如何扭轉R中

Question

一些我想編寫一個函數來扭轉任何數字的順序。這是我的，但它不起作用。請幫幫我！

n=123 
rev_number=function(n){ 
m=strsplit(as.character(n),"") 
if (m==rev(m)) print("reversed number") 
} 

所需的輸出是n=321

Answer 1

我覺得像扭轉整數應該留在整數世界，而不是進入的字符串操作世界。看起來在R中沒有這種任務的內置函數，所以我們可以創建一個，例如使用Rcpp包。這裏有一個例子

library(Rcpp) 
cppFunction('int Reverse_CPP(int x) { 
    int reverse = 0; 
    while(x != 0) { 
     int remainder = x%10; 
     reverse = reverse*10 + remainder; 
     x/= 10; 
    } 
    return reverse ; 
}') 

Reverse_CPP(1234) 
# [1] 4321 

---

而這裏的向量化版本

cppFunction('IntegerVector Reverse_CPP2(IntegerVector x) { 
    int n = x.size(); 
    IntegerVector out(n); 
    IntegerVector xx = clone(x); // Will need this if you don"t want to modify x in place 

    for (int i = 0; i < n; ++i){ 
    int reverse = 0; 
    while(xx[i] != 0) { 
     int remainder = xx[i]%10; 
     reverse = reverse*10 + remainder; 
     xx[i]/= 10; 
    } 
    out[i] = reverse; 
    } 

    return out; 

}') 

Reverse_CPP2(c(12345, 21331, 4324234, 4243)) 
# [1] 54321 13312 4324234 3424 

注意，我不得不添加IntegerVector xx = clone(x);，從而大大減緩功能（見@alexis_laz評論）作爲RCPP將修改原x作爲參考。你並不需要，如果你是通過裸載體，或者如果你如果原來的矢量被modifyied

---

一些基準對其他矢量字符串處理函數不在乎

Stringi <- function(x) as.integer(stringi::stri_reverse(x)) 

Base <- function(x) { 
    as.integer(vapply(lapply(strsplit(as.character(x), "", fixed = TRUE), rev), 
        paste, collapse = "", FUN.VALUE = character(1L))) 
} 


library(microbenchmark) 
set.seed(123) 
x <- sample(1e3L:1e7L, 1e5, replace = TRUE) 

microbenchmark(
       Base(x), 
       Stringi(x), 
       Reverse_CPP2(x) 
) 

# Unit: milliseconds 
#   expr  min   lq  mean  median   uq   max neval cld 
#   Base(x) 855.985729 913.602215 994.60640 976.836206 1025.482170 1867.448511 100 c 
#  Stringi(x) 86.244426 94.882566 105.58049 102.962924 110.334702 179.918461 100 b 
# Reverse_CPP2(x) 1.842699 1.865594 2.06674 1.947703 2.076983 6.546552 100 a 

Answer 2

對於整數n > 9this function可用於：

reverse_int <- function(n) { 
    t1 <- floor(log10(n)) 
    t2 <- 0 
    for (i in t1:1) t2 <- t2 + floor(n/10^i) * 10^(t1-i) 
    return(n*10^t1 - 99*t2) 
} 
reverse_int(678754) 
#[1] 457876 

注意，功能不是v ectorized;它只需要一個參數n作爲輸入。

Answer 3

的R功能基於整數除法與10連續權力這對迴文數有關學校的項目上來扭轉號碼。

Reverse_number <- function(x){ 
    n <- trunc(log10(x)) # now many powers of 10 are we dealing with 
    x.rem <- x # the remaining numbers to be reversed 
    x.out <- 0 # stores the output 
    for(i in n:0){ 
    x.out <- x.out + (x.rem %/% 10^i)*10^(n-i) # multiply and add 
    x.rem <- x.rem - (x.rem %/% 10^i)*10^i # multiply and subtract 
    } 
return(x.out) 
}