R计数次数字出现在列表元素中

Question

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，我想计算列表中每个元素的出现次数，而不是整个列表中出现的总次数。

我认为做到这一点的方法是使用stringr包中的str_count()函数和*ply()函数之一的组合，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”最终将成为单词矢量中的一个单词来计算出现次数。

我的理想输出会是这样的：

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

有什么建议？

Answer 1

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 

如果矢量w包含要计算字数：

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 

Answer 2

事情是这样的：

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 

Answer 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 

不e termco创建一个实际上是data.frames列表的类。

原料=原始频率计数（数字） 道具=计数（数字） RNP =原比例组合（字符）

使用斯科特的例子的比例：

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) 

Answer 4

你总是能为了简单，坚持grep在基础包...

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)