按日期范围和 ID 连接，面板数据答案

【问题标题】：Joint by date range and ID ,panel data按日期范围和 ID 连接，面板数据
【发布时间】：2019-01-18 05:44:42
【问题描述】：

我有基本的基金数据，我想按日期范围和基金ID添加经理姓名

我尝试了模糊右连接 x = fuzzy_right_join(manager, fundret, by = c("fundName" = "fundName", "date"= "managerStartdate", "date" = "managerENDdate"), match_fun = list('==', '>', '<')) 但是有错误

Error in mf(rep(u_x, n_y), rep(u_y, each = n_x), ...) :could not find function "mf

我尝试了一些 if /else 函数，但仍然不起作用......

library(lubridate)
manager=data.frame(managerName=c("A","B","C","D","E","F","G","H","I","J","K","L","M","N"),
                   fundName=c("fundA","fundA","fundA","fundA","fundA","fundA","fundA","fundA","fundA","fundA","fundA","fundB","fundB","fundC"),
                   managerStartdate=ymd(c("20040227","20041218","20140317","20140317","20120405","20151119","20140620","20150901","20100118","20150107","20051029","20140321","20130320","20140321")),
                   mangerENDdate=ymd(c("20051029","20050412","20151119","20170113","20130628","20170224","20150107","20170328","20140620","NA","20100118","20180912","20140430","20180912")))

fundret=data.frame(
                   fundName=c("fundA","fundA","fundA","fundA","fundA","fundA","fundB","fundC"),
                   date=ymd(c("20020628","20041003","20120508","20140321","20160101","20170201","20140707","20170909")),
                   ret=c(0.05,0.3,0.44444,0.00023,-0.0058,0.00006,-0.5,0.7))

这是我想联合的

wantedresult=data.frame(
  fundName=c("fundA","fundA","fundA","fundA","fundA","fundA","fundB","fundC"),
  date=ymd(c("20020628","20041003","20120508","20140321","20160101","20170201","20140707","20170909")),
  ret=c(0.05,0.3,0.44444,0.00023,-0.0058,0.00006,-0.5,0.7),
  managerName=c("NA","A","I&E","C&D","D&J","J","L","N"))

感谢您的帮助

经理：https://i.stack.imgur.com/sbfk7.jpg

基金：https://i.stack.imgur.com/aQyj1.jpg

想要的结果：https://i.stack.imgur.com/DbWap.jpg

【问题讨论】：

您的数据难以阅读，而且您的问题的重点没有得到理解。你想在这里加入什么？
@TimBiegeleisen 感谢您的建议，我会在帖子中添加我的数据图片

标签： r panel date-range fuzzyjoin

【解决方案1】：

如果你在 match_fun 参数中使用反引号而不是引号，它应该可以工作。

match_fun = list(backtick==backtick, backtick>backtick, backtick

反引号 = `

【讨论】：