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r ifelse条件用于计算多个数据帧

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我有3个数据帧,df1 =时间间隔,df2 = ID列表,df3 =具有关联日期的ID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")

代码应该比较df2 $ ID == df3 $ ID . 如果为真,如果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 . 这在我手动指定特定单元格时有效,但这不是我想要的,因为每个dfs都有更多(不均匀的行) .

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(当删除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

有任何想法吗?我不知道如何合并dfs使它们具有相同的长度 . 请注意,df2可以是任何行长度,不会影响代码 . 当df1和df3的行数不同时会出现问题 .

1 回答

  • 0

    >< 被矢量化:

    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 data.table 库中的 between 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
    

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