问题描述
是否有一种惯用的方法来确定 LazySeq 是否包含一个元素?从 Clojure 1.5 开始，调用 contains? 会抛出 IllegalArgumentException:
Is there an idiomatic way of determining if a LazySeq contains an element? As of Clojure 1.5 calling contains? throws an IllegalArgumentException:
IllegalArgumentException contains? not supported on type: clojure.lang.LazySeq
clojure.lang.RT.contains (RT.java:724)
在 1.5 之前，据我所知，它总是返回 false.
Before 1.5, as far as I know, it always returned false.
我知道在 LazySeq 上调用 contains? 可能永远不会返回，因为它可以是无限的.但是如果我知道它不是并且不在乎它是否被急切地评估怎么办?
I know that calling contains? on a LazySeq may never return as it can be infinite. But what if I know it isn't and don't care if it is evaluated eagerly?
我想到的是:
(defn lazy-contains? [col key]
(not (empty? (filter #(= key %) col))))
但是感觉不太对.有没有更好的办法?
But it doesn't feel quite right. Is there a better way?
推荐答案
首先，lazy seqs 在检查成员资格方面效率不高.考虑使用集合而不是惰性序列.
First, lazy seqs are not efficient for checking membership. Consider using a set instead of a lazy seq.
如果一套不切实际，你的解决方案还不错.一些可能的改进:
If a set is impractical, your solution isn't bad. A couple of possible improvements:
非空"有点别扭.仅使用 seq 就足以获得您的用户可以在 if 中使用的 nil-or-truthy 值.如果您想要 true 或 false，您可以将其包装在布尔值中.
"Not empty" is a bit awkward. Just using seq is enough to get a nil-or-truthy value that your users can use in an if.You can wrap that in boolean if you want true or false.
因为你只关心第一个匹配，所以你可以用 some 代替 filter 和 seq.
Since you only care about the first match, you can use some instead of filter and seq.
写等式谓词的一种方便的方法是使用文字集，如#{key}，但如果 key 为 nil，则无论是否找到 nil，都将始终返回 nil.
A convenient way to write an equality predicate is with a literal set, like #{key}, though if key is nil this will always return nil whether nil is found our not.
共同为您提供:
(defn lazy-contains? [col key]
(some #{key} col))
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