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如何将列表(包含多个元素)转换为字符串而不转换为“c(“xxx”、“xxx”、“xxx”)”R

data.table split sorting list r

rane · 技术社区 · 6 年前

library(data.table)

# Target string to convert

DATE_DATA <- c("2015-01-02;2015-01-07;2021-05-02;2019-02-05",
"2017-08-02;2000-01-22;2003-03-07;2017-10-09",
"2013-08-02;2022-06-02;2012-03-15")

# Dataset
DT <- data.table(NAME = c("JOE","MARY","PAUL"),DATE = c(DATE_DATA))

预期结果--在新的列调用“period”中转换日期列,如下所示: 拆分+排序递减=F+唯一年份

#  period
1: 2015,2019,2021
2: 2000,2003,2017
3: 2012,2013,2022

# 1st approach -- RESULT : created column with class -- "list"

DT[,period:= lapply(strsplit(DT$DATE,";"),
                                 function(x) sort(unique(str_sub(x,1,4)),
                                                  decreasing = FALSE))]

# 2nd approach -- RESULT : created column with class -- "character" but value
#                          turn to "c("xxx", "xxx", "xxx")" , not expected 
#                          "xxx,xxx,xxx"

DT[,period:= as.character(paste(lapply(strsplit(DT$DATE,";"),
                             function(x) sort(unique(str_sub(x,1,4)),
                                              decreasing = FALSE)),collapse = ","))]

3 回复 | 直到 6 年前

Ronak Shah 6 年前

对于每个 DATE 我们可以分开列在“;”,将它们转换为日期,使用 format toString .

DT$Period <- sapply(DT$DATE, function(x) 
         toString(sort(unique(format(as.Date(strsplit(x, ";")[[1]]), "%Y")))))
DT

#   NAME                                        DATE           Period
#1:  JOE 2015-01-02;2015-01-07;2021-05-02;2019-02-05 2015, 2019, 2021
#2: MARY 2017-08-02;2000-01-22;2003-03-07;2017-10-09 2000, 2003, 2017
#3: PAUL            2013-08-02;2022-06-02;2012-03-15 2012, 2013, 2022

我们可以减少 as.Date 和使用 lubridate 提供相同输出的包。

library(lubridate)
DT$Period <- sapply(DT$DATE, function(x) 
                   toString(sort(unique(year(strsplit(x, ";")[[1]])))))

我不是一个 data.table 但我认为你在尝试中缺少的是分组( by 日期 列中,需要指定 unique 通过 争论。

DT[,period:= paste(sapply(strsplit(DATE,";"),
  function(x) sort(unique(substr(x,1,4)),)),collapse = ","), by = 1:nrow(DT)]

DT

#   NAME                                        DATE         period
#1:  JOE 2015-01-02;2015-01-07;2021-05-02;2019-02-05 2015,2019,2021
#2: MARY 2017-08-02;2000-01-22;2003-03-07;2017-10-09 2000,2003,2017
#3: PAUL            2013-08-02;2022-06-02;2012-03-15 2012,2013,2022

akrun 6 年前

我们可以用 gsub 和 scan

DT[,  Period := toString(sort(unique(scan(text=gsub("-\\d+", 
               "", DATE), what = numeric(), sep=";")))), NAME]
DT
#   NAME                                        DATE           Period
#1:  JOE 2015-01-02;2015-01-07;2021-05-02;2019-02-05 2015, 2019, 2021
#2: MARY 2017-08-02;2000-01-22;2003-03-07;2017-10-09 2000, 2003, 2017
#3: PAUL            2013-08-02;2022-06-02;2012-03-15 2012, 2013, 2022

tidyverse ,在这里我们通过在 ; summarise “期间”作为 sort 预计起飞时间 year 已转换的 Date 班级( ymd select 将列按适当顺序排列(如果需要)

library(tidyverse)
DT %>% 
   separate_rows(DATE, sep = ";") %>% 
   group_by(NAME) %>% 
   summarise(Period = toString(sort(unique(year(ymd(DATE)))))) %>% 
   right_join(DT) %>%
   select(names(DT), everything())
# A tibble: 3 x 3
#  NAME  DATE                                        Period                
#  <chr> <chr>                                       <chr>                 
#1 JOE   2015-01-02;2015-01-07;2021-05-02;2019-02-05 2015, 2019, 2021
#2 MARY  2017-08-02;2000-01-22;2003-03-07;2017-10-09 2000, 2003, 2017
#3 PAUL  2013-08-02;2022-06-02;2012-03-15            2012, 2013, 2022

ira 6 年前

我不确定最快的方法是什么,但相对容易阅读和理解的方法是:

DT[, period:=sapply(strsplit(DATE, ";"), 
     function(x) paste(sort(unique(year(as.Date(x)))), collapse = ","))]

结果输出为:

   NAME                                        DATE         period
1:  JOE 2015-01-02;2015-01-07;2021-05-02;2019-02-05 2015,2019,2021
2: MARY 2017-08-02;2000-01-22;2003-03-07;2017-10-09 2000,2003,2017
3: PAUL            2013-08-02;2022-06-02;2012-03-15 2012,2013,2022

strsplit(DATE, ";")