我有以下数据框:
id <- c(1,1,2,3,3)
date <- c("23-01-08","01-11-07","30-11-07","17-12-07","12-12-08")
df <- data.frame(id,date)
df$date2 <- as.Date(as.character(df$date), format = "%d-%m-%y")
id date date2
1 23-01-08 2008-01-23
1 01-11-07 2007-11-01
2 30-11-07 2007-11-30
3 17-12-07 2007-12-17
3 12-12-08 2008-12-12
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现在我需要创建第四列并为其中的每一个插入最大事务日期id.决赛桌应如下:
id date date2 max
1 23-01-08 2008-01-23 2008-01-23
1 01-11-07 2007-11-01 0
2 30-11-07 2007-11-30 2007-11-30
3 17-12-07 2007-12-17 0
3 12-12-08 2008-12-12 2008-12-12
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如果你能帮助我,我将感激不尽.
sea*_*avi 21
id<-c(1,1,2,3,3)
date<-c("23-01-08","01-11-07","30-11-07","17-12-07","12-12-08")
df<-data.frame(id,date)
df$date2<-as.Date(as.character(df$date), format = "%d-%m-%y")
# aggregate can be used for this type of thing
d = aggregate(df$date2,by=list(df$id),max)
# And merge the result of aggregate
# with the original data frame
df2 = merge(df,d,by.x=1,by.y=1)
df2
id date date2 x
1 1 23-01-08 2008-01-23 2008-01-23
2 1 01-11-07 2007-11-01 2008-01-23
3 2 30-11-07 2007-11-30 2007-11-30
4 3 17-12-07 2007-12-17 2008-12-12
5 3 12-12-08 2008-12-12 2008-12-12
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编辑:由于您希望当日期与最大日期不匹配时,最后一列为"空",您可以尝试下一行.
df2[df2[,3]!=df2[,4],4]=NA
df2
id date date2 x
1 1 23-01-08 2008-01-23 2008-01-23
2 1 01-11-07 2007-11-01 <NA>
3 2 30-11-07 2007-11-30 2007-11-30
4 3 17-12-07 2007-12-17 <NA>
5 3 12-12-08 2008-12-12 2008-12-12
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当然,清理colnames等总是很好,但我会留给你.
另一种方法是使用plyr包:
library(plyr)
ddply(df, "id", summarize, max = max(date2))
# id max
#1 1 2008-01-23
#2 2 2007-11-30
#3 3 2008-12-12
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现在这不是你所追求的格式,因为它只显示每id一次.永远不要害怕,我们可以使用transform而不是summarize:
ddply(df, "id", transform, max = max(date2))
# id date date2 max
#1 1 01-11-07 2007-11-01 2008-01-23
#2 1 23-01-08 2008-01-23 2008-01-23
#3 2 30-11-07 2007-11-30 2007-11-30
#4 3 12-12-08 2008-12-12 2008-12-12
#5 3 17-12-07 2007-12-17 2008-12-12
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和@ seandavi的回答一样,这会重复max每个的日期id.如果你想要重复更改NA,这样的事情将完成工作:
within(ddply(df, "id", transform, max = max(date2)), max[max != date2] <- NA)
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dplyr在有人看的情况下添加解决方案:
library(dplyr)
df %>%
group_by(id) %>%
mutate(max = if_else(date2 == max(date2), date2, as.Date(NA)))
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结果:
# A tibble: 5 x 4
# Groups: id [3]
id date date2 max
<dbl> <fctr> <date> <date>
1 1 23-01-08 2008-01-23 2008-01-23
2 1 01-11-07 2007-11-01 NA
3 2 30-11-07 2007-11-30 2007-11-30
4 3 17-12-07 2007-12-17 NA
5 3 12-12-08 2008-12-12 2008-12-12
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