Question

我正在尝试根据每个人第一年的订单过滤订单表。

我的数据采用以下格式，其中每一行代表一个订单，但是我添加了客户级列来表示其首次下订单日期（Recruitment Date）以及自自招募以来，每个客户的第一笔订单（1st Year Since Recruitment）和第二年；最后一栏是当前订单的付款金额。

Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   1876202 obs. of  6 variables:
 $ Brand_Acc                 : chr  "B000000001" "B000000002" "B000000002" "B000000002" ...
 $ salesdate                 : Date, format: "2008-03-10" "2008-02-19" "2008-07-14" "2010-08-25" ...
 $ Recruitment Date          : Date, format: "2008-03-10" "2008-02-19" NA NA ...
 $ 1st Year Since Recruitment: Date, format: "2009-03-10" "2009-02-19" NA NA ...
 $ 2nd Year Since Recruitment: Date, format: "2010-03-10" "2010-02-19" NA NA ...
 $ TotalDiscount             : num  97.9 349.9 184.9 284.9 348.9 ...

我想返回一个数据框，以捕获每个客户第一年的订单额。

我尝试了以下方法：

df %>%
  group_by(Brand_Acc) %>%
  filter(salesdate, between(`Recruitment Date`, `1st Year Since Recruitment`))

但是我得到这个错误：

Error in filter_impl(.data, quo) : Evaluation error: argument "right" is missing, with no default.

正确的方法是什么？

编辑显示前5行的内容：

dput(df)
structure(list(Brand_Acc = c("B000000001", "B000000002", "B000000002", 
"B000000002", "B000000006"), salesdate = structure(c(13948, 13928, 
14074, 14846, 13934), class = "Date"), ordertype = c("Recruitment", 
"Recruitment", "Conversion", "Active Order", "Recruitment"), 
    actv_channel = c("MainMail", "MainMail", "Outbound-Other", 
    "MainMail", "MainMail"), TotalDiscount = c(97.87, 349.88, 
    184.94, 284.94, 348.9), campaignparentid = c("9017", "9017", 
    "9035", "9557", "9017"), BrandAccount_Brand = c("wp", "wp", 
    "wp", "wp", "wp"), recrtype = c("STNRD", "STNRD", "STNRD", 
    "STNRD", "STNRD"), POA_CODE = structure(c(1937L, 2302L, 2302L, 
    2302L, 466L), .Label = c("0", "200", "800", "801", "804"), class = "factor"), 
    `Recruitment Date` = structure(c(13948, 13928, NA, NA, 13934
    ), class = "Date"), `1st Year Since Recruitment` = structure(c(14313, 
    14294, NA, NA, 14300), class = "Date"), `2nd Year Since Recruitment` = structure(c(14678, 
    14659, NA, NA, 14665), class = "Date"), `3rd Year Since Recruitment` = structure(c(15043, 
    15024, NA, NA, 15030), class = "Date")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -5L))
> ```

Answer 1

令人惊讶的是，between的{{1}}和left参数中没有向量化，这是您可能期望的，因为它将自身描述为right和{ <=。我们只需要采取很长的路要走：

>=

^{由reprex package（v0.2.1）于2019-02-13创建}

library(tidyverse) df <- structure(list(Brand_Acc = c("B000000001", "B000000002", "B000000002", "B000000002", "B000000006"), salesdate = structure(c(13948, 13928, 14074, 14846, 13934), class = "Date"), ordertype = c("Recruitment", "Recruitment", "Conversion", "Active Order", "Recruitment"), actv_channel = c("MainMail", "MainMail", "Outbound-Other", "MainMail", "MainMail"), TotalDiscount = c(97.87, 349.88, 184.94, 284.94, 348.9), campaignparentid = c("9017", "9017", "9035", "9557", "9017"), BrandAccount_Brand = c("wp", "wp", "wp", "wp", "wp"), recrtype = c("STNRD", "STNRD", "STNRD", "STNRD", "STNRD"), POA_CODE = structure(c(1937L, 2302L, 2302L, 2302L, 466L), .Label = c("0", "200", "800", "801", "804"), class = "factor"), `Recruitment Date` = structure(c(13948, 13928, NA, NA, 13934), class = "Date"), `1st Year Since Recruitment` = structure(c(14313, 14294, NA, NA, 14300), class = "Date"), `2nd Year Since Recruitment` = structure(c(14678, 14659, NA, NA, 14665), class = "Date"), `3rd Year Since Recruitment` = structure(c(15043, 15024, NA, NA, 15030), class = "Date")), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -5L)) df %>% group_by(Brand_Acc) %>% filter(salesdate >= `Recruitment Date` & salesdate <= `1st Year Since Recruitment`) #> # A tibble: 3 x 13 #> # Groups: Brand_Acc [3] #> Brand_Acc salesdate ordertype actv_channel TotalDiscount #> <chr> <date> <chr> <chr> <dbl> #> 1 B0000000… 2008-03-10 Recruitm… MainMail 97.9 #> 2 B0000000… 2008-02-19 Recruitm… MainMail 350. #> 3 B0000000… 2008-02-25 Recruitm… MainMail 349. #> # … with 8 more variables: campaignparentid <chr>, #> # BrandAccount_Brand <chr>, recrtype <chr>, POA_CODE <fct>, `Recruitment #> # Date` <date>, `1st Year Since Recruitment` <date>, `2nd Year Since #> # Recruitment` <date>, `3rd Year Since Recruitment` <date>中也存在语法错误，尽管现在不相关了：

between

按日期分组和过滤日期

1 个答案: