R:如何将这个特殊文件保存为csv文件?

时间:2011-04-17 11:44:10

标签: r csv save

我的输入文件从普通的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)

这是一个以因子为所有条目的文件,如下所示:

> str(x)
'data.frame':   8 obs. of  3 variables:
 $ models: Factor w/ 2 levels "aa","bb": 1 1 1 1 2 2 2 2
 $ cores : Factor w/ 4 levels "c1","c2","c3",..: 1 2 3 4 1 2 3 4
 $ time  : Factor w/ 8 levels "uuu","vvv|www",..: 7 8 3 6 2 4 5 1

为了使用命令“strsplit”拆分最后一列,我参考之前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)

结果就是我所需要的。

> df2
    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

当我输入str(df2)时,我发现所有条目都是chr:

的列表
'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,我很难再将这个最终结果保存为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?请帮助。

2 个答案:

答案 0 :(得分:3)

你在做什么似乎非常愚蠢 - 为什么要把一些东西写到CSV只是为了再次读回来? - 但鉴于df2 大致你想要它,你需要unlist() df2中的三个组件并作为数据框投回。

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

这给了我们:

> out
    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(out)
'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 ...

可以再读出来:

> 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

<强>更新x直接转到所需的输出而不是将其读取到CSV并再次返回然后处理y会更简单。例如,这从x直接转到与上面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)))

给出了:

> 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

<强>除了: 顺便说一句,当您从R控制台复制代码时,我们不会轻易粘贴您的代码,因此它包含提示(+)。相反,你可以完成dput(x)并将其粘贴到你的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
))
然后我们都可以完成:

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
))

与创建df2的调用相同。这本来是更可取的:

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)
            }))))

对我们来说,重建你拥有的对象和你尝试过的东西是一件简单的事情。

答案 1 :(得分:0)

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)

编辑 - 示例数据

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)