整理自统计之都论坛
方法一 使用strsplit函数
a <- "aggcacggaaaaacgggaataacggaggaggacttggcacggcattacacggagg" b <- strsplit(as.character(a),"ag") length(b[[1]]) - 1 ##子字符串"ag"的出现个数
a <- "aggcacggaaaaacgggaataacggaggaggacttggcacggcattacacggagg"
b <- strsplit(as.character(a),"ag")
regexpr("ag",a)
gregexpr("ag",a)
gregexpr("a.g",a)
attr(gregexpr("a.g",a)[[1]], "match.length") #提取子模式长度library(stringr)
str_count("1212345", c("12", "23", "00"))
一个使用实例
计算a*g中间有0-5个任意字母的频率
library(stringr)
str <- "aggcacggaaaaacgggaataacggaggaggacttggcacggcattacacggagg"
n <- 0:5
patterns <- sapply(n,function(i) {
paste0("a\\w{",i,"}g")
})
counts <- str_count(str,patterns)
names(counts) <- n
counts
使用R完成字符串的子字符串频率统计,码迷,mamicode.com
原文地址:http://blog.csdn.net/yucan1001/article/details/24742871