R：如何将此特殊文件保存为csv文件？_R_Csv_Save

R：如何将此特殊文件保存为csv文件？

r csv

R：如何将此特殊文件保存为csv文件？,r,csv,save,R,Csv,Save,我的输入文件从一个普通的csv表开始 x <- read.table(textConnection( + ' models cores time + aa c1 xxx|yyy + aa c2 xxx|zzz + aa c3 www + aa c4 xxx|vvv + bb c1 vvv|www + bb c2 www|qqq + bb c3 xxx|uuu + bb c4 uuu' ), header=TRUE) 为了使用命令“strsplit”拆分最后一列，我参考前

我的输入文件从一个普通的csv表开始

x <- read.table(textConnection(
+ ' models cores  time 
+ aa c1 xxx|yyy 
+ aa c2 xxx|zzz 
+ aa c3 www 
+ aa c4 xxx|vvv 
+ bb c1 vvv|www 
+ bb c2 www|qqq 
+ bb c3 xxx|uuu 
+ bb c4 uuu' ), header=TRUE)

为了使用命令“strsplit”拆分最后一列，我参考前面发布的问题完成了以下步骤

> write.csv(x, file="x.csv")
> y <- read.csv(file="x.csv",header=TRUE,stringsAsFactors=FALSE)
> str(y)

'data.frame':   8 obs. of  4 variables:
 $ X     : int  1 2 3 4 5 6 7 8
 $ models: chr  "aa" "aa" "aa" "aa" ...
 $ cores : chr  "c1" "c2" "c3" "c4" ...
 $ time  : chr  "xxx|yyy" "xxx|zzz" "www" "xxx|vvv" ...
Warning messages:
1: closing unused connection 4 (" models cores  time \naa c1 xxx|yyy \naa c2 xxx|zzz \naa c3 www \naa c4 xxx|vvv \nbb c1 vvv|www \nbb c2 www|qqq \nbb c3 xxx|uuu \nbb c4 uuu") 
2: closing unused connection 3 (" models cores  time \n4 1 0.000365 \n4 2 0.000259 \n4 3 0.000239 \n4 4 0.000220 \n8 1 0.000259 \n8 2 0.000249 \n8 3 0.000251 \n8 4 0.000258") 

> df2 <- as.data.frame(
+ t(
+ do.call(cbind,
+ lapply(1:nrow(y),function(x){
+ sapply(unlist(strsplit(y[x,4],"\\|")),c,y[x,2:3],USE.NAMES=FALSE)
+ })     )   ) )
> str(df2)

当我输入str（df2）时，我发现所有条目都是一个chr列表：

'data.frame':   14 obs. of  3 variables:
 $ V1    :List of 14
  ..$ : chr "xxx"...
$ models:List of 14
  ..$ : chr "aa"
  ..$ : chr "aa"
$ models:List of 14
  ..$ : chr "aa"
  ..$ : chr "aa"

但是，我很难再次将最终结果保存为csv表

> write.csv(df2, file="df2.csv")
Error in write.table(x, file, nrow(x), p, rnames, sep, eol, na, dec, as.integer(quote),  : 
  unimplemented type 'list' in 'EncodeElement'

如何以csv格式再次保存df2文件？请帮忙。

你正在做的事情似乎非常愚蠢-为什么要写一些东西给CSV只是为了再次阅读但是考虑到

df2

大致就是您想要的，您需要

unlist（）
out <- data.frame(lapply(df2, function(x) factor(unlist(x))))

可以读取和重新输入：
> write.csv(out, file="out.csv", row.names = FALSE)
> read.csv("out.csv")
    V1 models cores
1  xxx     aa    c1
2  yyy     aa    c1
3  xxx     aa    c2
4  zzz     aa    c2
5  www     aa    c3
6  xxx     aa    c4
7  vvv     aa    c4
8  vvv     bb    c1
9  www     bb    c1
10 www     bb    c2
11 qqq     bb    c2
12 xxx     bb    c3
13 uuu     bb    c3
14 uuu     bb    c4


更新：
直接从x
到所需的输出将更简单，而不是将其读出到CSV，然后再重新输入，然后处理y
。例如，这从x
直接到与上面的out
相同的结果：
V1 <- with(x, strsplit(as.character(time), "\\|"))
lens <- lapply(V1, length)

out2 <- data.frame(V1 = factor(unlist(V1)),
                   models = with(x, rep(models, times = lens)),
                   cores = with(x, rep(cores, times = lens)))


旁白：
顺便说一句，由于您只是从R控制台复制了代码，因此它包含提示（+
），因此我们无法轻松地粘贴代码。相反，您可以执行dput（x）
并将其粘贴到Q:
structure(list(models = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("aa", "bb"), class = "factor"), cores = structure(c(1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("c1", "c2", "c3", "c4"
), class = "factor"), time = structure(c(7L, 8L, 3L, 6L, 2L, 
4L, 5L, 1L), .Label = c("uuu", "vvv|www", "www", "www|qqq", "xxx|uuu", 
"xxx|vvv", "xxx|yyy", "xxx|zzz"), class = "factor")), .Names = c("models", 
"cores", "time"), class = "data.frame", row.names = c(NA, -8L
))

