12 × 7A B C D H Value Location<chr> <chr> <chr> <chr> <chr> <chr> <chr>1 A_1_01 A 1 inside 1 1 I2 A_1_01 A 1 inside 2 1 I3 A_1_01 A 1 inside 3 0 O4 A_1_01 A 1 inside 4 0 O5 A_1_01 A 1 eating 1 1 I6 A_1_01 A 1 eating 2 0 I7 A_1_01 A 1 eating 3 0 O8 A_1_01 A 1 eating 4 1 O9 A_1_01 A 1 sleeping 1 0 I10 A_1_01 A 1 sleeping 2 0 I11 A_1_01 A 1 sleeping 3 1 O12 A_1_01 A 1 sleeping 4 1 O
英文:
Try with pivot_longer after adding a suffix '_Value' to the digit only column names
library(stringr)library(tidyr)library(dplyr)df2 %>%rename_with(~ str_c(.x, "_Value"), matches("^\\d+$")) %>%pivot_longer(cols = contains("_"), names_to = c("H", ".value"),names_pattern = "(\\d+)_(.*)")
-output
# A tibble: 12 × 7A B C D H Value Location<chr> <chr> <chr> <chr> <chr> <chr> <chr>1 A_1_01 A 1 inside 1 1 I2 A_1_01 A 1 inside 2 1 I3 A_1_01 A 1 inside 3 0 O4 A_1_01 A 1 inside 4 0 O5 A_1_01 A 1 eating 1 1 I6 A_1_01 A 1 eating 2 0 I7 A_1_01 A 1 eating 3 0 O8 A_1_01 A 1 eating 4 1 O9 A_1_01 A 1 sleeping 1 0 I10 A_1_01 A 1 sleeping 2 0 I11 A_1_01 A 1 sleeping 3 1 O12 A_1_01 A 1 sleeping 4 1 O
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