似乎Smalltalk实现缺少一种算法,该算法返回String中子字符串的所有索引.最相似的元素仅返回元素的一个索引,例如:firstIndexesOf:in:,findSubstring:,findAnySubstring:variant.
It seems Smalltalk implementations misses an algorithm which return all the indices of a substring in a String. The most similar ones returns only one index of an element, for example : firstIndexesOf:in: , findSubstring:, findAnySubstring: variants.
在Ruby中有实现,但是第一个依靠Ruby hack,第二个依靠忽略重叠的Strings无效,最后一个使用Enumerator类,我不知道该类如何转换为Smalltalk.我想知道 Python实现是否是开始的最佳途径,因为考虑了两种情况(是否重叠)并且不使用正则表达式
There are implementations in Ruby but the first one relies on a Ruby hack, the second one does not work ignoring overlapping Strings and the last one uses an Enumerator class which I don't know how to translate to Smalltalk. I wonder if this Python implementation is the best path to start since considers both cases, overlapping or not and does not uses regular expressions.
我的目标是找到提供以下行为的程序包或方法:
My goal is to find a package or method which provides the following behavior:
'ABDCDEFBDAC' indicesOf: 'BD'. "#(2 8)"考虑重叠时:
'nnnn' indicesOf: 'nn' overlapping: true. "#(0 2)"不考虑重叠时:
'nnnn' indicesOf 'nn' overlapping: false. "#(0 1 2)"在Pharo中,当在Playground中选择了文本时,扫描仪会检测到该子字符串并突出显示匹配项.但是我找不到这个的String实现.
In Pharo, when a text is selected in a Playground, a scanner detects the substring and highlights matches. However I couldn't find a String implementation of this.
到目前为止,我的最大努力是在String(Pharo 6)中实现了该实现:
My best effort so far results in this implementation in String (Pharo 6):
indicesOfSubstring: subString | indices i | indices := OrderedCollection new: self size. i := 0. [ (i := self findString: subString startingAt: i + 1) > 0 ] whileTrue: [ indices addLast: i ]. ^ indices推荐答案
首先让我澄清一下Smalltalk集合是基于1的,而不是基于0的.因此,您的示例应阅读
Let me firstly clarify that Smalltalk collections are 1-based, not 0-based. Therefore your examples should read
'nnnn' indexesOf: 'nn' overlapping: false. "#(1 3)" 'nnnn' indexesOf: 'nn' overlapping: true. "#(1 2 3)"请注意,我也注意到了@lurker的观察(并且也对选择器进行了调整).
Note that I've also taken notice of @lurker's observation (and have tweaked the selector too).
现在,从您的代码开始,我将对其进行如下更改:
Now, starting from your code I would change it as follows:
indexesOfSubstring: subString overlapping: aBoolean | n indexes i | n := subString size. indexes := OrderedCollection new. "removed the size" i := 1. "1-based" [ i := self findString: subString startingAt: i. "split condition" i > 0] whileTrue: [ indexes add: i. "add: = addLast:" i := aBoolean ifTrue: [i + 1] ifFalse: [i + n]]. "new!" ^indexes确保您编写了一些单元测试(不要忘记练习边框!)
Make sure you write some few unit tests (and don't forget to exercise the border cases!)
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Smalltalk中子字符串的索引
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