如何根据第二个字符串删除字符串中的字符？

Question

考虑这两个字符串:

string1 <- "GCTCCC...CTCCATGAAGTA...CTTCACATCCGTGT.CCGGCCTGGCCGCGGAGAGCCC"
string_reference <- "GCTCCC...CTCCATGAAGTATTTCTTCACATCCGTGT.CCGGCCTGGCCGCGGAGAGCCC"

如何轻松删除"string1"中的点,但只删除"string_reference"中位于相同位置的点？

预期产量:

string1 = "GCTCCCCTCCATGAAGTA...CTTCACATCCGTGTCCGGCCTGGCCGCGGAGAGCCC"

Answer 1

我只是使用R的真正矢量化子集和逻辑比较方法......

# Split the strings
x <- strsplit( c( string1 , string_reference ) , "" )
# Compare and remove dots from string1 when dots also appear in the reference string at the same position
paste( x[[1]][ ! (x[[2]]== "." & x[[1]] == ".") ] , collapse = "" )
#[1] "GCTCCCCTCCATGAAGTA...CTTCACATCCGTGTCCGGCCTGGCCGCGGAGAGCCC"

Answer 2

与罗伯特相似,但是"矢量化"版本:

s1 <- unlist(strsplit(string1, ""))
s2 <- unlist(strsplit(string_reference, ""))
paste0(Filter(Negate(is.na), ifelse(s1 == s2 & s1 == ".", NA, s1)), collapse="")
# [1] "GCTCCCCTCCATGAAGTA...CTTCACATCCGTGTCCGGCCTGGCCGCGGAGAGCCC"

我引用"矢量化",因为矢量化发生在字符串向量的字符上.这假设您的字符串向量中只有一个元素.如果你的字符串向量中有多个元素,则必须遍历结果strsplit.

Answer 3

使用intersect找到重叠.的

cutpos <- do.call(intersect, 
        sapply(list(string_reference,string1), gregexpr, pattern=".", fixed=TRUE)
          )
paste(strsplit(string1,"",fixed=TRUE)[[1]][-cutpos],collapse="")
#[1] "GCTCCCCTCCATGAAGTA...CTTCACATCCGTGTCCGGCCTGGCCGCGGAGAGCCC"

上面的一小部分(@Arun提供):

attr(cutpos, 'match.length') <- rep(1L, length(cutpos))
attr(cutpos, 'useBytes') <- TRUE

do.call(paste0, c(regmatches(string1, list(cutpos), invert=TRUE), collapse=""))
## [1] "GCTCCCCTCCATGAAGTA...CTTCACATCCGTGTCCGGCCTGGCCGCGGAGAGCCC"