Question

如何有效地计算在另一个字符串中出现的一个字符串的实例数？

以下是我迄今为止的代码。它成功识别是否在另一个字符串中出现了一个字符串的任何实例。但是，我不知道如何将它从TRUE / FALSE关系扩展到计数关系。

x <- ("Hello my name is Christopher. Some people call me Chris")
y <- ("Chris is an interesting person to be around")
z <- ("Because he plays sports and likes statistics")

lll <- tolower(list(x,y,z))
dict <- tolower(c("Chris", "Hell"))

mmm <- matrix(nrow=length(lll), ncol=length(dict), NA)

for (i in 1:length(lll)) {
for (j in 1:length(dict)) {
    mmm[i,j] <- sum(grepl(dict[j],lll[i]))
}
}
mmm

它产生：

       [,1] [,2]
 [1,]    1    1
 [2,]    1    0
 [3,]    0    0

由于小写字符串“chris”在lll[1]中出现两次，我希望mmm[1,1]为2而不是1.

真实的例子是更高的维度...如果代码可以被矢量化而不是使用我的强力循环，那就太喜欢了。

Answer 1

两个快速提示：

避免双重for-loop，你不需要它;）
使用stringr包

library(stringr)

dict <- setNames(nm=dict)  # simply for neatness
lapply(dict, str_count, string=lll)
# $chris
# [1] 2 1 0
#
# $hell
# [1] 1 0 0

或作为矩阵：

#  sapply(dict, str_count, string=lll)
#      chris hell
# [1,]     2    1
# [2,]     1    0
# [3,]     0    0

Answer 2

而不是sum(grepl(dict[j],lll[i]))，请尝试sum(gregexpr(dict[j],lll[i])[[1]] > 0)

Answer 3

您也可以这样做：

count.matches <- function(pat, vec) sapply(regmatches(vec, gregexpr(pat, vec)), length)
mapply(count.matches, c('chris', 'hell'), list(lll))
#      chris hell
# [1,]     2    1
# [2,]     1    0
# [3,]     0    0

Answer 4

llll<-rep(lll,length(dict))
dict1<-rep(dict,each=length(lll))


 do.call(rbind,Map(function(x,y)list(y,sum(gregexpr(y,x)[[1]] > 0)), llll,dict1))
                                                        [,1]    [,2]
hello my name is christopher. some people call me chris "chris" 2   
chris is an interesting person to be around             "chris" 1   
because he plays sports and likes statistics            "chris" 0   
hello my name is christopher. some people call me chris "hell"  1   
chris is an interesting person to be around             "hell"  0   
because he plays sports and likes statistics            "hell"  0

然后，您可以使用reshape来获得所需内容。

Answer 5

这使用qdap包。 CRAN版本应该可以正常工作，但您可能需要dev version

library(qdap)

termco(c(x, y, z), 1:3, c('chris', 'hell'))

##   3 word.count     chris      hell
## 1 1         10 2(20.00%) 1(10.00%)
## 2 2          8 1(12.50%)         0
## 3 3          7         0         0


termco(c(x, y, z), 1:3, c('chris', 'hell'))$raw

##   3 word.count chris hell
## 1 1         10     2    1
## 2 2          8     1    0
## 3 3          7     0    0

R中的计数模式匹配

5 个答案:

或作为矩阵：