There are two ‘AA’ in ‘AAA’, one starting from 0 and the other starting from 1.
If ‘AA’ starting from 0 is deleted and inserted with ‘BANAN’, ‘AAA’ becomes
‘BANANA ‘.
If ‘AA’ starting from 1 is deleted and inserted with ‘PPLE’, ‘AAA’ becomes
‘APPLE’.
Depending on which one is chosen, ‘AAA’ can be edited to ‘BANANA’ or ‘APPLE ‘,
two different results.
I wrote a program which edits a part of a text. If the part to be edited occurs
more than once, it presents the positions and asks the user to choose which one
to be edited.
I tried with different algorithms. Best one so far would be using just find()
and collecting the results in a list.
> On Apr 25, 2018, at 11:57 PM, Wes Turner <wes.tur...@gmail.com> wrote:
>
>
>
>> On Wednesday, April 25, 2018, Steven D'Aprano <st...@pearwood.info> wrote:
>> On Wed, Apr 25, 2018 at 11:22:24AM -0700, Julia Kim wrote:
>> > Hi,
>> >
>> > There’s an error with the string method count().
>> >
>> > x = ‘AAA’
>> > y = ‘AA’
>> > print(x.count(y))
>> >
>> > The output is 1, instead of 2.
>>
>> Are you proposing that there ought to be a version of count that looks
>> for *overlapping* substrings?
>>
>> When will this be useful?
>
> "Finding a motif in DNA"
> http://rosalind.info/problems/subs/
>
> This is possible with re.find, re.finditer, re.findall, regex.findall(,
> overlapped=True), sliding window
> https://stackoverflow.com/questions/2970520/string-count-with-overlapping-occurrences
>
> n-grams can be by indices or by value.
> count = len(indices)
> https://en.wikipedia.org/wiki/N-gram#Examples
>
> https://en.wikipedia.org/wiki/String_(computer_science)#String_processing_algorithms
>
> https://en.wikipedia.org/wiki/Sequential_pattern_mining
>
>>
>>
>> --
>> Steve
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