Converting product into individual entries

Can I write [4*3 5 7*0 11 3*9] in the following format.

[3,3,3,3,5,0,0,0,0,0,0,0,11,9,9,9]

[Stephen23]

Given this vector:

V = [4*3,5,7*0,11,3*9]

V = 1×5

12 5 0 11 27

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how do you determine that specific vector? For example, there are infinite possible products that give 0, not just 7*0:

Why do you expect 0,0,0,0,0,0,0 and not 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 which has exactly the same product?

Why do you expect 3,3,3,3 and not 4,4,4 ? Or 6,6 ? Or 2,2,2,2,2,2 ? Or 1,1,1,1,1,1,1,1,1,1,1,1 ?

Why do you expect 9,9,9 and not 3,3,3,3,3,3,3,3,3 ?

So far you have not provided any explanation of how you selected those particular factors from vector V.

[Mahendra Yadav]

For any product, Let's say we have a*b.

Then "a" refers here to the number of times, the value "b" should appear.

Hence for a*b, we have

[b,b,b,b,b,b.....(a times)]

[Stephen23]

@Mahendra Yadav: Sure, I know what product is. But nowhere do you explain how were are supposed to determine which factors you want to use. For example, given the product 12 you could have:

• a=3 b=4
• a=4 b=3
• a=2 b=6
• a=6 b=2
• a=1 b=12
• a=12 b=1

So far you have given absolutely no explanation of how you want to select the specific factors given some random value.

Or perhaps you are just not explaining the form that your data actually have. I cannot guess this.

Answer 1

[Steven Lord]

If you have the individual terms and their replication factors already, see the repelem function. If you don't you may need to use the factor function first.

[Mahendra Yadav]

Yes, I'm aware about both of these functions. But how can I implement these in the above mentioned array.

Or

wheter any alternative solutions are possible?

[Steven Lord]

Do you have your data in this form?

option1 = [4*3 5 7*0 11 3*9]

option1 = 1×5

12 5 0 11 27

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Or do you have data in this form?

option2a = [4 1 7 1 3]

option2a = 1×5

4 1 7 1 3

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option2b = [3 5 0 11 9]

option2b = 1×5

3 5 0 11 9

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If you have option 1, how do you handle the ambiguity with 7*0 called out by Stephen23?

And since you seem to be trying to do something with prime factorizations, why is the first group [3 3 3 3] and not [2 2 2 2 2 2] (which you would write 6*2)? Both those groups sum to 12 which is the first element of option1, so how do you disambiguate? And why isn't the third group just empty [] (zero 7s) instead of [0 0 0 0 0 0 0] (seven 0s)?

With option 2:

x = repelem(option2b, option2a)

x = 1×16

3 3 3 3 5 0 0 0 0 0 0 0 11 9 9 9

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Answer 2

[Xiaotao]

a = [3*ones(1,4), 5, zeros(1,7), 11, 9*ones(1,3)]

a = 1×16

3 3 3 3 5 0 0 0 0 0 0 0 11 9 9 9

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[Mahendra Yadav]

Okay. It's good. But whether can it be done programmatically with any other similiar inputs.

[Xiaotao]

cnt=[4 1 7 1 3]; % counts
val=[3 5 0 11 9]; % values
a = arrayfun(@(c, v) v * ones(1, c), cnt, val, 'UniformOutput', false);
a = [a{:}] % Concatenate the results

a = 1×16

3 3 3 3 5 0 0 0 0 0 0 0 11 9 9 9

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[Stephen23]

The simpler MATLAB approach:

cnt = [4,1,7,1,3]; % counts
val = [3,5,0,11,9]; % values
out = repelem(val,cnt)

out = 1×16

3 3 3 3 5 0 0 0 0 0 0 0 11 9 9 9

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[Mahendra Yadav]

Thank You!

It worked