我有一个 df 列表。我想使用 Order 中的信息对 lst 中的每个 df 进行排序。我尝试使用 map 或 map2 来执行此操作。但是我的代码不起作用。有什么建议吗？
我想知道如何使用 map 或 lapply 实现此目标(如果可行的话)。谢谢。
## Sample list
lst1<- list(Test = structure(list(SubjectID = c("101-05-202", "101-06-203",
"102-10-401", "101-11-402", "105-12-403", "101-13-501", "101-14-502",
"101-15-503", "101-01-101", "101-05-202", "101-06-203", "101-11-402",
"101-13-501", "101-14-502", "101-15-503", "101-01-101", "101-05-202",
"101-06-203", "101-11-402", "101-13-501", "101-14-502", "101-15-503",
"101-01-101", "101-05-202", "101-06-203", "101-11-402", "101-13-501",
"101-14-502", "101-15-503", "101-05-202", "101-06-203", "101-11-402",
"101-13-501", "101-14-502", "101-15-503", "101-06-203", "101-11-402",
"101-13-501", "101-14-502", "101-15-503", "101-11-402", "101-13-501",
"101-14-502", "101-15-503", "101-06-203", "101-11-402", "101-14-502",
"101-06-203", "101-14-502", "101-06-203", "101-06-203", "101-06-203",
"101-06-203", "101-06-203", "101-06-203"), Test = c("ELA",
"ELA", "ART", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA, PE", "ELA",
"ELA", "ELA", "ELA, MATH", "ELA", "ELA", "ELA,PE", "ELA", "ELA",
"ELA", "ELA", "ELA", "ELA", "PE", "ELA", "ELA", "ELA", "ELA",
"ELA", "ELA", "ELA", "ELA", "ELA", "ELA,MATH", "ELA", "ELA",
"ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA",
"ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA", "ELA",
"ELA,PE", "ELA")), class = "data.frame", row.names = c(NA, -55L
)), Comments = structure(list(SubjectID = c("101-01-101", "101-02-102",
"106-03-103", "101-05-202", "103-04-201", "103-09-303", "105-12-403"
), COMYN = c("No", "No", "No", "No", "No", "No", "Yes")), class = "data.frame", row.names = c(NA,
-7L)), Enroll = structure(list(SubjectID = c("101-01-101", "101-02-102",
"106-03-103", "101-05-202", "103-04-201", "101-06-203", "102-07-301",
"105-08-302", "103-09-303", "102-10-401", "101-11-402", "105-12-403"
), EDT = c("09/03/2020", "09/11/2020", "10/08/2020", "12/08/2020",
"12/09/2020", "12/23/2020", "02/02/2021", "02/17/2021", "03/10/2021",
"04/19/2021", "04/28/2021", "06/03/2021")), row.names = c(NA,
-12L), class = c("spec_tbl_df", "tbl_df", "tbl", "data.frame"
), spec = structure(list(cols = list(SubjectID = structure(list(), class = c("collector_character",
"collector")), Group = structure(list(), class = c("collector_character",
"collector")), Site = structure(list(), class = c("collector_character",
"collector")), SubjectStatus = structure(list(), class = c("collector_character",
"collector")), Visit = structure(list(), class = c("collector_character",
"collector")), Form = structure(list(), class = c("collector_character",
"collector")), FormEntryDate = structure(list(), class = c("collector_logical",
"collector")), ICDT = structure(list(), class = c("collector_character",
"collector")), ICDT_CODED = structure(list(), class = c("collector_logical",
"collector")), ICPVDATE = structure(list(), class = c("collector_character",
"collector")), ICPVDATE_CODED = structure(list(), class = c("collector_double",
"collector")), Informed_C_CREATE_DTM = structure(list(), class = c("collector_character",
"collector")), Informed_C_LAST_UPD_DTM = structure(list(), class = c("collector_character",
"collector")), SubjectVisitFormID = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1), class = "col_spec")))
## Order file
Order<-structure(list(File = c("Test", "Comments", "Enroll"), Order = c("SubjectID",
"SubjectID", "SubjectID, mdy(EDT)")), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
## Order lst with Order
lst[Order$File] <- map2(lst[Order$File], Order$Order, ~ .x %>%
arrange(paste(.y)))
最佳答案
我们可能需要解析表达式
library(dplyr)
library(purrr)
library(stringr)
library(lubridate)
lst1[Order$File] <- map2(lst1[Order$File], Order$Order,
~ .x %>%
arrange(!!! rlang::parse_exprs(str_replace(.y, ",", ";"))))
关于r - 如何重新排序列表中的多个 df，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/70084226/