有时我看到Stack Overflow问题中发布的数据格式与此问题类似.这不是第一次,所以我决定提出一个问题,然后用一种方法回答问题,使发布的数据变得可口.
我将在此处发布数据集示例,以防问题被删除.
+------------+------+------+----------+--------------------------+
| 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 |
+------------+------+------+----------+--------------------------+
正如您所看到的,这不是发布数据的正确方法.当用户在评论中写道,
它必须花费一些时间来按照您在此处显示的方式格式化数据.不幸的是,这不是我们复制和粘贴的好格式.
我相信这说明了一切.提问者很有意思,并且需要一些工作和时间来尝试变得更好,但结果并不好.
R代码可以做什么来使该表可用,如果有的话？会不会有很多麻烦？
dww*_*dww 28
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
该gsub部分删除水平规则. drop删除由行末端的分隔符引起的额外列.
Rui*_*das 21
对问题的简短回答是肯定的,R代码可以解决这个问题,不会,它不会带来太多麻烦.
复制&粘贴表成R会话之后的第一个步骤是读取它在read.table设置header,sep,comment.char和strip.white参数.
提醒我参与的信用comment.char并strip.white转到@nicola,以及他的评论.
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)
但正如您所看到的,结果存在一些问题.
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
为了让分隔符开始和结束每个数据行,R相信那些分隔符标记了额外的列,这不是原始问题OP的含义.
所以第二步是只保留真正的列.我将按照数字对列进行子集化,轻松完成,它们通常是第一列和最后一列.
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
这不是太难,更好.
在这种情况下,仍然存在一个问题,即强制列Date到类Date.
dat$Date <- as.Date(dat$Date)
结果令人满意.
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
请注意,我没有设置或多或少的标准参数stringsAsFactors = FALSE.如果需要,这应该在运行时完成read.table.
整个过程只需3行基本R代码.
最后,dput格式的最终结果,就像它应该在第一位.
dat <-
structure(list(Date = structure(c(17683, 17685, 17685, 17685,
17685, 17686, 17683, 17684, 17685), class = "Date"), Emp1 = c("A",
"A", "A", "A", "A", "A", "B", "B", "B"), Case = c("A1", "A2",
"A3", "A4", "A5", "A6", "B1", "B2", "B3"), Priority = c(0, 0,
0, 1, 2, 0, 0, 0, 0), PriorityCountinLast7days = c(0, 1, 2, 1,
1, 3, 1, 2, 3)), row.names = c(NA, -9L), class = "data.frame")
Aks*_*elA 11
问题不在于需要多少行代码，两行或五行，差别不大。问题更多的是它是否会超出您在此处发布的示例。
我还没有在野外遇到过这种事情，但是我尝试构建了另一个我认为可能存在的示例。
从那以后，我又遇到了几个案例并将它们添加到测试套件中。
我还包含了一个使用方框图字符绘制的表格。这些天你不会遇到这么多，但为了完整起见，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 |
------------------------------------------------------------
"
我去参加一个活动
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
x3 来自此处（必须查看编辑历史记录）。
x4 来自这里
x6 来自这里
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)