有时我会看到以this question格式格式化的堆栈溢出问题中发布的数据。这不是第一次，所以我决定问一个问题，并以使发布的数据可口的方式回答该问题。
我将在此处发布数据集示例，以防万一问题被删除。
+------------+------+------+----------+--------------------------+
| Date | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A | A1 | 0 | 0 |
| 2018-06-03 | A | A2 | 0 | 1 |
| 2018-06-03 | A | A3 | 0 | 2 |
| 2018-06-03 | A | A4 | 1 | 1 |
| 2018-06-03 | A | A5 | 2 | 1 |
| 2018-06-04 | A | A6 | 0 | 3 |
| 2018-06-01 | B | B1 | 0 | 1 |
| 2018-06-02 | B | B2 | 0 | 2 |
| 2018-06-03 | B | B3 | 0 | 3 |
+------------+------+------+----------+--------------------------+
如您所见，这不是发布数据的正确方法。正如a user在评论中写道，
以某种方式格式化数据必须花费一些时间 在这里显示。不幸的是，这不是我们的好格式 复制并粘贴。
我相信这说明了一切。提问者的意图很强，花了一些时间和精力才能变得更好，但结果并不理想。
如果有的话，R代码可以做什么使该表可用？这会带来很多麻烦吗？
答案 0 :(得分：25)
使用data.table::fread：
x = '
+------------+------+------+----------+--------------------------+
| Date | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A | A1 | 0 | 0 |
| 2018-06-03 | A | A2 | 0 | 1 |
| 2018-06-03 | A | A3 | 0 | 2 |
| 2018-06-03 | A | A4 | 1 | 1 |
| 2018-06-03 | A | A5 | 2 | 1 |
| 2018-06-04 | A | A6 | 0 | 3 |
| 2018-06-01 | B | B1 | 0 | 1 |
| 2018-06-02 | B | B2 | 0 | 2 |
| 2018-06-03 | B | B3 | 0 | 3 |
+------------+------+------+----------+--------------------------+
'
fread(gsub('\\+.+\\n' ,'', x, perl = T), drop=c(1,7))
# Date Emp1 Case Priority PriorityCountinLast7days
# 1: 2018-06-01 A A1 0 0
# 2: 2018-06-03 A A2 0 1
# 3: 2018-06-03 A A3 0 2
# 4: 2018-06-03 A A4 1 1
# 5: 2018-06-03 A A5 2 1
# 6: 2018-06-04 A A6 0 3
# 7: 2018-06-01 B B1 0 1
# 8: 2018-06-02 B B2 0 2
# 9: 2018-06-03 B B3 0 3
gsub部分删除水平线。 drop删除了由行尾定界符引起的多余列。
答案 1 :(得分：18)
这个问题的简短答案是肯定的，R代码可以解决麻烦，而不是，它不会带来太多麻烦。
将表复制并粘贴到R会话后的第一步是使用read.table设置header，sep，comment.char和{{1 }}参数。
贷方使我想起了论点strip.white和comment.char到@nicola，以及他的评论。
strip.white但是您可以看到结果存在一些问题。
dat <- read.table(text = "
+------------+------+------+----------+--------------------------+
| Date | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A | A1 | 0 | 0 |
| 2018-06-03 | A | A2 | 0 | 1 |
| 2018-06-03 | A | A3 | 0 | 2 |
| 2018-06-03 | A | A4 | 1 | 1 |
| 2018-06-03 | A | A5 | 2 | 1 |
| 2018-06-04 | A | A6 | 0 | 3 |
| 2018-06-01 | B | B1 | 0 | 1 |
| 2018-06-02 | B | B2 | 0 | 2 |
| 2018-06-03 | B | B3 | 0 | 3 |
+------------+------+------+----------+--------------------------+
", header = TRUE, sep = "|", comment.char = "+", strip.white = TRUE)
