我应该如何称呼执行列表推导式的 REBOL 函数？
REBOL 没有内置的方法来执行列表推导。 然而，REBOL 有一个强大的工具（称为parse），可用于创建特定于域的语言（DSL）。 我使用 parse 来创建这样一个用于列表理解的迷你 DSL。 为了解释表达式，包含推导式的块被传递给一个函数，由于缺乏更好的术语，我将其称为“comprehend”。
示例：
comprehend [(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]
出于某种原因，comprehend 对我来说听起来不太正确，但像 eval 这样的东西太笼统了。
我还没有发现任何其他语言需要关键字或函数来进行列表理解。 无论它们存在于何处，它们都是纯粹的语法糖。 不幸的是我没有这个选择。 那么，既然我必须有一个函数，那么它的一个好的、简洁的、合乎逻辑的名称是什么？
REBOL has no built-in way to perform list comprehensions. However, REBOL has a powerful facility (known as parse) that can be used to create domain-specific languages (DSLs). I've used parse to create such a mini-DSL for list comprehensions. In order to interpret the expression, the block containing the comprehension is passed to a function, which for lack of a better term I've called comprehend.
Example:
comprehend [(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]
For some reason, comprehend doesn't sound right to me, but something like eval is too general.
I haven't found any other language that requires a keyword or function for list comprehensions. They are pure syntactic sugar wherever they exist. Unfortunately I don't have that option. So, seeing that I must have a function, what's a good, succinct, logical name for it?
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选择怎么样？选择 [(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]How about
select?select [(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]因为列表推导式可以被认为类似于映射，所以您可能会考虑将其称为“listmap”之类的名称。 或者，由于列表推导式基于集合构建器表示法，因此您可以将其称为“构建”或“构建列表”。
（免责声明：我对 REBOL 知之甚少，所以如果这些名字已经被占用，请原谅我）
Because list comprehensions can be thought of as analogous to map, you might think about calling it something like "listmap". Alternately, because list comprehensions are based on set-builder notation, you could call it something along the lines of "build" or "buildlist".
(Disclaimer: I know very little about REBOL, so forgive me if these names are already taken)
变形
transmogrify
do可能是合适的，因为列表推导式只是 Monad 推导式的一个实例，而do是 Haskell 中用于加糖 Monadic 计算的关键字，但我怀疑它对于一个人来说太模糊了。用户库。 我将列表理解函数命名为comp，但这只是您已有函数的缩写。 也许屈服？ 例如产生[(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]。 只是眯着眼睛假装 [ ] 不存在。docould be appropriate, as list comprehensions are just one instance of Monad comprehensions, anddois the keyword used in Haskell for sugared Monadic computations but I suspect it's too vague for a user library. I called my list comprehension functioncomp, but that's just an abbreviation of what you already have. Perhapsyielding? E.g.yielding [(a * b) for a in 1x100 for b in 4x10 where (all [odd? a odd? b])]. Just sort of squint and pretend the [ ] aren't there.