I'Ve a list; lis=['orange','apple','orange','apple','orange','apple','car','bus'] In this list, we see orange and apple

Question

I'Ve a list; lis=['orange','apple','orange','apple','orange','apple','car','bus'] In this list, we see orange and apple

repeating adjacently 3 times. Let's write since we see "same pattern" repeats 3 times , we can say fruits-RPT-3

Now, how can we have a good algorithm for this. fruits-RPT-{N} should be only printed when there are "adjacent patterns".My idea so far is; I could;
set X=0 for index 0 and y=1 for index 1 and from then add X+=2 and Y+=2 so that it will search for one after another. Seems it only works WITHOUT going to last indexes. Since len(lis) will be crossed if we add up +2. So, it's not effective as well. Then, what is an effective algorithm for this.?

Thanks for reading and help!

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28.12.2020

7 ответов

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Пользователь 6165a Автор вопроса

Gian
I think you don't even need to mess with dicts, yo...

It will give indexouterror

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28.12.2020

Gian

Пользователь 6165a
It will give indexouterror

not if you check where you are at on the list.

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28.12.2020

Пользователь 6165a Автор вопроса

Gian
I think you don't even need to mess with dicts, yo...

-1 and +1 , then +2 .... And?

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28.12.2020

Gian

Пользователь 6165a
-1 and +1 , then +2 .... And?

well, I think that's that. unless you want to check for 3 repeated sequences, and this kind of stuff.

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28.12.2020

Пользователь 6165a Автор вопроса

Gian
well, I think that's that. unless you want to chec...

3 is just a number, it can repeat N times or even None

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28.12.2020

Gian

Пользователь 6165a
3 is just a number, it can repeat N times or even ...

No, that's not what I meant. I meant, if you want to test for 3 elements being repeated N times adjacent, you need to go n + 3, for example.

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28.12.2020

Gian · Accepted Answer

Gian

I think you don't even need to mess with dicts, you can go through the list looking if element - 1 and element +1 are the same, you can then check element + 2 to see if they are also the same, and count that as a repetition. you can then jump you index by one and do it again. you also have to check if the index you're at is the beginning or end of the list.

0

28.12.2020

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I'Ve a list; lis=['orange','apple','orange','apple','orange','apple','car','bus'] In this list, we see orange and apple

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