Create sperate matrices for all unique values in the 3rd column of a cell array

I have a cell array of 666x3, in this array the first and second columns have completely unique values, but the 3rd column has identifying values. I want to make a new array for each identifying value. Keep in mind there are many different values in column 3 as well and i dont want to manually input them, I want it to be done automatically so any data I use, the same result will be shown. Thank you!

Example:

initialArray =

column1 column2 column3

240.747574230792 298.790970549445 127

239.818737359994 297.862133678646 127

241.676411101591 297.862133678646 126

240.283155795393 296.933296807848 126

235.174553006003 296.004459937050 125

233.781297699806 295.540041501651 125

220.777581508630 269.997027554699 125

required:

array1=

column1 column2 column3

240.747574230792 298.790970549445 127

239.818737359994 297.862133678646 127

array2=

column1 column2 column3

241.676411101591 297.862133678646 126

240.283155795393 296.933296807848 126

array3=

column1 column2 column3

235.174553006003 296.004459937050 125

233.781297699806 295.540041501651 125

220.777581508630 269.997027554699 125

Answer 1

[Matt J]

Well, I won't talk about creating separate arrays because it's bad, but here's what you could do which wouldn't be bad:

initialArray = {
240.747574230792            298.790970549445	127
239.818737359994            297.862133678646	127
241.676411101591	    297.862133678646	126
240.283155795393	    296.933296807848	126
235.174553006003	    296.004459937050	125
233.781297699806	    295.540041501651	125
220.777581508630	    269.997027554699	125};


initialArray=cell2mat(initialArray);

[~,~,G]=unique(initialArray(:,3),'stable');

array = splitapply(@(x){x}, initialArray, G);

It's essentially what you asked for, except array is a cell array whose contents are the array chunks that you asked for.

array{:}

ans = 2×3

240.7476 298.7910 127.0000 239.8187 297.8621 127.0000

ans = 2×3

241.6764 297.8621 126.0000 240.2832 296.9333 126.0000

ans = 3×3

235.1746 296.0045 125.0000 233.7813 295.5400 125.0000 220.7776 269.9970 125.0000

[Talha Majeed]

Thanks for the answer, doesnt give me what I want but i found an answer using accumarray:

[~,~,X] = unique(initialArray(:,3));

C = accumarray(X,1:size(initialArray,1),[],@(r){finalArray(r,:)})

[Matt J]

It does give you what you want.

[Talha Majeed]

You're right it does, I was looking at the wrong thing, thanks for you help!