要使分隔符在每个数据行的开头和结尾处都使R相信这些分隔符会标记额外的列，这不是原始问题的OP的含义。
所以第二步是只保留 real 列。我将按其编号对列进行分组，这很容易做到，它们通常是第一列和最后一列。
dat
X Date Emp1 Case Priority PriorityCountinLast7days X.1
1 NA 2018-06-01 A A1 0 0 NA
2 NA 2018-06-03 A A2 0 1 NA
3 NA 2018-06-03 A A3 0 2 NA
4 NA 2018-06-03 A A4 1 1 NA
5 NA 2018-06-03 A A5 2 1 NA
6 NA 2018-06-04 A A6 0 3 NA
7 NA 2018-06-01 B B1 0 1 NA
8 NA 2018-06-02 B B2 0 2 NA
9 NA 2018-06-03 B B3 0 3 NA
不太难，要好得多。
在这种情况下，仍然存在将列dat <- dat[-c(1, ncol(dat))]
dat
Date Emp1 Case Priority PriorityCountinLast7days
1 2018-06-01 A A1 0 0
2 2018-06-03 A A2 0 1
3 2018-06-03 A A3 0 2
4 2018-06-03 A A4 1 1
5 2018-06-03 A A5 2 1
6 2018-06-04 A A6 0 3
7 2018-06-01 B B1 0 1
8 2018-06-02 B B2 0 2
9 2018-06-03 B B3 0 3
强制转换为类Date的问题。
Date结果令人满意。
dat$Date <- as.Date(dat$Date)
请注意，我尚未设置或多或少的标准参数str(dat)
'data.frame': 9 obs. of 5 variables:
$ Date : Date, format: "2018-06-01" "2018-06-03" ...
$ Emp1 : Factor w/ 2 levels "A","B": 1 1 1 1 1 1 2 2 2
$ Case : Factor w/ 9 levels "A1","A2","A3",..: 1 2 3 4 5 6 7 8 9
$ Priority : int 0 0 0 1 2 0 0 0 0
$ PriorityCountinLast7days: int 0 1 2 1 1 3 1 2 3
。如果需要，应在运行stringsAsFactors = FALSE时执行此操作。
整个过程仅需3行基本R代码。
最后，最终结果采用read.table格式，就像它应该放在首位一样。
dput答案 2 :(得分：5)
md_table <- scan(text = "
+------------+------+------+----------+--------------------------+
| Date | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A | A1 | 0 | 0 |
| 2018-06-03 | A | A2 | 0 | 1 |
| 2018-06-03 | A | A3 | 0 | 2 |
| 2018-06-03 | A | A4 | 1 | 1 |
| 2018-06-03 | A | A5 | 2 | 1 |
| 2018-06-04 | A | A6 | 0 | 3 |
| 2018-06-01 | B | B1 | 0 | 1 |
| 2018-06-02 | B | B2 | 0 | 2 |
| 2018-06-03 | B | B3 | 0 | 3 |
+------------+------+------+----------+--------------------------+",
what = "", sep = "", comment.char = "+", quiet = TRUE)
## it is clear that there are 5 columns
mat <- matrix(md_table[md_table != "|"], ncol = 5, byrow = TRUE)
# [,1] [,2] [,3] [,4] [,5]
# [1,] "Date" "Emp1" "Case" "Priority" "PriorityCountinLast7days"
# [2,] "2018-06-01" "A" "A1" "0" "0"
# [3,] "2018-06-03" "A" "A2" "0" "1"
# [4,] "2018-06-03" "A" "A3" "0" "2"
# [5,] "2018-06-03" "A" "A4" "1" "1"
# [6,] "2018-06-03" "A" "A5" "2" "1"
# [7,] "2018-06-04" "A" "A6" "0" "3"
# [8,] "2018-06-01" "B" "B1" "0" "1"
# [9,] "2018-06-02" "B" "B2" "0" "2"
#[10,] "2018-06-03" "B" "B3" "0" "3"
## a data frame with all character columns
dat <- setNames(data.frame(mat[-1, ], stringsAsFactors = FALSE), mat[1, ])
