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如何在R中更改为年-月-日格式_R_Date - Fatal编程技术网
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如何在R中更改为年-月-日格式

r date

如何在R中更改为年-月-日格式,r,date,R,Date,我的数据如下所示：“19970325”“19970325”“19970422”“19970516” 我想转换为：1997-03-251977-03-251997-04-22…. 我发现 df[，2]这对我很有效 as.Date(as.character(df1$date), format='%Y%m%d') #[1] "1997-01-01" "1997-01-18" "1997-08-02" "1997-12-12" "1997-01-01" #[6] "1997-01-13" "199

我的数据如下所示：
“19970325”“19970325”“19970422”“19970516”

我想转换为：
1997-03-251977-03-251997-04-22….

我发现 
df[，2]这对我很有效


 as.Date(as.character(df1$date), format='%Y%m%d')
 #[1] "1997-01-01" "1997-01-18" "1997-08-02" "1997-12-12" "1997-01-01"
 #[6] "1997-01-13" "1997-01-01" "1997-01-01" "1997-01-01" "1997-01-01"
 #[11] "1997-01-11" "1997-03-15" "1997-04-16" "1997-04-24"


更新
如果需要删除重复的ID，一个选项是
duplicated
。下面的代码给出了每个重复ID的第一行


 df1[!duplicated(df1$id),]
 #    id     date Amount
 #1    4 19970101  29.33
 #5   21 19970101  63.34
 #7   50 19970101   6.79
 #8   71 19970101  13.97
 #9   86 19970101  23.94
 #10 111 19970101  35.99


假设，如果需要获得每个id的“金额”之和，请使用一个“按组聚合”函数


  aggregate(Amount~id, df1, sum)


数据

df1我会使用
lubridate
（
install.library（“lubridate”）
）




尝试
v1错误：根据示例，在“df[，2]中出现意外的数字常量，前提是我没有收到任何错误。我尝试从表中转换列（我使用了CDNOW_示例）。我不知道
CDNOW_示例
是什么。请使用
dput
的输出用小示例更新您的帖子，即
dput（droplevels（head（yourdata）））


  aggregate(Amount~id, df1, sum)



 df1 <- structure(list(id = c(4L, 4L, 4L, 4L, 21L, 21L, 50L, 71L, 86L, 
 111L, 111L, 111L, 111L, 111L), date = c(19970101L, 19970118L, 
 19970802L, 19971212L, 19970101L, 19970113L, 19970101L, 19970101L, 
 19970101L, 19970101L, 19970111L, 19970315L, 19970416L, 19970424L
 ), Amount = c(29.33, 29.73, 14.96, 26.48, 63.34, 11.77, 6.79, 
 13.97, 23.94, 35.99, 32.99, 77.96, 59.3, 134.98)), .Names = c("id", 
 "date", "Amount"), class = "data.frame", row.names = c("1", "2", 
 "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14"))



library(lubridate)
ymd("20110604")

## [1] "2011-06-04 UTC"