R count 次单词出现在列表元素中答案

【问题标题】：R count times word appears in element of listR count 次单词出现在列表元素中
【发布时间】：2012-07-09 07:58:27
【问题描述】：

我有一个由单词组成的列表。

> head(splitWords2)
[[1]]
 [1] "Some"        "additional"  "information" "that"        "we"          "would"       "need"        "to"          "replicate"   "the"        
[11] "experiment"  "is"          "how"         "much"        "vinegar"     "should"      "be"          "placed"      "in"          "each"       
[21] "identical"   "container"   "or"          "what"        "tool"        "use"         "measure"     "mass"        "of"          "four"       
[31] "different"   "samples"     "and"         "distilled"   "water"       "rinse"       "after"       "taking"      "them"        "out"        

[[2]]
 [1] "After"       "reading"     "the"         "expirement"  "I"           "realized"    "that"        "additional"  "information" "you"        
[11] "need"        "to"          "replicate"   "expireiment" "is"          "one"         "amant"       "of"          "vinegar"     "poured"     
[21] "in"          "each"        "container"   "two"         "label"       "containers"  "before"      "start"       "yar"         "and"        
[31] "three"       "write"       "a"           "conclusion"  "make"        "sure"        "results"     "are"         "accurate"

我有一个单词向量，我想计算列表的每个元素中出现的次数，而不是整个列表中出现的总数。

我认为这样做的方法是结合stringr 包中的str_count() 函数和*ply() 函数之一，但我无法使其工作。

numWorder1 <- sapply(ifelse(str_count(unlist(splitWords2), ignore.case("we" ) )> 0, 1, 0))

其中“我们”最终将是来自单词向量的单词，以计算出现次数。

我的理想输出是这样的：

lineNum       count
   1           0
   2           1
   3           1
   4           0
  ...         ...

有什么建议吗？

【问题讨论】：

标签： r list apply stringr

【解决方案1】：

类似这样的：

wordlist <- list(
    c("the","and","it"),
    c("we","and","it")
)
require(plyr); require(stringr)
> ldply(wordlist, function(x) str_count(x, "we"))
  V1 V2 V3
1  0  0  0
2  1  0  0

【讨论】：

【解决方案2】：

对于一个特定的词：

words <- list(a = c("a","b","c","a","a","b"), b = c("w","w","q","a"))
$a
[1] "a" "b" "c" "a" "a" "b"

$b
[1] "w" "w" "q" "a"
wt <- data.frame(lineNum = 1:length(words))
wt$count <- sapply(words, function(x) sum(str_count(x, "a")))
  lineNum count
1       1     3
2       2     1

如果向量w 包含您要计数的单词：

w <- c("a","q","e")
allwords <- lapply(w, function(z) data.frame(lineNum = 1:length(words), 
            count = sapply(words, function(x) sum(str_count(x, z)))))
names(allwords) <- w
$a
  lineNum count
a       1     3
b       2     1

$q
  lineNum count
a       1     0
b       2     1

$e
  lineNum count
a       1     0
b       2     0

【讨论】：

【解决方案3】：

library(qdap)

#create a fake data set like yours:
words <- list(first = c("a","b","c","a","a","bc", "dBs"), 
    second = c("w","w","q","a"))
## termco functions require sentence like structure in a data frame so covert:
words2 <- list2df(lapply(words, paste, collapse = " "), "wl", "list")[2:1]


## trailing and leading spaces are important in match terms
## both a trailing and leading space will match exactly that trerm
termco(text.var=words2$wl, grouping.var=words2$list, match.list=c(" a "))
termco(words2$wl, words2$list, match.list=c(" b ", " a "))

## notice no space at the end of b finds and case of b + any.chunk
termco(words2$wl, words2$list, match.list=c(" b", " a "))

## no trailing/leading spaces means find any words containing the chunk b
termco(words2$wl, words2$list, match.list=c("b", " a "))

#ignores case
termco(words2$wl, words2$list, match.list=c("b", " a "), ignore.case=T)

## Last use yields:
## 
##     list word.count  term(b) term( a )
## 1  first          7 3(42.86)  2(28.57)
## 2 second          4        0     1(25)
## Also:


## transpose like function that transposes a raw matrix 
with(words2, termco2mat(termco(wl, list, match.list=c("b", " a "))))

## Which yields raw.score(percentage):
## 
##   first second
## b     2      0
## a     2      1

请注意，termco 创建了一个实际上是 data.frames 列表的类。

raw = 原始频率计数（数字） prop = 计数比例（数字） rnp = 原始和比例组合（字符）

以斯科特为例：

words <- list(
    first=c("the","and","it", "we're"),
    second=c("we","and","it")
)
words2 <- data.frame(list=names(words), 
    wl=unlist(lapply(words, paste, collapse=" ")))

termco(words2$wl, words2$list, match.list=c(" we ", " we"))
termco(words2$wl, words2$list, match.list=c(" we ", " we"), short.term = FALSE)

【讨论】：

【解决方案4】：

为简单起见，您始终可以坚持使用基本包中的 grep...

LinesList <- list ( "1"=letters[1:10], "2"=rep(letters[1:3],3) )
CountsA <- grep("[a]", LinesList) # find 'a' in each element of list
length(CountsA) <- length(LinesList) # gives NAs if not counted
data.frame( lineNum = names(LinesList), count = CountsA)

【讨论】：