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 Wednesday, April 25, 2018, Steven D'Aprano <
steve@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"
This is possible with re.find, re.finditer, re.findall, regex.findall(, overlapped=True), sliding window
n-grams can be by indices or by value.
count = len(indices)
--
Steve
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