ddply错误的含义:'names'属性[9]的长度必须与vector [1]的长度相同

Question

我正在通过黑客机器学习,我被困在这一行:

from.weight <- ddply(priority.train, .(From.EMail), summarise, Freq = length(Subject))

这会产生以下错误:

Error in attributes(out) <- attributes(col) : 
  'names' attribute [9] must be the same length as the vector [1]

这是一个追溯():

> traceback()
11: FUN(1:5[[1L]], ...)
10: lapply(seq_len(n), extract_col_rows, df = x, i = i)
9: extract_rows(x$data, x$index[[i]])
8: `[[.indexed_df`(pieces, i)
7: pieces[[i]]
6: function (i) 
   {
       piece <- pieces[[i]]
       if (.inform) {
           res <- try(.fun(piece, ...))
           if (inherits(res, "try-error")) {
               piece <- paste(capture.output(print(piece)), collapse = "\n")
               stop("with piece ", i, ": \n", piece, call. = FALSE)
           }
       }
       else {
           res <- .fun(piece, ...)
       }
       progress$step()
       res
   }(1L)
5: .Call("loop_apply", as.integer(n), f, env)
4: loop_apply(n, do.ply)
3: llply(.data = .data, .fun = .fun, ..., .progress = .progress, 
       .inform = .inform, .parallel = .parallel, .paropts = .paropts)
2: ldply(.data = pieces, .fun = .fun, ..., .progress = .progress, 
       .inform = .inform, .parallel = .parallel, .paropts = .paropts)
1: ddply(priority.train, .(From.EMail), summarise, Freq = length(Subject))

priority.train对象是一个数据框,这里有更多信息:

> mode(priority.train)
[1] "list"
> names(priority.train)
[1] "Date"       "From.EMail" "Subject"    "Message"    "Path"      
> sapply(priority.train, mode)
       Date  From.EMail     Subject     Message        Path 
     "list" "character" "character" "character" "character" 
> sapply(priority.train, class)
$Date
[1] "POSIXlt" "POSIXt" 

$From.EMail
[1] "character"

$Subject
[1] "character"

$Message
[1] "character"

$Path
[1] "character"

> length(priority.train)
[1] 5
> nrow(priority.train)
[1] 1250
> ncol(priority.train)
[1] 5
> str(priority.train)
'data.frame':   1250 obs. of  5 variables:
 $ Date      : POSIXlt, format: "2002-01-31 22:44:14" "2002-02-01 00:53:41" "2002-02-01 02:01:44" "2002-02-01 10:29:23" ...
 $ From.EMail: chr  "removed@removed.ca" "removed@removed.net" "removed@removed.ca" "removed@removed.net" ...
 $ Subject   : chr  "please help a newbie compile mplayer :-)" "re: please help a newbie compile mplayer :-)" "re: please help a newbie compile mplayer :-)" "re: please help a newbie compile mplayer :-)" ...
 $ Message   : chr  "    \n Hello,\n   \n         I just installed redhat 7.2 and I think I have everything \nworking properly.  Anyway I want to in"| __truncated__ "Make sure you rebuild as root and you're in the directory that you\ndownloaded the file.  Also it might complain of a few depen"| __truncated__ "Lance wrote:\n\n>Make sure you rebuild as root and you're in the directory that you\n>downloaded the file.  Also it might compl"| __truncated__ "Once upon a time, rob wrote :\n\n>  I dl'd gcc3 and libgcc3, but I still get the same error message when I \n> try rpm --rebuil"| __truncated__ ...
 $ Path      : chr  "../03-Classification/data/easy_ham/01061.6610124afa2a5844d41951439d1c1068" "../03-Classification/data/easy_ham/01062.ef7955b391f9b161f3f2106c8cda5edb" "../03-Classification/data/easy_ham/01063.ad3449bd2890a29828ac3978ca8c02ab" "../03-Classification/data/easy_ham/01064.9f4fc60b4e27bba3561e322c82d5f7ff" ...
Warning messages:
1: In encodeString(object, quote = "\"", na.encode = FALSE) :
  it is not known that wchar_t is Unicode on this platform
2: In encodeString(object, quote = "\"", na.encode = FALSE) :
  it is not known that wchar_t is Unicode on this platform

我会发布一个样本,但内容有点长,我不认为这里的内容是相关的.

这里也发生了同样的错误:

> ddply(priority.train, .(Subject))
Error in attributes(out) <- attributes(col) : 
  'names' attribute [9] must be the same length as the vector [1]

有没有人知道这里发生了什么？错误似乎是由priority.train之外的其他对象生成的,因为它的names属性显然有9个元素.

我很感激任何帮助.谢谢!

问题解决了

由于@ user1317221_G使用dput功能的提示,我发现了这个问题.问题在于Date字段,此时此列表包含9个字段(sec,min,hour,mday,mon,year,wday,yday,isdst).为了解决这个问题,我只是将日期转换为字符向量,使用ddply然后将日期转换回日期:

> tmp <- priority.train$Date
> priority.train$Date <- as.character(priority.train$Date)
> from.weight <- ddply(priority.train, .(From.EMail), summarise, Freq = length(Subject))
> priority.train$Date <- tmp
> rm(tmp)

Answer 1

我通过将格式从POSIXlt转换为POSIXct来解决我遇到的问题,正如Hadley在上面建议的那样 - 一行代码:

    mydata$datetime<-strptime(mydata$datetime, "%Y-%m-%d %H:%M:%S") # original conversion from datetime string : > class(mydata$datetime) [1] "POSIXlt" "POSIXt" 
    mydata$datetime<-as.POSIXct(mydata$datetime) # convert to POSIXct to use in data frames / ddply

Answer 2

你可能已经看过这个并没有帮助.我想我们可能还没有答案,因为人们无法重现你的错误.

一个dput或更小head(dput())可能有助于此.但是这里有一个替代方案base:

x <- data.frame(A=c("a","b","c","a"),B=c("e","d","d","d"))

ddply(x,.(A),summarise, Freq = length(B))
  A Freq
1 a    2
2 b    1
3 c    1

 tapply(x$B,x$A,length)
a b c 
2 1 1 

这tapply对你有用吗？

x2 <- data.frame(A=c("removed@removed.ca", "removed@removed.net"),
                 B=c("please help a newbie compile mplayer :-)", 
                     "re: please help a newbie compile mplayer :-)"))

tapply(x2$B,x2$A,length)
removed@removed.ca removed@removed.net 
              1                   1 

ddply(x2,.(A),summarise, Freq = length(B))
                    A Freq
1  removed@removed.ca    1
2 removed@removed.net    1

你也可以尝试更简单:

table(x2$A)

 removed@removed.ca removed@removed.net 
              1                   1