我有一个CSV文件，其格式很糟糕，我无法更改（此处简化）：
Inc,a_One,a_Two,a_Three,b_One,b_Two,b_Three
1,1,1.5,"5 Things",2,2.5,"10 Things"
2,5,5.5,"10 Things",6,6.5,"20 Things"
Inc,a_One,a_Two,a_Three,b_One,b_Two,b_Three
3,9,9.5,"15 Things",10,10.5,"30 Things"
我想要的输出是一个新的CSV，包含：
inc,label,one,two,three
1,"a",1,1.5,"5 Things"
2,"a",5,5.5,"10 Things"
3,"a",9,9.5,"15 Things"
1,"b",2,2.5,"10 Things"
2,"b",6,6.5,"20 Things"
3,"b",10,10.5,"30 Things"
基本上：
a_One和b_One值应合并到同一列中。 Inc值（在不同的地方可能有多个这样的行）。有警告：
Inc等属性的列。通常，Inc表示没有a_或b_前缀的任何列。我有一个正则表达式来删除这些前缀。到目前为止，我已经完成了这个：
> wip_path <- 'C:/path/to/horrible.csv'
> rawwip <- read.csv(wip_path, header = FALSE, fill = FALSE)
> rawwip
V1 V2 V3 V4 V5 V6 V7
1 Inc a_One a_Two a_Three b_One b_Two b_Three
2 1 1 1.5 5 Things 2 2.5 10 Things
3 2 5 5.5 10 Things 6 6.5 20 Things
4 Inc a_One a_Two a_Three b_One b_Two b_Three
5 3 9 9.5 15 Things 10 10.5 30 Things
> skips <- which(rawwip$V1==rawwip[1,1])
> skips
[1] 1 4
> filwip <- rawwip[-skips,]
> filwip
V1 V2 V3 V4 V5 V6 V7
2 1 1 1.5 5 Things 2 2.5 10 Things
3 2 5 5.5 10 Things 6 6.5 20 Things
5 3 9 9.5 15 Things 10 10.5 30 Things
> rawwip[1,]
V1 V2 V3 V4 V5 V6 V7
1 Inc a_One a_Two a_Three b_One b_Two b_Three
但是当我尝试对这些字符串应用tolower（）时，我得到：
> tolower(rawwip[1,])
[1] "4" "4" "4" "4" "4" "4" "4"
这是非常意外的。
所以我的问题是：
1）如何访问rawwip[1,]中的标题字符串，以便我可以使用tolower()和其他字符串操作函数重新格式化它们？
2）一旦我完成了这项工作，使用共享名称堆叠列的最有效方法是什么，同时保留每行的inc值？
请记住，将有超过一千个重复列可以过滤到大约20个共享列名。我不会提前知道每个可堆叠列的位置。这需要在脚本中确定。
答案 0 :(得分：3)
您可以使用基本reshape()功能。例如，输入
dd<-read.csv(text='Inc,a_One,a_Two,a_Three,b_One,b_Two,b_Three
1,1,1.5,"5 Things",2,2.5,"10 Things"
2,5,5.5,"10 Things",6,6.5,"20 Things"
inc,a_one,a_two,a_three,b_one,b_two,b_three
3,9,9.5,"15 Things",10,10.5,"30 Things"')
你可以做到
dx <- reshape(subset(dd, Inc!="inc"),
varying=Map(function(x) paste(c("a","b"), x, sep="_"), c("One","Two","Three")),
v.names=c("One","Two","Three"),
idvar="Inc",
timevar="label",
times = c("a","b"),
direction="long")
dx
获取
Inc label One Two Three
1.a 1 a 1 1.5 5 Things
2.a 2 a 5 5.5 10 Things
3.a 3 a 9 9.5 15 Things
1.b 1 b 2 2.5 10 Things
2.b 2 b 6 6.5 20 Things
3.b 3 b 10 10.5 30 Things
因为您的输入数据很乱（嵌入式标题），所以这会创建所有内容作为因素。您可以尝试使用
转换为正确的数据类型dx[]<-lapply(lapply(dx, as.character), type.convert)
答案 1 :(得分：0)
我建议从我的＆＃34; splitstackshape＆＃34;中read.mtable merged.stack和library(splitstackshape) # for merged.stack
library(SOfun) # for read.mtable
的组合。封装
这是方法。我假设您的数据存储在名为＆＃34; somedata.txt＆＃34;的文件中。在你的工作目录中。
我们需要的包裹：
merged.stack首先，抓住名字的矢量。在我们处理它时，从＆＃34; a_one＆＃34;更改名称结构。到＆＃34; one_a＆＃34; - 对于reshape和theNames <- gsub("(.*)_(.*)", "\\2_\\1",
tolower(scan(what = "", sep = ",",
text = readLines("somefile.txt", n = 1))))
来说，这是一种更方便的格式。
read.mtable其次，使用list来读取数据。我们通过识别以字母开头的所有行来创建数据块。如果它与您的实际数据不匹配，您可以使用更具体的正则表达式。
这将创建data.frame do.call(rbind, ...)个，因此我们使用data.frame将其组合在一个theData <- read.mtable("somefile.txt", "^[A-Za-z]", header = FALSE, sep = ",")
theData <- setNames(do.call(rbind, theData), theNames)
中：
theData
# inc one_a two_a three_a one_b two_b three_b
# Inc,a_One,a_Two,a_Three,b_One,b_Two,b_Three.1 1 1 1.5 5 Things 2 2.5 10 Things
# Inc,a_One,a_Two,a_Three,b_One,b_Two,b_Three.2 2 5 5.5 10 Things 6 6.5 20 Things
# inc,a_one,a_two,a_three,b_one,b_two,b_three 3 9 9.5 15 Things 10 10.5 30 Things
这就是数据现在的样子：
merged.stack从这里开始，您可以使用来自＆＃34; splitstackshape＆＃34; .... {/ p>的merged.stack(theData, var.stubs = c("one", "two", "three"), sep = "_")
# inc .time_1 one two three
# 1: 1 a 1 1.5 5 Things
# 2: 1 b 2 2.5 10 Things
# 3: 2 a 5 5.5 10 Things
# 4: 2 b 6 6.5 20 Things
# 5: 3 a 9 9.5 15 Things
# 6: 3 b 10 10.5 30 Things
reshape ...或reshape(theData, direction = "long", idvar = "inc",
varying = 2:ncol(theData), sep = "_")
# inc time one two three
# 1.a 1 a 1 1.5 5 Things
# 2.a 2 a 5 5.5 10 Things
# 3.a 3 a 9 9.5 15 Things
# 1.b 1 b 2 2.5 10 Things
# 2.b 2 b 6 6.5 20 Things
# 3.b 3 b 10 10.5 30 Things
：
meteor