Repeated values in matrix

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Rachel Ramirez
Rachel Ramirez on 13 Jan 2021
Commented: Rachel Ramirez on 20 Jan 2021
I have no idea how to approach this problem and needed some guidance. Any help would be appreciated.
Say I have a matrix/array similar to Table_1. How can I go about finding the first 5 numbers that repeat at least twice only on the FIRST column? Once found a new matrix would be generated where all values from the original matrix would be included up until the 5th value from the FIRST column that repeats at least twice is found similar to Table_2. I have no idea if this is even possible to do. Keep in mind this is just an example, matrix could be larger or smaller so I would prefer not to set a value.
Please refer to images. Thank you.
I would like to keep them as array/matrix and not use cells.
Rachel Ramirez
Rachel Ramirez on 13 Jan 2021
Edited: Rachel Ramirez on 13 Jan 2021
oh yes! that was an error from my part. it would stop at [7 6]. Here is an updated version. I was actually thinking, I could either include all values that are in b/w even though they don't meet the condition of repeating or just generate matrix that meets the condition. Does it make sense?

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Accepted Answer

Prudhvi Peddagoni
Prudhvi Peddagoni on 19 Jan 2021
Edited: Prudhvi Peddagoni on 19 Jan 2021
You can define a variable called freq, to store the frequency of the a number. Frequency of number is stored in index. You will go through the matrix until atleast 5 of these indices has a value greater than 2.
freq = zeros(100,1); %assuming the numbers in the matrix won't be greater than 100
for i = 1:length(M)
freq(M(i,1)) = freq(M(i,1))+1;
if nnz(freq>=2) >= 5
%now we need to remove numbers which have frequency lessthan 2
freq(freq<2) = 0;
Now you need to iterate through the original matrix . Let x be the number present in the first column. You will include this row in the final matrix if freq(x) is not equal to zero. You iterate through the original matrix until you reach the row (variable i).
Hope this helps.
Rachel Ramirez
Rachel Ramirez on 20 Jan 2021
Thank you for the explanation. I tested out and seems to be working well. I'm going to see how to get the x values from the first column.

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