Okay, let's say; [['a', 'b'], 1, 2, ['a', 'b'], 1, 2] Is

Question

Okay, let's say; [['a', 'b'], 1, 2, ['a', 'b'], 1, 2] Is

there a "better" way to add *adjacent* numbers in the list and replace adjacent indexes with result of addition?

Expected:

[['a', 'b'], 3, ['a', 'b'], 3]

Problem is, list can be of any length and cannot guarantee *adjacent* numbers always. I mean, the solution is only for adjacent numbers

0

01.01.2021

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Пользователь 6165a Автор вопроса

drunktimelord
well just loop over it

Do you have a pseudo code?

0

01.01.2021

drunktimelord

Пользователь 6165a
Do you have a pseudo code?

new = [] while i < len(array): if array[i] is integer and array[i+1] is integer: new.append(array[i] + array[i+1]) i += 2 else: new.append(array[i]) i += 1

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01.01.2021

Пользователь 6165a Автор вопроса

drunktimelord
new = [] while i < len(array): if array[i] is ...

Here, i+1 eventually cause indexOUTerror

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01.01.2021

drunktimelord

Пользователь 6165a
Here, i+1 eventually cause indexOUTerror

then handle 0th index separately and start i at 1

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01.01.2021

Пользователь 6165a Автор вопроса

drunktimelord
then handle 0th index separately and start i at 1

Who knows 0th index is not number

0

01.01.2021

drunktimelord

Пользователь 6165a
Who knows 0th index is not number

new = [] if array[0] is int and array[1] is int: new.append(array[0] + array[1]) i = 2 else: new.append(array[0]) i = 1 while i < len(array): if array[i] is integer and array[i-1] is integer: new.append(array[i] + array[i-1]) i += 2 else: new.append(array[i-1]) i += 1

0

01.01.2021

drunktimelord · Accepted Answer

drunktimelord

well just loop over it

0

01.01.2021

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Okay, let's say; [['a', 'b'], 1, 2, ['a', 'b'], 1, 2] Is

7 ответов

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