Mailing List Archive

On 2020-11-11 01:26, Bischoop wrote:
>
> Can anybody help?Here's the code that gives me a hedeache for second day https://bpa.st/XLOA
> If I have letter_list = ["a","a",,"b","b","c","c"] it's working
> correctly but if I have three or more of any letters for
> example:letter_list = ["a","a","a","b","b","c","c"], I'm ending up with
> some pairs somewhere.
> I took another way approach today: https://bpa.st/E7HQ, was thinking
> about iterating and checking if neighbour characters won't make a pair
> but I've end up with error:
> if x != letter_list[i+1] and letter_list[i-1]:
> IndexError: list index out of range
> andin addition: got 4 "a" (had 3only) and missing 1 "b" :-/
>
Points to note in your first code:

1. Modifying a list while iterating over it is a bad idea.

2. You're modifying the list that you're passing in and also returning
it. That's a bad idea. Either modify it in place or modify and return a
copy.

3. The line:

m = (letter_list.index(x))

returns the position of the first occurrence of x.

Points to note in your second code:

1. (See above)

2. (See above)

3. (See above)

4. 'i' goes from 0 to len(letter_list)-1, so i+1 goes from 1 to
len(letter_list), but the maximum index permitted by letter_list is
len(letter_list)-1, hence letter_list[i+1] will raise IndexError on the
last iteration.

5. 'i' goes from 0 to len(letter_list)-1, so i-1 goes from -1 to
len(letter_list)-2. letter_list[-1] returns the last (final) letter in
the list, and it's treated as a true.

6. This:

x != letter_list[i+1] and letter_list[i-1]

is checking whether:

x != letter_list[i+1]

is true and also whether:

letter_list[i-1]

is true.
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On 2020-11-11, MRAB <python@mrabarnett.plus.com> wrote:
>>
> Points to note in your first code:
>
> 1. Modifying a list while iterating over it is a bad idea.
>
> 2. You're modifying the list that you're passing in and also returning
> it. That's a bad idea. Either modify it in place or modify and return a
> copy.
>
> 3. The line:
>
> m = (letter_list.index(x))
>
> returns the position of the first occurrence of x.
>
> Points to note in your second code:
>
> 1. (See above)
>
> 2. (See above)
>
> 3. (See above)
>
> 4. 'i' goes from 0 to len(letter_list)-1, so i+1 goes from 1 to
> len(letter_list), but the maximum index permitted by letter_list is
> len(letter_list)-1, hence letter_list[i+1] will raise IndexError on the
> last iteration.
>
> 5. 'i' goes from 0 to len(letter_list)-1, so i-1 goes from -1 to
> len(letter_list)-2. letter_list[-1] returns the last (final) letter in
> the list, and it's treated as a true.
>
> 6. This:
>
> x != letter_list[i+1] and letter_list[i-1]
>
> is checking whether:
>
> x != letter_list[i+1]
>
> is true and also whether:
>
> letter_list[i-1]
>
> is true.

I see now I overcomplicated it, what is a good idea then?
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On 12/11/2020 00:45, Bischoop wrote:
> I see now I overcomplicated it, what is a good idea then?


This is a variation on a data-compression technique called
RLE=Run-Length Encoding (excepting that in this case there is no need to
count the repetitions).

Web.Ref:
https://stackabuse.com/run-length-encoding/
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