38

我正在为黑客学习机器学习，我被困在这条线上：

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]

这是一个回溯（）：

> 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

我会发布一个示例，但内容有点长，我认为内容与这里无关。

同样的错误也发生在这里：

> 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？该错误似乎是由与priority.train 不同的对象生成的，因为它的names 属性显然有9 个元素。

我会很感激任何帮助。谢谢！

问题解决了

由于@user1317221_G 使用 dput 函数的提示，我发现了问题。问题在于 Date 字段，此时它是一个包含 9 个字段（sec、min、hour、mday、mon、year、wday、yday、isdst）的列表。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)
4

8 回答 8

44

正如 Hadley 建议的那样，我通过将格式从 POSIXlt 转换为 POSIXct 解决了这个问题 - 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
于 2013-02-02T03:58:02.390 回答
6

您可能已经看到了这一点，但它并没有帮助。我想我们可能还没有答案，因为人们无法重现您的错误。

Adput或更小head(dput())可能对此有所帮助。但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

你也可以更简单地尝试：

table(x2$A)

 removed@removed.ca removed@removed.net 
              1                   1
于 2013-01-04T09:15:20.410 回答
4

我有一个非常相似的问题，但不确定它是否相同。我收到以下错误。 

Error in attributes(out) <- attributes(col) : 
  'names' attribute [20388] must be the same length as the vector [128]

我在列表模式下没有任何变量，所以 Mota 的解决方案不适用于我的情况。我对问题进行排序的方式是删除 plyr 1.8 并手动安装 plyr 1.7。然后错误消失了。我也尝试重新安装 plyr 1.8 并复制了问题。

HTH。

于 2013-01-04T17:58:46.940 回答
3

我也遇到了 ddply 的类似问题，并给出了以下代码/错误：

    test <- ddply(test, "catColumn", function(df) df[1:min(nrow(df), 3),])
    Error: 'names' attribute [11] must be the same length as the vector [2]

数据框“测试”中有很多分类变量。

如下将分类变量转换为字符变量使 ddply 命令起作用：

    test <- data.frame(lapply(test, as.character), stringsAsFactors=FALSE)
于 2014-07-30T09:13:41.287 回答
2

一旦您了解干扰的是一个日期列，您也可以在运行命令时简单地忽略该列而不是转换它...

所以

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

可以变成

from.weight <- ddply(priority.train[,c(1:7,9:10)], .(From.EMail), summarise, Freq = length(Subject))

例如，如果 POSIXlt 日期恰好在数据框的第 8 列中。报告的错误的奇怪之处在于它可能与您尝试分组的内容或您正在寻找的输出信息无关......

于 2015-07-14T17:40:27.290 回答
1

我在使用ddply和修复时遇到了同样的问题doBy

library(doBy) 
bylength = function(x){length(x)} 
newdt = bylength(X ~From.EMail + To.EMail, data = dt, FUN = bylength)
于 2015-03-24T21:18:38.673 回答
0

我也遇到了同样的问题，我通过只保留 ddply 所需的数据并使用 as.character 将过滤器变量和所有所需的 Text 变量转换为字符来解决它

有效

于 2016-02-11T09:56:04.153 回答
0

没有数据我无法对此进行测试，但请尝试使用dplyr而不是plyr. 像这样的东西应该返回预期的输出。您必须将其强制返回数据框，因为dplyr输出将是一个小标题。

from.weight <- priority.train %>%
               group_by(From.EMail) %>%
               summarise(Freq = length(Subject)) %>%
               ungroup() %>%
               as.data.frame()
于 2022-02-24T17:31:48.340 回答