我正在为黑客进行机器学习，我被困在这一行：
from.weight <- ddply(priority.train, .(From.EMail), summarise, Freq = length(Subject))
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？错误似乎是由一个不同于priority.train的对象生成的，因为它的names属性显然有9个元素。
我很感激任何帮助。谢谢！
问题已解决
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)
答案 0 :(得分：43)
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$datetime) # convert to POSIXct to use in data frames / ddply
答案 1 :(得分：6)
你可能已经seen this但它没有帮助。我想我们可能还没有答案，因为人们无法重现你的错误。
dput或更小head(dput())可能有助于此。但是这里有一个使用base的替代方案：
x <- data.frame(A=c("a","b","c","a"),B=c("e","d","d","d"))
ddply(x,.(A),summarise, Freq = length(B))
A Freq
1 a 2
2 b 1
3 c 1
tapply(x$B,x$A,length)
a b c
2 1 1
这tapply对您有用吗？
x2 <- data.frame(A=c("removed@removed.ca", "removed@removed.net"),
B=c("please help a newbie compile mplayer :-)",
"re: please help a newbie compile mplayer :-)"))
tapply(x2$B,x2$A,length)
removed@removed.ca removed@removed.net
1 1
ddply(x2,.(A),summarise, Freq = length(B))
A Freq
1 removed@removed.ca 1
2 removed@removed.net 1
您还可以尝试更简单：
table(x2$A)
removed@removed.ca removed@removed.net
1 1
答案 2 :(得分：4)
我有一个非常类似的问题，虽然不确定它是否是一个相同的问题。我收到了以下错误。
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的解决方案对我的情况不起作用。我对问题进行排序的方法是删除plyr 1.8并手动安装plyr 1.7。然后错误就消失了。我也尝试重新安装plyr 1.8并重复了这个问题。
HTH。
答案 3 :(得分：3)
我也遇到了与ddply类似的问题，并在下面给出了代码/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”中有相当多的分类变量。
将分类变量转换为字符变量，如下所示使ddply命令起作用：
test <- data.frame(lapply(test, as.character), stringsAsFactors=FALSE)
答案 4 :(得分：2)
一旦你理解了一个干扰你的日期列，你也可以在运行命令时将该列留下而不是转换它......
所以
from.weight <- ddply(priority.train[,c(1:7,9:10)], .(From.EMail), summarise, Freq = length(Subject))
可以成为
Semaphore例如，如果POSIXlt日期恰好位于数据帧的第8列。报告错误的奇怪之处在于，它可能与您正在分组的内容或您正在寻找的输出信息无关...
答案 5 :(得分：1)
使用ddply并使用doBy
library(doBy)
bylength = function(x){length(x)}
newdt = bylength(X ~From.EMail + To.EMail, data = dt, FUN = bylength)
答案 6 :(得分：0)
我也面临同样的问题，我通过仅保留ddply所需的数据并使用as.character将所有必需的Text变量转换为字符来解决它
它有效