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