rebol - 在一个系列！删除最后一个元素的最佳方法是什么

Question

删除 Rebol 系列中最后一个元素的最简洁的方法是什么？

到目前为止我发现的选项是

s: "abc"
head remove back tail s

和

s: "abc"
take/last s

Answer 1

定义“最佳”。最高性能？最清晰？您希望表达式最后返回什么，或者您不在乎？您提供的两个返回不同的结果......一个是删除后系列的头部，另一个是删除的项目。

如果要删除后的系列头部，则需要take/last s后跟s以获取该表达式。比较：

>> delta-time [
    loop 10000 [
        s: copy "abc"
        take/last s
        s
    ]
]
== 0:00:00.012412

>> delta-time [
    loop 10000 [
        s: copy "abc"
        head remove back tail s
    ]
]
== 0:00:00.019222

如果您希望表达式评估为已删除的项目，则需要与take/last s诸如...之类的复杂内容进行比较，also (last s) (remove back tail s)因为还将运行第一个表达式，然后运行第二个...返回第一个表达式的结果：

>> delta-time [
    loop 10000 [
        s: copy "abc"
        take/last s
    ]
]
== 0:00:00.010838

>> delta-time [
    loop 10000 [
        s: copy "abc"
        also last s remove back tail s
    ]
]
== 0:00:00.024859

等等

If you don't care about the result, I'm going with take/last. If you do care about the result and want the head of the series, I'm going with take/last s followed by s. To me that reads better than head remove back tail s, although it's an aesthetic choice. It's still faster, at least on my netbook.

If you want the tail of the series at the end of the expression, remove back tail s is surprisingly similar in performance to take/last s followed by tail s. I'd say the latter is more explicit, and probably preferable, in case the reader forgets the return convention of REMOVE.

And also last s remove back tail s looks terrible, but is a reminder about also, which is pretty useful and easy to forget it's there. FWIW, it performs about the same as using an intermediate variable.

Answer 2

Here I wrote a REMOVE-LAST function,

remove-last: func [
    "Removes value(s) from tail of a series."
    series [series! port! bitset! none!]
    /part range [number!] "Removes to a given length."
] [
    either part [
        clear skip tail series negate range
    ] [
        remove back tail series
    ]
]

Example use:

b: [a b c d e f g]
remove-last b ;== [], 'g removed, tail of the series return.
head remove-last/part b 2 ;== [a b c d], 'e and 'f removed

It returns the tail of the series to be able to use in following situation;

b: [a b c d e f g]
head insert remove-last b 'x  ;== [a b c d e f x]