从单列文本文件中读取R中的数据

Question

名为myfile.txt的文件设置如下：

Column1
Column2
Column3
...
Column10
Row1
1
2
3
4
Row2
5
6
7
8
...

行最终会变为100并且我在使用read.table命令时遇到问题。我不是一个经验丰富的R用户，所以我只需要解决这个问题并完成它。

我认为col.names看起来像：

read.table("myfile.txt", col.names = 1:10)

但那不起作用

Answer 1

示例myfile.txt：

Column1
Column2
Column3
Column4
Row1
1
2
3
4
Row2
5
6
7
8

阅读文件并创建一个矩阵：

lin <- scan("myfile.txt", "") # read lines

lin2 <- grep("Column|Row", lin, value = TRUE, invert = TRUE) # values

matrix(as.numeric(lin2), ncol = sum(grepl("Column", lin)), byrow = TRUE)
  # create matrix

     [,1] [,2] [,3] [,4]
[1,]    1    2    3    4
[2,]    5    6    7    8

---

如果第一行未命名为Column...但包含实际列名，则可以使用以下方法：

lin <- scan("myfile.txt", "") # read lines

idx <- grep("^Row", lin) # index of lines containing 'Row'

lin2 <- lin[-(c(seq(1, idx[1] - 1), idx))] # actual values

matrix(as.numeric(lin2), nrow = length(idx), 
       dimnames = list(NULL, lin[seq(1, idx[1] - 1)]), byrow = TRUE)

      Column1 Column2 Column3 Column4
 [1,]       1       2       3       4
 [2,]       5       6       7       8

Answer 2

里卡多给出了一个提示，这是一种让它发挥作用的方法：

x <- read.table(text="Column1
Column2
Column3
Column10
Row1
1
2
3
4
Row2
5
6
7
8")

现在插入换行符：

(combined <- paste(x[[1]], collapse='\n'))
[1] "Column1\nColumn2\nColumn3\nColumn10\nRow1\n1\n2\n3\n4\nRow2\n5\n6\n7\n8"

拆分行\ d + \ n：

(comb.split <- strsplit(combined, 'Row\\d+\\n'))
[[1]]
[1] "Column1\nColumn2\nColumn3\nColumn10\n" "1\n2\n3\n4\n"                          "5\n6\n7\n8"                           

将这些元素拆分为换行符：

(split.list <- strsplit(comb.split[[1]], '\\n'))
[[1]]
[1] "Column1"  "Column2"  "Column3"  "Column10"

[[2]]
[1] "1" "2" "3" "4"

[[3]]
[1] "5" "6" "7" "8"

强制数字（如果适用）：

(numeric.list <- lapply(split.list[-1], as.numeric))
[[1]]
[1] 1 2 3 4

[[2]]
[1] 5 6 7 8

创建数据框：

dat <- do.call(rbind, numeric.list)
colnames(dat) <- split.list[[1]]
dat
     Column1 Column2 Column3 Column10
[1,]       1       2       3        4
[2,]       5       6       7        8

这里确实丢失了行名。如果您知道它们是什么，可以使用rownames(dat)<- names添加它们。

Answer 3

1。样本数据

X <- read.table(text=
"Column1
Column2
Column3
Column10
Row1
1
2
3
4
Row2
5
6
7
8
Row99
1
2
3
4
Row100
5
6
7
8", stringsAsFactors=FALSE)

2。读入并处理

# Some string that does not appear naturally in your data
dummyCollapse <- "\n"  # eg:  "zQz5Nsddfdfjjj"
## Make sure to properly escape the escape char.
dummyFind <- "\\n"  

flat  <- paste(unlist(X), collapse=dummyCollapse)
splat <- strsplit(flat, paste0("Row\\d+", dummyFind))

## strsplit returns a list.  You likely want just the first element
splat <- splat[[1]]

## weed out colnames 
cnms <- splat[[1]]  # now, the first element is the coloumn names from the weird data structure
# split them, also on dummyFind
cnms <- strsplit(cnms, dummyFind) 
# again, only want the first element
cnms <- cnms[[1]]  

## Weed out the rows
rows <- tail(splat, -1)

# split on dummy find
rows <- strsplit(rows, dummyFind)
## NOTE: This time, do NOT take just the first element from strsplit. You want them all

## Combine the data into a matrix
MyData <- do.call(cbind, rows)
## Coerce to data.frame, if you'd like
MyData <- as.data.frame(MyData)
## Add in Column names
colnames(MyData) <- cnms

3。结果

> MyData
  Column1 Column2 Column3 Column10
1       1       5       1        5
2       2       6       2        6
3       3       7       3        7
4       4       8       4        8

4。发送电子邮件给whomever以这种格式向您发送数据并点击它们

This should help