Question

我想将一个块从block: [ a: 1 b: 2 ]转换为[a 1 b 2]。有比这更简单的方法吗？

map-each word block [ either set-word? word [ to-word word ] [ word ] ]

Answer 1

保持简单：

>> block: [a: 1 b: 2]
== [a: 1 b: 2]
>> forskip block 2 [block/1: to word! block/1]
== b
>> block
== [a 1 b 2]

Answer 2

我有同样的问题所以我写了这个函数。也许有一些我不知道的简单解决方案。

flat-body-of: function [
    "Change all set-words to words"
    object [object! map!]
][
    parse body: body-of object [
        any [
            change [set key set-word! (key: to word! key)] key 
            | skip
        ]
    ]
    body 
]

Answer 3

这些会创建新的块，但相当简洁。对于已知的set-word/value对：

collect [foreach [word val] block [keep to word! word keep val]]

否则，你可以像你的情况一样使用'

collect [foreach val block [keep either set-word? val [to word! val][val]]]

我建议你的map-each本身也很简洁。

Answer 4

我喜欢DocKimbel的答案，但为了另一种选择...

for i 1 length? block 2 [poke block i to word! pick block i]

Answer 5

来自Graham Chiu的回答：

在R2中你可以这样做：

>> to block! form [ a: 1 b: 2 c: 3]
== [a 1 b 2 c 3]

Answer 6

或使用PARSE：

block: [ a: 1 b: 2 ]
parse block [some [m: set-word! (change m to-word first m) any-type!]]

如何将块中的set-words转换为单词

6 个答案: