Question

我在列表中有多行，例如＆＃39;

[1,]  "Home"
[2,]  "A"
[3,]  "B"
[4,]  "C"
[5,]  "Home"
[6,]  "D"
[7,]  "E"
[8,]  "Home"
[9,]  "F"
[10,] "G"
[11,] "H"
[12,] "I"

这些行动态地......在＆＃34; Home＆＃34;可以有两个，三个，四个，五个或更多子类....所以绑定不起作用...我有超过5000行和＆＃34; Home＆＃34;在每个子类别的开头都很常见..

我希望它看起来像这样。

       [,1]   [,2] [,3] [,4] [,5]

[1,]  "Home"  "A"  "B"  "C"   
[2,]  "Home"  "D"  "E"
[3,]  "Home"  "F"  "G"  "H"  "I"

OR

我还使用transpose将所有行转换为列并且我使用了转置。

   [,1]    [,2] [,3] [,4]  [,5]   [,6]  [,7]  [,8]   [,9] [,10] [,11] [,12]

   "Home"  "A"  "B"  "C"  "Home"   "D"   "E"  "Home"  "F"  "G"   "H"   "I"

任何解决方案对我来说都可以使用＆＃34; Home＆＃34;
的字符串匹配将行转换为列要么使用＆＃34; Home＆＃34;将列隐藏到行中字符串匹配（转置一个）....

数据

x <- c("Home", "A", "B", "C", "Home", "D", "E", "Home", "F", "G", "H", "I")
x <- matrix(x)

问题已经解决了......谢谢你的回复...... 我通过其他方式做到了...通过在循环中运行它并在每个节点结束后添加行

List <- c() 

#loop start
nodes <- html_nodes(file,".class a b c ") %>% html_text()
List[[length(List)+1]] = nodes      
#loop ends

library(stringi)
catdf <- stri_list2matrix(List, byrow = TRUE)
catdf <- as.data.frame(catdf)

Answer 1

# create the data
x <- as.matrix(c("Home", "A", "B", "C", "Home", "D", "E", "Home", "F" ,"G" ,"H" ,"I"))

# split the data into a list of vectors, wherever "Home" is found
rowstarts <- x == "Home"
rowlist <- split(x, cumsum(rowstarts))

然后，我们可以使用plyr的{{1}}函数将列表绑定到单个数据框中：

ldply

并将所有内容放在一起，简短的单行：

> plyr::ldply(rowlist, rbind)[-1]
     1 2 3    4    5
1 Home A B    C <NA>
2 Home D E <NA> <NA>
3 Home F G    H    I

Answer 2

与C Braun的答案类似，但保留在{"data " : null}：

base

Answer 3

我无法想到任何内置函数来执行此操作，但您可以使用base R函数创建自己的函数：

vector.split <- function(x, sep = "Home") {
    bool.sep <- x == sep
    split(x[!bool.sep], cumsum(bool.sep)[!bool.sep])
}

通过匹配R中的字符串将行转换为列

3 个答案: