R 中的计数模式匹配

问题描述

如何有效地计算一个字符串在另一个字符串中出现的实例数?

How would one efficiently count the number of instances of one character string which occur within another character string?

以下是我迄今为止的代码.它成功地识别出一个字符串的任何实例是否出现在另一个字符串中.但是，我不知道如何将其从 TRUE/FALSE 关系扩展到计数关系.

Below is my code to date. It successfully identifies if any instance of the one string occurs in the other string. However, I do not know how to extend it from a TRUE/FALSE relationship to a counting relationship.

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.

Since the lower-case string "chris" appears twice in the lll[1] I would like mmm[1,1] to be 2 instead of 1.

真实的例子是更高的维度......所以如果代码可以被向量化而不是使用我的暴力循环，我会很高兴.

Real example is much higher dimension...so would love if code could be vectorized instead of using my brute force for loops.

推荐答案

两个快速提示:

1. 避免双重 for 循环，你不需要它;)
2. 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

或者作为矩阵:

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

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