2

我有一个清单：

s <- c('peel', 'peer', 'pear', 'tggc', 'gcgt')

我想将每个字符串与列表中的每个其他字符串进行比较，我使用以下命令：

z <- Map(utf8ToInt, s)
dmat <- outer(z, z, FUN=Vectorize(function(x, y) sum(bitwXor(x, y) > 0)))

但是，我想根据位置输出字符差异（而不是字符匹配）的数量：

例如"tggc"与字符串比较时"gcgt"应该输出为3.

4

3 回答 3

3

!只需按照以下方式使用简单的否定：

s <- c('peel', 'peer', 'pear', 'tggc', 'gcgt')
z <- Map(utf8ToInt, s)
dmat <- outer(z, z, FUN = Vectorize(function(x, y) sum(!bitwXor(x, y))))
dmat

或者使用简单的相等比较，因为您已将字符映射到整数。

dmat <- outer(z, z, FUN = Vectorize(function(x, y) sum(x == y)))

两者都给出输出：

     peel peer pear tggc gcgt
peel    4    3    2    0    0
peer    3    4    3    0    0
pear    2    3    4    0    0
tggc    0    0    0    4    1
gcgt    0    0    0    1    4

注意：如果你有固定的字符串长度，你也可以使用减法，但是上面的方法可以避免你显式传递这个，这增加了一点通用性。

于 2022-01-14T18:06:45.847 回答
1

如果性能是一个问题：

s <- c('peel', 'peer', 'pear', 'tggc', 'gcgt')
z <- mapply(utf8ToInt, s)
n <- length(s)
n1 <- 1:(n - 1L)
replace(matrix(nrow = n, ncol = n),
        sequence(n1, seq(n + 1L, by = n, length.out = n - 1L)),
        colSums(z[, sequence(n1)] == z[, rep.int(2:n, n1)]))
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]   NA    3    2    0    0
#> [2,]   NA   NA    3    0    0
#> [3,]   NA   NA   NA    0    0
#> [4,]   NA   NA   NA   NA    1
#> [5,]   NA   NA   NA   NA   NA

# benchmarking with a larger character vector
s <- mapply(FUN = function(x) paste0(sample(letters[1:4]), collapse = ""), 1:100)
microbenchmark::microbenchmark(bitwXor = {z <- Map(utf8ToInt, s)
                                          outer(z, z, FUN = Vectorize(function(x, y) sum(!bitwXor(x, y))))},
                               logical = {z <- Map(utf8ToInt, s)
                                          outer(z, z, FUN = Vectorize(function(x, y) sum(x == y)))},
                               mat = {z <- mapply(utf8ToInt, s)
                                      n <- length(s)
                                      n1 <- 1:(n - 1L)
                                      replace(matrix(nrow = n, ncol = n),
                                              sequence(n1, seq(n + 1L, by = n, length.out = n - 1L)),
                                              colSums(z[, sequence(n1)] == z[, rep.int(2:n, n1)]))})
#> Unit: microseconds
#>     expr     min      lq      mean   median       uq     max neval
#>  bitwXor 23846.1 24875.6 26207.230 26120.95 27134.35 33842.8   100
#>  logical 16645.5 17514.8 19020.051 18383.35 19875.15 32716.8   100
#>      mat   387.4   455.0   511.322   482.70   544.05  1224.4   100

# confirm that the results are the same
z <- Map(utf8ToInt, s)
mat1 <- outer(z, z, FUN = Vectorize(function(x, y) sum(!bitwXor(x, y))))
mat2 <- outer(z, z, FUN = Vectorize(function(x, y) sum(x == y)))
z <- mapply(utf8ToInt, s)
n <- length(s)
n1 <- 1:(n - 1L)
mat3 <- replace(matrix(nrow = n, ncol = n), sequence(n1, seq(n + 1L, by = n, length.out = n - 1L)), colSums(z[, sequence(n1)] == z[, rep.int(2:n, n1)]))
all.equal(mat1[upper.tri(mat1)], mat2[upper.tri(mat2)])
#> [1] TRUE
all.equal(mat1[upper.tri(mat1)], mat3[upper.tri(mat3)])
#> [1] TRUE
于 2022-01-14T21:38:50.727 回答
0

一个可能的解决方案：

library(tidyverse)

sample <- c('peel','peer','pear','tggc','gcgt')

sample %>% 
  expand.grid(sample) %>% 
  rowwise %>% 
  mutate(cmp = mapply(function(x,y) 
                { x != y}, x=str_split(Var1, ""), y=str_split(Var2, "")) %>% sum)

#> # A tibble: 25 × 3
#> # Rowwise: 
#>    Var1  Var2    cmp
#>    <fct> <fct> <int>
#>  1 peel  peel      0
#>  2 peer  peel      1
#>  3 pear  peel      2
#>  4 tggc  peel      4
#>  5 gcgt  peel      4
#>  6 peel  peer      1
#>  7 peer  peer      0
#>  8 pear  peer      1
#>  9 tggc  peer      4
#> 10 gcgt  peer      4
#> # … with 15 more rows
于 2022-01-14T18:09:23.443 回答