# Date Emp1 Case Priority PriorityCountinLast7days
#1 2018-06-01 A A1 0 0
#2 2018-06-03 A A2 0 1
#3 2018-06-03 A A3 0 2
#4 2018-06-03 A A4 1 1
#5 2018-06-03 A A5 2 1
#6 2018-06-04 A A6 0 3
#7 2018-06-01 B B1 0 1
#8 2018-06-02 B B2 0 2
#9 2018-06-03 B B3 0 3
## or maybe just use `type.convert` on some columns?
dat[] <- lapply(dat, type.convert)
答案 3 :(得分：4)
问题不在于需要多少行代码，两到五行，相差不大。问题是，它是否还能超出您在此处发布的示例的范围？
我没有在野外遇到过这种事情，但是我尝试构建另一个我认为可能存在的示例。
此后，我又遇到了一些案例，并将它们添加到测试套件中。
我还提供了一个使用box-drawing characters绘制的表格。这些天你不会碰到太多，但是为了完整起见，它在这里。
x1 <- "
+------------+------+------+----------+--------------------------+
| Date | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A | A1 | 0 | 0 |
| 2018-06-03 | A | A2 | 0 | 1 |
| 2018-06-02 | B | B2 | 0 | 2 |
| 2018-06-03 | B | B3 | 0 | 3 |
+------------+------+------+----------+--------------------------+
"
x2 <- "
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Date | Emp1 | Case | Priority | PriorityCountinLast7days
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
2018-06-01 | A | A|1 | 0 | 0
2018-06-03 | A | A|2 | 0 | 1
2018-06-02 | B | B|2 | 0 | 2
2018-06-03 | B | B|3 | 0 | 3
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
"
x3 <- "
Maths | English | Science | History | Class
0.1 | 0.2 | 0.3 | 0.2 | Y2
0.9 | 0.5 | 0.7 | 0.4 | Y1
0.2 | 0.4 | 0.6 | 0.2 | Y2
0.9 | 0.5 | 0.2 | 0.7 | Y1
"
x4 <- "
Season | Team | W | AHWO
-------------------------------------
1 | 2017/2018 | TeamA | 2 | 1.75
2 | 2017/2018 | TeamB | 1 | 1.85
3 | 2017/2018 | TeamC | 1 | 1.70
4 | 2016/2017 | TeamA | 1 | 1.49
5 | 2016/2017 | TeamB | 3 | 1.51
6 | 2016/2017 | TeamC | 2 | N/A
"
x5 <- "
A B C
┌───┬───┬───┐
A │ 5 │ 1 │ 4 │
├───┼───┼───┤
B │ 2 │ 5 │ 3 │
├───┼───┼───┤
C │ 3 │ 4 │ 4 │
└───┴───┴───┘
"
x6 <- "
------------------------------------------------------------
|date |Material |Description |
|----------------------------------------------------------|
|10/04/2013 |WM.5597394 |PNEUMATIC |
|11/07/2013 |GB.D040790 |RING |
------------------------------------------------------------
------------------------------------------------------------
|date |Material |Description |
|----------------------------------------------------------|
|08/06/2013 |WM.4M01004A05 |TOUCHEUR |
|08/06/2013 |WM.4M010108-1 |LEVER |
------------------------------------------------------------
"
我去参加一项功能
f <- function(x=x6, header=TRUE, rem.dup.header=header,
na.strings=c("NA", "N/A"), stringsAsFactors=FALSE, ...) {
# read each row as a character string
x <- scan(text=x, what="character", sep="\n", quiet=TRUE)
# keep only lines containing alphanumerics
x <- x[grep("[[:alnum:]]", x)]
# remove vertical bars with trailing or leading space
x <- gsub("\\|? | \\|?", " ", x)
# remove vertical bars at beginning and end of string
x <- gsub("\\|?$|^\\|?", "", x)
# remove vertical box-drawing characters
x <- gsub("\U2502|\U2503|\U2505|\U2507|\U250A|\U250B", " ", x)
if (rem.dup.header) {
dup.header <- x == x[1]
dup.header[1] <- FALSE
x <- x[!dup.header]
}
# read the result as a table
read.table(text=paste(x, collapse="\n"), header=header,
na.strings=na.strings, stringsAsFactors=stringsAsFactors)
}
lapply(c(x1, x2, x3, x4, x5, x6), f)
输出
[[1]]
Date Emp1 Case Priority PriorityCountinLast7days
1 2018-06-01 A A1 0 0
2 2018-06-03 A A2 0 1
3 2018-06-02 B B2 0 2
4 2018-06-03 B B3 0 3
[[2]]
Date Emp1 Case Priority PriorityCountinLast7days
1 2018-06-01 A A|1 0 0
2 2018-06-03 A A|2 0 1
3 2018-06-02 B B|2 0 2
4 2018-06-03 B B|3 0 3
[[3]]
Maths English Science History Class
1 0.1 0.2 0.3 0.2 Y2
2 0.9 0.5 0.7 0.4 Y1
3 0.2 0.4 0.6 0.2 Y2
4 0.9 0.5 0.2 0.7 Y1
[[4]]
Season Team W AHWO
1 2017/2018 TeamA 2 1.75
2 2017/2018 TeamB 1 1.85
3 2017/2018 TeamC 1 1.70
4 2016/2017 TeamA 1 1.49
5 2016/2017 TeamB 3 1.51
6 2016/2017 TeamC 2 NA
[[5]]
A B C
A 5 1 4
B 2 5 3
C 3 4 4
[[6]]
date Material Description
1 10/04/2013 WM.5597394 PNEUMATIC
2 11/07/2013 GB.D040790 RING
3 08/06/2013 WM.4M01004A05 TOUCHEUR
4 08/06/2013 WM.4M010108-1 LEVER
答案 4 :(得分：2)
好吧，关于这个特定的数据集，我在RStudio中使用了导入功能，但是我事先又采取了另一步骤。
|个字符替换为, Import使用read.csv到RStudio的记事本文件（使用,分隔的列）。但是，如果您要使用R一步来完全理解它，那么我不知道。
答案 5 :(得分：-3)
按照建议，您可以使用 dput 将数据框的内容保存到文件中，在文本编辑器中打开文件并粘贴其内容。仅限前10行的mtcar数据集示例：
dput(mtcars %>% head(10), file = 'reproducible.txt')
reproducible.txt的内容可用于制作数据框/ tibble，如下所示。在这种情况下，数据的格式是机器可读的，但乍一看（不粘贴到R中）很难被人理解。
df <- structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3,
24.4, 22.8, 19.2), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6), disp = c(160,
160, 108, 258, 360, 225, 360, 146.7, 140.8, 167.6), hp = c(110,
110, 93, 110, 175, 105, 245, 62, 95, 123), drat = c(3.9, 3.9,
3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92), wt = c(2.62,
2.875, 2.32, 3.215, 3.44, 3.46, 3.57, 3.19, 3.15, 3.44), qsec = c(16.46,
17.02, 18.61, 19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3), vs = c(0,
0, 1, 1, 0, 1, 0, 1, 1, 1), am = c(1, 1, 1, 0, 0, 0, 0, 0, 0,
0), gear = c(4, 4, 4, 3, 3, 3, 3, 4, 4, 4), carb = c(4, 4, 1,
1, 2, 1, 4, 2, 2, 4)), .Names = c("mpg", "cyl", "disp", "hp",
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4",
"Mazda RX4 Wag", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout",
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280"), class = "data.frame")