Question

我有3个数据帧，df1 =一个时间间隔，df2 = ID列表，df3 = ID及相关日期列表。

df1 <- structure(list(season = structure(c(2L, 1L), .Label = c("summer", 
    "winter"), class = "factor"), mindate = structure(c(1420088400, 
    1433131200), class = c("POSIXct", "POSIXt")), maxdate = structure(c(1433131140, 
    1448945940), class = c("POSIXct", "POSIXt")), diff = structure(c(150.957638888889, 
    183.040972222222), units = "days", class = "difftime")), .Names = c("season", 
    "mindate", "maxdate", "diff"), row.names = c(NA, -2L), class = "data.frame")

df2 <- structure(list(ID = c(23796, 23796, 23796)), .Names = "ID", row.names = c(NA, 
    -3L), class = "data.frame")

df3 <- structure(list(ID = c("23796", "123456", "12134"), time = structure(c(1420909920, 
1444504500, 1444504500), class = c("POSIXct", "POSIXt"), tzone = "US/Eastern")), .Names = c("ID", 
"time"), row.names = c(NA, -3L), class = "data.frame")

如果df2 $ ID == df3 $ ID，则代码应进行比较。如果为true，并且df3 $ time> = df1 $ mindate和df3 $ time <= df1 $ maxdate，则df1 $ maxdate-df3 $ time，否则df1 $ maxdate-df1 $ mindate。我尝试使用ifelse函数。当我手动指定特定的单元格时此方法有效，但这不是我想要的，因为我为每个dfs有更多（行不均匀）。

df1$result <- ifelse(df2[1,1] == df3[1,1] & df3[1,2] >= df1$mindate & df3[1,2] <= df1$maxdate, 
                     difftime(df1$maxdate,df3[1,2],units="days"),
                     difftime(df1$maxdate,df1$mindate,units="days")

编辑：所需的输出是（删除df2的最后一行时）：

 season    mindate             maxdate          diff   result
1 winter 2015-01-01 2015-05-31 23:59:00 150.9576 days 141.9576
2 summer 2015-06-01 2015-11-30 23:59:00 183.0410 days 183.0410

有什么想法吗？我看不到如何合并df以使其具有相同的长度。注意，df2可以是任何行长，并且不影响代码。当df1和df3的行数不同时，就会出现问题。

Answer 1

>和<是矢量化的：

transform(df1,result=ifelse(df3$ID%in%df2$ID & df3$time>mindate & df3$time <maxdate, difftime(maxdate,df3$time),difftime(maxdate,mindate)))
  season             mindate             maxdate          diff   result
1 winter 2014-12-31 21:00:00 2015-05-31 20:59:00 150.9576 days 141.9576
2 summer 2015-05-31 21:00:00 2015-11-30 20:59:00 183.0410 days 183.0410

您还可以使用between库中的data.table函数

library(data.table)
transform(df1,result=ifelse(df3$ID%in%df2$ID&df3$time%between%df1[2:3],
               difftime(maxdate,df3$time),difftime(maxdate,mindate)))

  season             mindate             maxdate          diff   result
1 winter 2014-12-31 21:00:00 2015-05-31 20:59:00 150.9576 days 141.9576
2 summer 2015-05-31 21:00:00 2015-11-30 20:59:00 183.0410 days 183.0410

r在多个数据帧上进行计算的其他条件

1 个答案: