R:分裂字符串&根据拆分分配变量

时间:2013-04-09 15:00:28

标签: r

我有一个单一的语义标签字段&语义标签类型。每个标签类型/标签都是逗号分隔的,而每个标签类型和标签都是逗号分隔的。标签是冒号分开的(见下文)。

ID | Semantic Tags

1  |   Person:mitch mcconnell, Person:ashley judd, Position:senator

2  |   Person:mitch mcconnell, Position:senator, ProvinceOrState:kentucky, topicname:politics 

3  |   Person:mitch mcconnell, Person:ashley judd, Organization:senate, Organization:republican 

4  |   Person:ashley judd, topicname:politics

5  |   URL:www.huffingtonpost.com, Company:usa today, Person:chuck todd, Company:msnbc

我想分割每种标签类型(结肠前的术语)&标签(冒号后的术语)分为两个独立的字段:"标签类型" &安培; "标签&#34 ;.生成的文件应如下所示:

ID | Tag Type  |  Tag

1  |  Person   |  mitch McConnell

1  |  Person   |  ashley judd  

1  |  Position |  senator

2  |  Person   |  mitch McConnell

2  |  Position |  senator

2  |  State    |  kentucky

这是我到目前为止的代码......

tag<-strsplit(as.character(emtable$Symantic.Tags),","))
tagtype<-strsplit(as.character(tag),":")

但在那之后,我迷路了!我相信我需要使用lapply或sapply,但我不确定它在哪里...

如果在网站上以其他形式回答了我的道歉 - 我是R&amp; S的新手。这对我来说仍然有点复杂。

提前感谢任何人的帮助。

2 个答案:

答案 0 :(得分:4)

这是另一种(略有不同)的方法:

## dat <- readLines(n=5)
## Person:mitch mcconnell, Person:ashley judd, Position:senator
## Person:mitch mcconnell, Position:senator, ProvinceOrState:kentucky, topicname:politics
## Person:mitch mcconnell, Person:ashley judd, Organization:senate, Organization:republican
## Person:ashley judd, topicname:politics
## URL:www.huffingtonpost.com, URL:http://www.regular-expressions.info

dat3 <- lapply(strsplit(dat, ","), function(x) gsub("^\\s+|\\s+$", "", x))
#dat3 <- lapply(dat2, function(x) x[grepl("Person|Position", x)]) 
dat3 <- lapply(dat3, strsplit, ":(?!/)", perl=TRUE) #break on : not folled by /
dat3 <- data.frame(ID=rep(seq_along(dat3), sapply(dat3, length)),
    do.call(rbind, lapply(dat3, function(x) do.call(rbind, x)))
)

colnames(dat3)[-1] <- c("Tag Type", "Tag")

##    ID        Tag Type                    Tag
## 1   1          Person        mitch mcconnell
## 2   1          Person            ashley judd
## 3   1        Position                senator
## 4   2          Person        mitch mcconnell
## 5   2        Position                senator
## 6   2 ProvinceOrState               kentucky
## 7   2       topicname               politics
## 8   3          Person        mitch mcconnell
## 9   3          Person            ashley judd
## 10  3    Organization                 senate
## 11  3    Organization             republican
## 12  4          Person            ashley judd
## 13  4       topicname               politics
## 14  5             URL www.huffingtonpost.com
## 15  5         Company              usa today
## 16  5          Person             chuck todd
## 17  5         Company                  msnbc

彻底解释:

## dat <- readLines(n=5)
## Person:mitch mcconnell, Person:ashley judd, Position:senator
## Person:mitch mcconnell, Position:senator, ProvinceOrState:kentucky, topicname:politics
## Person:mitch mcconnell, Person:ashley judd, Organization:senate, Organization:republican
## Person:ashley judd, topicname:politics
## URL:www.huffingtonpost.com, URL:http://www.regular-expressions.info

dat3 <- lapply(strsplit(dat, ","), function(x) gsub("^\\s+|\\s+$", "", x))
#dat3 <- lapply(dat2, function(x) x[grepl("Person|Position", x)]) 
dat3 <- lapply(dat3, strsplit, ":(?!/)", perl=TRUE) #break on : not folled by /

# Let the explanation begin...

# Here I have a short list of the variables from the rows
# of the original dataframe; in this case the row numbers:

seq_along(dat3)      #row variables

# then I use sapply and length to figure out hoe long the
# split variables in each row (now a list) are

sapply(dat3, length) #n times

# this tells me how many times to repeat the short list of 
# variables.  This is because I stretch the dat3 list to a vector
# Here I rep the row variables n times

rep(seq_along(dat3), sapply(dat3, length))

# better assign that for later:

ID <- rep(seq_along(dat3), sapply(dat3, length))

#============================================
# Now to explain the next chunk...
# I take dat3

dat3

# Each element in the list 1-5 is made of a new list of 
# Vectors of length 2 of Tag_Types and Tags.
# For instance here's element 5 a list of two  lists 
# with character vectors of length 2 

## [[5]]
## [[5]][[1]]
## [1] "URL"  "www.huffingtonpost.com"
## 
## [[5]][[2]]
## [1] "URL"  "http://www.regular-expressions.info"

# Use str to look at this structure:

dat3[[5]]
str(dat3[[5]])

## List of 2
##  $ : chr [1:2] "URL" "www.huffingtonpost.com"
##  $ : chr [1:2] "URL" "http://www.regular-expressions.info"

# I use lapply (list apply) to apply an anynomous function:
# function(x) do.call(rbind, x) 
#
# TO each of the 5 elements.  This basically glues the list 
# of vectors together to make a matrix.  Observe just on elenet 5:

do.call(rbind, dat3[[5]])

##      [,1]  [,2]                                 
## [1,] "URL" "www.huffingtonpost.com"             
## [2,] "URL" "http://www.regular-expressions.info"

# We use lapply to do that to all elements:

lapply(dat3, function(x) do.call(rbind, x))

# We then use the do.call(rbind on this list and we have a 
# matrix

do.call(rbind, lapply(dat3, function(x) do.call(rbind, x)))

# Let's assign that for later:

the_mat <- do.call(rbind, lapply(dat3, function(x) do.call(rbind, x)))

#============================================    
# Now we put it all together with data.frame:

data.frame(ID, the_mat)

答案 1 :(得分:3)

DF
##   ID                                                                                  Semantic.Tags
## 1  1                                   Person:mitch mcconnell, Person:ashley judd, Position:senator
## 2  2        Person:mitch mcconnell, Position:senator, ProvinceOrState:kentucky, topicname:politics 
## 3  3      Person:mitch mcconnell, Person:ashley judd, Organization:senate, Organization:republican 
## 4  4                                                         Person:ashley judd, topicname:politics
## 5  5                URL:www.huffingtonpost.com, Company:usa today, Person:chuck todd, Company:msnbc


ll <- lapply(strsplit(DF$Semantic.Tags, ","), strsplit, split = ":")

f <- function(x) do.call(rbind, x)

f(lapply(ll, f))
##       [,1]               [,2]                    
##  [1,] "     Person"      "mitch mcconnell"       
##  [2,] " Person"          "ashley judd"           
##  [3,] " Position"        "senator"               
##  [4,] "     Person"      "mitch mcconnell"       
##  [5,] " Position"        "senator"               
##  [6,] " ProvinceOrState" "kentucky"              
##  [7,] " topicname"       "politics "             
##  [8,] "     Person"      "mitch mcconnell"       
##  [9,] " Person"          "ashley judd"           
## [10,] " Organization"    "senate"                
## [11,] " Organization"    "republican "           
## [12,] "     Person"      "ashley judd"           
## [13,] " topicname"       "politics"              
## [14,] "     URL"         "www.huffingtonpost.com"
## [15,] " Company"         "usa today"             
## [16,] " Person"          "chuck todd"            
## [17,] " Company"         "msnbc"