Question

示例：

s <- "aaabaabaa"
p <- "aa"

我想返回4而不是3（即将初始"aa"中"aaa"个实例的数量计为2，而不是1）。

有没有解决方案？或者有没有办法计算R？

Answer 1

我相信

find_overlaps <- function(p,s) {
    gg <- gregexpr(paste0("(?=",p,")"),s,perl=TRUE)[[1]]
    if (length(gg)==1 && gg==-1) 0 else length(gg)
}


find_overlaps("aa","aaabaabaa")  ## 4
find_overlaps("not_there","aaabaabaa") ## 0 
find_overlaps("aa","aaaaaaaa")  ## 7

会做你想要的，更明确地表达为“查找字符串中重叠子串的数量”。

这是Finding the indexes of multiple/overlapping matching substrings

的一个小变化

Answer 2

substring在这里可能很有用，可以通过拍摄每一对连续的字符。

( ss <- sapply(2:nchar(s), function(i) substring(s, i-1, i)) )
## [1] "aa" "aa" "ab" "ba" "aa" "ab" "ba" "aa"
sum(ss %in% p)
## [1] 4

Answer 3

我需要一个相关的更一般性问题的答案。以下是我提出的推广Ben Bolker解决方案的方法：

my.data <- read.table(text = '
  my.string   my.cov
     1.2...        1
     .21111        2
     ..2122        3
     ...211        2
     112111        4
     212222        1
', header = TRUE, stringsAsFactors = FALSE)

desired.result.2ch <- read.table(text = '
  my.string   my.cov   n.11   n.12   n.21   n.22
     1.2...        1      0      0      0      0
     .21111        2      3      0      1      0
     ..2122        3      0      1      1      1
     ...211        2      1      0      1      0
     112111        4      3      1      1      0
     212222        1      0      1      1      3
', header = TRUE, stringsAsFactors = FALSE)

desired.result.3ch <- read.table(text = '
  my.string   my.cov   n.111   n.112   n.121   n.122   n.222   n.221   n.212   n.211
     1.2...        1       0       0       0       0       0       0       0       0
     .21111        2       2       0       0       0       0       0       0       1
     ..2122        3       0       0       0       1       0       0       1       0
     ...211        2       0       0       0       0       0       0       0       1
     112111        4       1       1       1       0       0       0       0       1
     212222        1       0       0       0       1       2       0       1       0
', header = TRUE, stringsAsFactors = FALSE)

find_overlaps <- function(s, my.cov, p) {
    gg <- gregexpr(paste0("(?=",p,")"),s,perl=TRUE)[[1]]
    if (length(gg)==1 && gg==-1) 0 else length(gg)
}

p <- c('11', '12', '21', '22', '111', '112', '121', '122', '222', '221', '212', '211')

my.output <- matrix(0, ncol = (nrow(my.data)+1), nrow = length(p))

for(i in seq(1,length(p))) {
    my.data$p <- p[i]
    my.output[i,1] <- p[i]
    my.output[i,(2:(nrow(my.data)+1))] <-apply(my.data, 1, function(x) find_overlaps(x[1],  x[2],  x[3]))
    apply(my.data, 1, function(x) find_overlaps(x[1],  x[2],  x[3]))
}

my.output
desired.result.2ch
desired.result.3ch

pre.final.output <- matrix(t(my.output[,2:7]), ncol=length(p), nrow=nrow(my.data))

final.output <- data.frame(my.data[,1:2], t(apply(pre.final.output, 1, as.numeric)))
colnames(final.output) <- c(colnames(my.data[,1:2]), paste0('x', p))
final.output

#  my.string my.cov x11 x12 x21 x22 x111 x112 x121 x122 x222 x221 x212 x211
#1    1.2...      1   0   0   0   0    0    0    0    0    0    0    0    0
#2    .21111      2   3   0   1   0    2    0    0    0    0    0    0    1
#3    ..2122      3   0   1   1   1    0    0    0    1    0    0    1    0
#4    ...211      2   1   0   1   0    0    0    0    0    0    0    0    1
#5    112111      4   3   1   1   0    1    1    1    0    0    0    0    1
#6    212222      1   0   1   1   3    0    0    0    1    2    0    1    0

Answer 4

整洁，我认为更具可读性的解决方案是

library(tidyverse)
PatternCount <- function(text, pattern) {
    #Generate all sliding substrings
    map(seq_len(nchar(text) - nchar(pattern) + 1), 
        function(x) str_sub(text, x, x + nchar(pattern) - 1)) %>%
    #Test them against the pattern
    map_lgl(function(x) x == pattern) %>%
    #Count the number of matches
    sum
}

PatternCount("aaabaabaa", "aa")
# 4

计算字符串中重叠子串的数量

4 个答案: