我一直在寻找一种优雅而高效的方法,在Ruby中将字符串分成给定长度的子字符串。
到目前为止,我能想到的最好的结果是:
def chunk(string, size)
(0..(string.length-1)/size).map{|i|string[i*size,size]}
end
>> chunk("abcdef",3)
=> ["abc", "def"]
>> chunk("abcde",3)
=> ["abc", "de"]
>> chunk("abc",3)
=> ["abc"]
>> chunk("ab",3)
=> ["ab"]
>> chunk("",3)
=> []
您可能希望chunk("", n)
返回[""]
而不是[]
。如果是这样,只需添加以下内容作为该方法的第一行:
return [""] if string.empty?
你能推荐更好的解决方案吗?
编辑
感谢Jeremy Ruten这个优雅而高效的解决方案:编辑:效率不高!
def chunk(string, size)
string.scan(/.{1,#{size}}/)
end
编辑
string.scan解决方案将512k分成1k块10000次需要大约60秒,而最初基于切片的解决方案只需要2.4秒。
发布于 2009-04-16 01:26:05
使用String#scan
>> 'abcdefghijklmnopqrstuvwxyz'.scan(/.{4}/)
=> ["abcd", "efgh", "ijkl", "mnop", "qrst", "uvwx"]
>> 'abcdefghijklmnopqrstuvwxyz'.scan(/.{1,4}/)
=> ["abcd", "efgh", "ijkl", "mnop", "qrst", "uvwx", "yz"]
>> 'abcdefghijklmnopqrstuvwxyz'.scan(/.{1,3}/)
=> ["abc", "def", "ghi", "jkl", "mno", "pqr", "stu", "vwx", "yz"]
发布于 2011-02-05 04:05:00
这是另一种方法:
"abcdefghijklmnopqrstuvwxyz".chars.to_a.each_slice(3).to_a.map {|s| s.to_s }
=> "abc","def","ghi","jkl","mno","pqr","stu","vwx","yz“
发布于 2015-07-26 22:00:09
我认为这是最有效的解决方案,如果你知道你的字符串是区块大小的倍数
def chunk(string, size)
(string.length / size).times.collect { |i| string[i * size, size] }
end
而对于部件
def parts(string, count)
size = string.length / count
count.times.collect { |i| string[i * size, size] }
end
https://stackoverflow.com/questions/754407
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