R count times word appears in element of list

Question

I have a list comprised of words.

> head(splitWords2)
[[1]]
 [1] \"Some\"        \"additional\"  \"information\" \"that\"        \"we\"          \"would\"         


        

Answer 1

Something like this:

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 2

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

Note that termco creates a class that is actually a list of data.frames.

raw = raw frequency counts (numeric) prop = proportion of counts (numeric) rnp = raw and proportion combined (character)

Using Scott's example:

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 3

You could always stick to grep in the base package for simplicity...

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)

Answer 4

For one specific word:

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

If vector w contains words that you want to count:

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