那么我们都可以简单地做到：
x <- structure(list(models = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("aa", "bb"), class = "factor"), cores = structure(c(1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("c1", "c2", "c3", "c4"
), class = "factor"), time = structure(c(7L, 8L, 3L, 6L, 2L, 
4L, 5L, 1L), .Label = c("uuu", "vvv|www", "www", "www|qqq", "xxx|uuu", 
"xxx|vvv", "xxx|yyy", "xxx|zzz"), class = "factor")), .Names = c("models", 
"cores", "time"), class = "data.frame", row.names = c(NA, -8L
))

这样，我们就可以简单地重建您拥有的对象和您尝试过的对象。
fun\u转换您之前提到的Q是什么？请提供链接，因为代码中不清楚生成df2
的代码试图做什么，因此我解决了您所问的问题，但由于生成df2
的代码正在做什么，您似乎遇到了各种各样的问题。除上述评论外，我已经用一个解决方案更新了我的答案，该解决方案直接从x
到所需的输出，而不需要上述中间步骤和y
。这里的x
是什么（最后一行）。我得到：strsplit（.x$time，split=“\\\\；”）中的错误：非字符参数
使用在我的答案中提供和输出的x
。啊，我想你需要在函数的第一行中使用as.character（）
环绕.x$time
。我已经添加了适合我的函数的数据示例。问题是它的名字是“y”而不是“x”。我明白了，所以你有一个与OP的Q不同的x
。请注意，要读取的代码x
与OP的代码相同，但read.table（）
调用中的as.is=TRUE
除外。因此，您的字符变量是@sally has factors，因此当我尝试您的答案时，它在x
上不起作用。在Qy中有一些不同之处，增加了一列。@Gavin我已将变量名称更改为与@sally问题中相同的名称。我更喜欢使用“as.is”而不是“stringsAsFactors”。
fun_transform <- function(.x){
    time_split <- strsplit(.x$time,split="\\|")
    n_rec <- sapply(time_split,length)
    ind <- rep(seq(nrow(.x)),n_rec)
    cbind(.x[ind,1:2],time=unlist(time_split,use.names=FALSE))
}

df2 <- fun_transform(y)

> out2
    V1 models cores
1  xxx     aa    c1
2  yyy     aa    c1
3  xxx     aa    c2
4  zzz     aa    c2
5  www     aa    c3
6  xxx     aa    c4
7  vvv     aa    c4
8  vvv     bb    c1
9  www     bb    c1
10 www     bb    c2
11 qqq     bb    c2
12 xxx     bb    c3
13 uuu     bb    c3
14 uuu     bb    c4
> str(out2)
'data.frame':   14 obs. of  3 variables:
 $ V1    : Factor w/ 7 levels "qqq","uuu","vvv",..: 5 6 5 7 4 5 3 3 4 4 ...
 $ models: Factor w/ 2 levels "aa","bb": 1 1 1 1 1 1 1 2 2 2 ...
 $ cores : Factor w/ 4 levels "c1","c2","c3",..: 1 1 2 2 3 4 4 1 1 2 ...
> all.equal(out, out2)
[1] TRUE

structure(list(models = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("aa", "bb"), class = "factor"), cores = structure(c(1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("c1", "c2", "c3", "c4"
), class = "factor"), time = structure(c(7L, 8L, 3L, 6L, 2L, 
4L, 5L, 1L), .Label = c("uuu", "vvv|www", "www", "www|qqq", "xxx|uuu", 
"xxx|vvv", "xxx|yyy", "xxx|zzz"), class = "factor")), .Names = c("models", 
"cores", "time"), class = "data.frame", row.names = c(NA, -8L
))

x <- structure(list(models = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("aa", "bb"), class = "factor"), cores = structure(c(1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("c1", "c2", "c3", "c4"
), class = "factor"), time = structure(c(7L, 8L, 3L, 6L, 2L, 
4L, 5L, 1L), .Label = c("uuu", "vvv|www", "www", "www|qqq", "xxx|uuu", 
"xxx|vvv", "xxx|yyy", "xxx|zzz"), class = "factor")), .Names = c("models", 
"cores", "time"), class = "data.frame", row.names = c(NA, -8L
))

write.csv(x, file="x.csv")
y <- read.csv(file="x.csv", header=TRUE, stringsAsFactors=FALSE)
df2 <- data.frame(
 t(do.call(cbind,
       lapply(1:nrow(y),function(x){
            sapply(unlist(strsplit(y[x,4],"\\|")),c,y[x,2:3],
                     USE.NAMES=FALSE)
            }))))

fun_transform <- function(.x){
    time_split <- strsplit(.x$time,split="\\|")
    n_rec <- sapply(time_split,length)
    ind <- rep(seq(nrow(.x)),n_rec)
    cbind(.x[ind,1:2],time=unlist(time_split,use.names=FALSE))
}

df2 <- fun_transform(y)

txt <- textConnection(
        ' models cores  time 
         aa c1 xxx|yyy 
         aa c2 xxx|zzz 
         aa c3 www 
         aa c4 xxx|vvv 
         bb c1 vvv|www 
         bb c2 www|qqq 
         bb c3 xxx|uuu 
         bb c4 uuu' )

y <- read.table(txt, header=TRUE,as.is=TRUE)
close(txt)