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
>
和<
被矢量化:您还可以使用
data.table
库中的between
函数