如何在Racket中的列表中找到元素的索引? [英] How do I find the index of an element in a list in Racket?

问题描述

这当然是微不足道的工具，但我觉得Racket肯定内置了某些功能.我的直觉是正确的吗?如果是，函数是什么?

This is trivial implement of course, but I feel there is certainly something built in to Racket that does this. Am I correct in that intuition, and if so, what is function?

推荐答案

奇怪的是，Racket中没有内置的过程可以在列表中查找元素的从0开始的索引(相反的过程是否存在，它称为

Strangely, there isn't a built-in procedure in Racket for finding the 0-based index of an element in a list (the opposite procedure does exist, it's called list-ref). However, it's not hard to implement efficiently:

(define (index-of lst ele)
  (let loop ((lst lst)
             (idx 0))
    (cond ((empty? lst) #f)
          ((equal? (first lst) ele) idx)
          (else (loop (rest lst) (add1 idx))))))

但是中有 类似的过程，称为

But there is a similar procedure in srfi/1, it's called list-index and you can get the desired effect by passing the right parameters:

(require srfi/1)

(list-index (curry equal? 3) '(1 2 3 4 5))
=> 2

(list-index (curry equal? 6) '(1 2 3 4 5))
=> #f

更新

从球拍6.7开始，

As of Racket 6.7, index-of is now part of the standard library. Enjoy!

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