检查一个字符串的所有字符是否存在于 r 中的另一个字符串中 [英] check if all characters of one string exist in another string in r

问题描述

我正在尝试比较 PRABHAKAR SHARMA 和 SHARMA KUMAR PRABHAKAR 之类的字符串.目的是检查较短字符串的所有字符是否存在于另一个字符串中.如果是这种情况，我应该得到 100% 匹配，否则会得到一个表示匹配字符百分比的百分比.

I am trying to compare strings like PRABHAKAR SHARMA and SHARMA KUMAR PRABHAKAR. the intention is to check if all the characters of the shorter string exist in the other string. If that is the case, I should get a 100% match otherwise a percentage representing the percentage of characters that matched.

我尝试在 RecordLinkage 包中使用 levenshteinSim，但它给出了一个数字，对应于将一个字符串更改为另一个字符串所需的更改次数.

I tried using levenshteinSim in RecordLinkage package but it gives a number corresponding to the number of changes required to change one string to another.

install.packages("RecordLinkage")
require(RecordLinkage)
levenshteinSim("PRABHAKAR SHARMA","SHARMA KUMAR PRABHAKAR")

#[1] 0.3636364

在这种情况下，我想要 100% 匹配.此外，这必须复制超过 1,000,000 条记录.

I want a 100% match in such a case. Also, this has to be replicated for over 1,000,000 records.

推荐答案

这是一种方法

s1 <- "PRABHAKAR SHARMA"
s2 <- "SHARMA KUMAR PRABHAKAR"

compare <- function(s1, s2) {
    c1 <- unique(strsplit(s1, "")[[1]])
    c2 <- unique(strsplit(s2, "")[[1]])
    length(intersect(c1,c2))/length(c1)
}

compare(s1,s2)
#1

不过可能有点慢.它也将空格字符视为字符.使用 Vectorize 应用于列:

It may be a little slow, though. And it considers the space character as character, too. Use Vectorize to apply on a column:

dat <- data.frame(small=c("a", "b"), big=c("aa", "cc"), stringsAsFactors=FALSE)
vcomp <- Vectorize(compare)
dat <- transform(dat, comp=vcomp(small, big))

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