use*_*642 11 string scheme list intermediate-language racket
比如我有(list "a" "1" "b" "2" "c" "3").
现在我想把这个列表变成一个"a1b2c3".
我怎么做?
谢谢.
Wes*_*Wes 12
(apply string-append (list "a" "1" "b" "2" "c" "3"))或者(string-append* "" (list "a" "1" "b" "2" "c" "3"))应该工作.请参阅:http://docs.racket-lang.org/reference/strings.html
如果你想要一个程序来做这个,你可以写 (define (strings->string sts) (apply string-append sts))
不要重新发明轮子!在Racket中,有一个专门用于此的程序,它的名字叫string-join:
(string-join '("a" "1" "b" "2" "c" "3") "")
=> "a1b2c3"
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引用文档:
(string-join strs
[sep
#:before-first before-first
#:before-last before-last
#:after-last after-last]) ? string?
strs : (listof string?)
sep : string? = " "
before-first : string? = ""
before-last : string? = sep
after-last : string? = ""
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将字符串追加到strs中的每对字符串之间
strs插入sep.before-last,before-first和after-last类似于以下内容的输入add-between:它们分别在最后两个字符串,前缀字符串和后缀字符串之间指定备用分隔符.
无论如何,这里有一些带有和不带有分隔符的实现(即在每对字符串之间插入一个字符串,例如空格或逗号)。
函数fold和fold-right来自 SRFI 1。
当连接非常多或非常长的字符串时,使用字符串端口可能会更快。否则不太可能有太大的速度差异。
(define (string-join strings)
(fold-right string-append "" strings))
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(define (string-join strings)
(let loop ((strings strings) (so-far ""))
(if (null? strings)
so-far
(loop (cdr strings) (string-append so-far (car strings))))))
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(define (string-join strings)
(parameterize ((current-output-port (open-output-string)))
(for-each write-string strings)
(get-output-string (current-output-port))))
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(define (string-join strings delimiter)
(if (null? strings)
""
(fold (lambda (s so-far) (string-append so-far delimiter s))
(car strings)
(cdr strings))))
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(define (string-join strings delimiter)
(if (null? strings)
""
(let loop ((strings (cdr strings)) (so-far (car strings)))
(if (null? strings)
so-far
(loop (cdr strings)
(string-append so-far delimiter (car strings)))))))
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(define (string-join strings delimiter)
(if (null? strings)
""
(parameterize ((current-output-port (open-output-string)))
(write-string (car strings))
(for-each (lambda (s)
(write-string delimiter)
(write-string s))
(cdr strings))
(get-output-string (current-output-port)))))
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