Question

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

现在我需要创建第四列，并在其中插入每个id的最大交易日期。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

如果你能帮助我，我将感激不尽。

Answer 1

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

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

当然，清理colnames等总是很好，但我会留给你。

Answer 2

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

现在这不是您所使用的格式，因为它只显示每个id一次。不要害怕，我们可以使用transform代替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

与@ seandavi的回答一样，这会重复每个max的{{1}}日期。如果您想将重复项更改为id，则可以执行以下操作：

NA

Answer 3

如果有人正在寻找，请添加dplyr解决方案：

library(dplyr)

df %>%
  group_by(id) %>%
  mutate(max = if_else(date2 == max(date2), date2, as.Date(NA)))

<强>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

Answer 4

library(sqldf)
tables<- '(SELECT * FROM df
           )
           AS t1,
           (SELECT id,max(date2) date2 FROM df GROUP BY id
           )
           AS t2'

out<-fn$sqldf("SELECT t1.*,t2.date2 mdate FROM $tables WHERE  t1.id=t2.id")
out$mdate<-as.Date(out$mdate)
out$mdate[out$date2!=out$mdate]<-NA
#  id     date      date2      mdate
#1  1 01-11-07 2007-11-01       <NA>
#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       <NA>

Answer 5

您不能将0用作日期值，因此您需要放弃将其保留为日期或接受NA值：

# Date values:
df$maxdt <- ave(df$date2, df$id, 
                    FUN=function(x) ifelse( x == max(x), as.character(x), NA) ) 
str(ave(df$date2, df$id, FUN=function(x) ifelse( x == max(x), as.character(x), NA) ) )
# Date[1:5], format: "2008-01-23" NA "2007-11-30" NA "2008-12-12"

ifelse机制执行一些奇怪的类型检查，只使用x作为上面的第二个参数，但仍然返回Date-class向量。去搞清楚！下面是字符向量选项。

# Character values:
df$maxdt <- ave(as.character(df$date2), df$id, 
                   FUN=function(x) ifelse( x == max(x), x,  "0") )
ave(as.character(df$date2), df$id, FUN=function(x) ifelse( x == max(x), x,  "0") )
[1] "2008-01-23" "0"          "2007-11-30" "0"          "2008-12-12"

Answer 6

当我想查看列的最小/最大日期时，我发现这对您有帮助

最大值：<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <input type="text" class="someClass"> <input type="text" class="otherClass"> <input type="text" id="whatever">
最小值：head(df %>% distinct(date) %>% arrange(desc(date)))

最大值将按降序对日期列进行排序，从而使您可以看到最大值。分钟将以升序排序，以便您查看分钟。

您需要为此使用head(df %>% distinct(date) %>% arrange(date))软件包。

在多行中查找单个列中的最大日期

6 个答案: