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for some vector [25 15 23 20 .. n ] i want to find if the adjacent values exceed 1.2 of each other

aaaaaaaa
6-Contributor

for some vector [25 15 23 20 .. n ] i want to find if the adjacent values exceed 1.2 of each other

Hi,

 

I want to write a code where for some vector [25 15 23 20 .. n ] i want to find if the ratios of the adjacent values exceed 1.2 of each other. The output result would be like "Yes" or "No". how do i go about doing that?

 

For example: for vector [15 25 20], because 25/15>1.2, I want the answer as "Yes".

But I would also want the same answer "Yes" if it was [25 15 20] because in this case also for the adjacent values 25/15>1.2.

 

Thanks.

 

Edit1: added "ratios of the"

Edit2: added example

1 ACCEPTED SOLUTION

Accepted Solutions

Here is a variant which returns just a single answer:

Werner_E_0-1581355629135.png

 

 

View solution in original post

7 REPLIES 7

Maybe a custom command like the following will do the job:

Werner_E_0-1581347964974.png

 

In case you are using Prime, use "else" instead of "otherwise".

 

aaaaaaaa
6-Contributor
(To:Werner_E)

I'm so sorry that i mistyped the question. 

 

For example: for vector [15 25 20], because 25/15>1.2, I want the answer as "Yes".

But I would also want the same answer "Yes" if it was [25 15 20] because in this case also for the adjacent values 25/15>1.2.

 

Thanks for your reply

 

OK, so its the Ratio, not the difference.

And what do you expect as a result? Just one single "Yes" or "No" for the whole Vector if the Ratio of any adjacent pair exceeds 1.2, or a "Yes" or "No" for every triple (or pair?) in the Vector?

If the latter - what about the ends? Do you expect an  n x 1 vector, an  (n-1) x 1 vector or an  (n-2) x 1 vector as a result?

Here is a variant which returns just a single answer:

Werner_E_0-1581355629135.png

 

 

aaaaaaaa
6-Contributor
(To:Werner_E)

Just one single "Yes" if the Ratio of all adjacent pair does not exceeds 1.2, and just a single "No" if the Ratio of any adjacent pair does not exceeds 1.2.

 

Sorry for my bad english. 

 

Thank you.

OK, so you would have to exchange "Yes" and "No" in my last attempt.

Otherwise it should do what you demand.

aaaaaaaa
6-Contributor
(To:Werner_E)

Thanks I was looking for the algorithm on how i should work it out;; your last attempt helped me a bunch!! thanks a lot!

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