7

我遇到了这个问题。

我有一个数据框（日期），其中一些文档 ID 和日期存储在字符向量中：

  Doc     Dates
1 12345 c("06/01/2000","08/09/2002")
2 23456 c("07/01/2000", 09/08/2003", "07/01/2000")
3 34567 c("09/06/2004", "09/06/2004", "12/30/2006")
4 45678 c("06/01/2000","08/09/2002")

我正在尝试删除 Dates 中的重复元素以获得此结果：

  Doc     Dates
1 12345 c("06/01/2000","08/09/2002")
2 23456 c("07/01/2000", 09/08/2003")
3 34567 c("09/06/2004", "12/30/2006")
4 45678 c("06/01/2000","08/09/2002")

我努力了：

R>unique(dates$dates)

但它会按日期删除重复的行：

  Doc     Dates
1 12345 c("06/01/2000","08/09/2002")
2 23456 c("07/01/2000", 09/08/2003")
3 34567 c("09/06/2004", "12/30/2006")

有关如何仅删除 Dates 中的重复元素而不是按 Dates 删除重复行的任何帮助？

* *已更新数据

# Match some text string (dates) from some text:

df1$dates <- as.character(strapply(df1[[2]], "((\\D\\d{1,2}(/|-)\\d{1,2}(/|-)\\d{2,4})|    ([^/]\\d{1,2}(/|-)\\d{2,4})|((JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV){1}[\\s|-]{0,2}\\d{1,4}(\\D[\\s|-]{0,}\\d{2,4}){0,}))"))

# Drop first 2 columns from dataframe
df2<-df1[ -c(1,2)]

# List data
>df2
872                     7/23/2007
873 c(" 11/4/2007", " 11/4/2007")
874   c(" 4/2/2008", " 8/2/2007")
880                    11/14/2006

> class(df2)
[1] "data.frame"

> class(df2$dates)
[1] "character"

> dput(df2)
structure(list(dates = c("NULL", "NULL", " 7/23/2007", "c(\" 11/4/2007\", \" 11/4/2007\")", 
"c(\" 4/2/2008\", \" 8/2/2007\")", "NULL", "NULL", "NULL", "NULL", 
"NULL", " 11/14/2006")), .Names = "dates", class = "data.frame", row.names = 870:880)

所以我的问题是如何摆脱第 873 行中的重复日期？

4

4 回答 4

1

尝试这个：

within(dates, Dates <- lapply(Dates, unique))
于 2013-07-03T16:00:48.813 回答
1

我解决了从字符向量中删除重复值的问题 - 包装 lapply(strapply(), unique)：

df1$date <- as.character(lapply((strapply(df1[[2]], "((\\D\\d{1,2}(/|-)\\d{1,2}(/|-    )\\d{2,4})|(\\s\\d{1,2}(/|-)\\d{2,4})|((JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV){1}[\\s|-]{0,2}\\d{1,4}(\\D[\\s|-]{0,}\\d{2,4}){0,}))")),unique))

感谢你的帮助。

于 2013-07-10T14:59:42.710 回答
0

我会gsub外出约会，c(然后)对于每一行，我会unique使用strsplit,

未经测试，但可能类似于： sapply(dates$dates, function(x){ new.x=gsub("c(|)","",x) new.x=strsplit(new.x, ",") unique(new.x) })

于 2013-07-03T16:01:34.877 回答
0

你可能正在寻找这样的东西。

 df

     Doc                                       Dates
 1 12345                c("06/01/2000","08/09/2002")
 2 23456 c("07/01/2000", "09/08/2003", "07/01/2000")
 3 34567 c("09/06/2004", "09/06/2004", "12/30/2006")
 4 45678                c("06/01/2000","08/09/2002")

 Eval and Parse
 x <- t(sapply(df[,"Dates"],function(x){unique(eval(parse(text = x)))}))
 df$Dates <- paste(x[,1],x[,2],sep=",")

 df
      Doc                 Dates
  1 12345 06/01/2000,08/09/2002
  2 23456 07/01/2000,09/08/2003
  3 34567 09/06/2004,12/30/2006
  4 45678 06/01/2000,08/09/2002


 Same can be achieved using Regex:

 paste(unique(unlist(strsplit(gsub("c\\(|\\)","",'c("24/07/2012","22/01/2012","24/07/2012")'),","))),sep = "")

 [1] "\"24/07/2012\"" "\"22/01/2012\""

 Haven't tried on data but works
于 2016-08-05T09:22:25.190 回答