使用 data.table 生成输出 table 时出错

Question

关于第二个代码的调整，我需要一些帮助。代码旨在做同样的事情。第一个代码正确生成输出 table。然而，在第二个代码中，它使用了 data.table 函数。第二个代码取自这里：在问题中完成的示例中，结果是正确的，但是当我测试我的新数据库时，它没有给出预期的结果。第二个代码的输出 table 也必须是 coef = 14 和 Result = 1

你能帮我解决一下吗？

图书馆

library(dplyr)
library(tidyverse)
library(lubridate)
library(data.table)

数据库

df1<-structure(list(Id = 8, date1 = structure(1649376000, tzone = "UTC", class = c("POSIXct", 
        "POSIXt")), date2 = structure(1649376000, tzone = "UTC", class = c("POSIXct", 
        "POSIXt")), Week = "Friday", DT = "0", Category = "ABC", 
            GR = 1, DayR1 = 0, DayM000 = 13, DayM001 = 13, 
            DayM002 = 14, DayM003 = 14, DayM004 = 13, DayM005 = 13, DayM006 = 13, 
            DayM007 = 12, DayM008 = 12, DayM009 = 12, coef = 14), class = "data.frame", row.names = c(NA, 
        -1L))
    
  Id      date1      date2   Week DT Category GR DayR1 DayM000 DayM001 DayM002 DayM003 DayM004 DayM005 DayM006 DayM007 DayM008 DayM009 coef
1  8 2022-04-08 2022-04-08 Friday  0      ABC  1     0      13      13      14      14      13      13      13      12      12      12   14

第一个代码（结果正确）

 df1%>% mutate(across(starts_with("Day"), ~coef - .),
                       across(contains("date"), ymd),
                       datedif = parse_number(as.character(date2-date1)))%>%
    rename_with(~str_replace(.,'(?<=[A-Z])0+(?=.)', ""), starts_with('Day')) %>%
    rowwise %>%
    mutate(Result= if (str_c('DayM', datedif) %in% names(.)) get(str_c('DayM', datedif)) else coef) %>%
    ungroup() %>%
  select(coef, Result)%>%data.frame()

 coef Result
1   14   1

第二个代码（使用data.table函数）。结果错误

dr_names <- grep("^Day", names(df1), value = TRUE)
date_names <- grep("date", names(df1), value = TRUE)
setDT(df1)[, (dr_names) := lapply(.SD, function(x) coef - x), .SDcols = dr_names
    ][, (date_names) := lapply(.SD, as.IDate), .SDcols = date_names
    ][, datedif := date2 - date1]
setnames(df1, dr_names, sub("([A-Z])0+", "\1", dr_names))

   df1[,  .(coef, Result = fcoalesce(as.matrix(.SD)[cbind(.I,         
   match(paste0('DayM', datedif), names(.SD)))], coef)),       .SDcols = patterns("^DayM\d+")]%>%data.frame()

  coef Result
1   14     14

Answer 1

问题出在带有 setnames 的 sub 代码中。该代码匹配大写字母 ([A-Z]) 后的一个或多个 0 (0+)，并删除替换中的那些。在之前的数据集中，它之所以有效，是因为名称模式不同。在这里，名称以 'dr_names'

的数字结尾

> dr_names
 [1] "DayR1"   "DayM000" "DayM001" "DayM002" "DayM003" "DayM004" "DayM005" "DayM006" "DayM007" "DayM008" "DayM009"

因此，对于 'DayM000'，它 returns DayM 无法与 paste0("DayM", datedif) returns DayM0 正确匹配.更改模式以保留最后一位将修复它

setnames(df1, dr_names, sub("([A-Z])0+(.)", "\1\2", dr_names))

现在，我们运行代码

df1[,  .(coef, Result = fcoalesce(as.matrix(.SD)[cbind(.I,         
    match(paste0('DayM', datedif), names(.SD)))], coef)),  
       .SDcols = patterns("^DayM\d+")]%>%
   data.frame()
  coef Result
1   14      1

使用 data.table 生成输出 table 时出错

Error using data.table for output table generation

r

data.table