Adjusting if loop to calculate probability for each column of dynamic array

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ajk1 2015-4-15

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评论： Michael Haderlein 2015-4-16

Hi, I am trying to adjust this for loop such that for each column of the iteration that is stored in list_val_table it counts the number of non-zero elements in each column and divides it by the total number of elements in that column. For instance if after the first iteration the result is a 5x1 array with the first element 5 and the remaining four elements 0's then the value 1/5 is stored.

if length(o_ue)>0

    list_val=val_set(o_ue)+length(o_ue);
    list_val_table(1:numel(list_val),it_3)=list_val;
end

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pfb 2015-4-15

not very clear. Your code forms a vector (or scalar) "list_val" from the vector (or scalar) "o_ue", which has length L.

Then it assigns "list_val" to the first L elements of one column of "list_val_table".

For this to work, I think "o_ue" should be a column of integers greater than one and less than the length of the matrix. Does "o_ue" contain the indices of the nonzero elements? Of "list_val_table" perhaps?

A lot of guesswork here

I fail to see the part with the zeros and ones, and the first element.

Also, where should be 1/5 stored?

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Answer 1

Michael Haderlein 2015-4-15

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I cannot find a for loop in your code. Anyway, there's also no need for a loop, this can easily be vectorized.

num_of_zeros=sum(o_ue==0,1); %count the number of zeros in every column
frac_of_zeros=num_of_zeros/size(o_ue,1);

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ajk1 2015-4-15

Thank you, yes I wanted help with the fraction of nonzero elements but from Michael's response I have adjusted it to work with the rest of my program. Thank you for both of your help.

Michael Haderlein 2015-4-16

Oops, looks like I haven't read your question too carefully. Anyway, good you could get the correct solution though. @pfb: You're right, the direction is not necessary, but I usually give it explicitly when applied on arrays of dimension > 1. It's just my preference.

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