在 Clojure 中，我如何将所有内容映射到常量值？

发布于 2025-01-03 07:27:01 字数 185 浏览 2 评论 0原文

例如

(map #(+ 10 %1) [ 1 3 5 7 ])

将向所有内容添加 10

假设我想将所有内容映射到常量 1。我已经尝试过

(map #(1) [ 1 3 5 7 ])

但我不明白编译器错误。

原文

For example

(map #(+ 10 %1) [ 1 3 5 7 ])

Will add 10 to everything

Suppose I want to map everything to the constant 1. I have tried

(map #(1) [ 1 3 5 7 ])

But I don't understand the compiler error.

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为你鎻心 2025-01-10 07:27:01

(map #(1) [ 1 3 5 7 ])

不起作用有两个原因：

#(1) 是一个零参数匿名函数，因此它不能与 map 一起使用（与一个输入序列一起使用时需要一个单参数函数）。
即使它有正确的数量，它也不会工作，因为它试图将常量 1 作为函数调用，如 (1) - 尝试 (#(1)) 例如，如果您想查看此错误。

以下是一些可行的替代方案：

; use an anonymous function with one (ignored) argument
(map (fn [_] 1) [1 3 5 7])

; a hack with do that ignores the % argument 
(map #(do % 1) [1 3 5 7])

; use a for list comprehension instead
(for [x [1 3 5 7]] 1)

; use constantly from clojure.core
(map (constantly 1) [1 3 5 7])

在上述版本中，我认为使用 constantly 或for 应该是首选 - 这些更清晰、更惯用。

(map #(1) [ 1 3 5 7 ])

Won't work for two reasons:

#(1) is a zero-argument anonymous function, so it won't work with map (which requires a one-argument function when used with one input sequence).
Even if it had the right arity, it wouldn't work because it is trying to call the constant 1 as a function like (1) - try (#(1)) for example if you want to see this error.

Here are some alternatives that will work:

; use an anonymous function with one (ignored) argument
(map (fn [_] 1) [1 3 5 7])

; a hack with do that ignores the % argument 
(map #(do % 1) [1 3 5 7])

; use a for list comprehension instead
(for [x [1 3 5 7]] 1)

; use constantly from clojure.core
(map (constantly 1) [1 3 5 7])

Of the above, I think the versions using constantly or for should be preferred - these are clearer and more idiomatic.

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