Groupcount function sort data in alphabetical order

Question

Gianluca Regina 2022-6-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1736605-groupcount-function-sort-data-in-alphabetical-order

评论： Jan 2022-6-9

采纳的回答： Jan

event_id.mat

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Dear MatLab users,

I have a cell array (attached) containing the IDs of my events. I used the groupcount function to see how many instances and how many events I had, and everything seemed to work fine. However, I just found out that the resulting variable containing the events sorted the data in alphabetical order. Thus, the counting also refers to the modified order, which is something I do not want. I managed to find the instances manually, but I also need the list in the original order. Is there a way to do using the groupcount function?

[ N , EVENTS ] = groupcounts(event_id); % EVENTS are in alphabetical order

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Answer 1

Jan 2022-6-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1736605-groupcount-function-sort-data-in-alphabetical-order#answer_981770

编辑：Jan 2022-6-8

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% groupcounts sorts the input:
C =  {'C', 'C', 'C',  'A', 'A', 'E', 'E', 'E', 'E', 'B'}.';
[N, EVENTS] = groupcounts(C)
N = 4×1
     2
     1
     3
     4
EVENTS = 4×1 cell array
    {'A'}
    {'B'}
    {'C'}
    {'E'}
% Let N and EVENTS have the same order as in C:
[~, iC] = unique(C);   % [EDITED, bug fixed]
[~, q]  = sort(iC);
[sN, sEVENTS] = groupcounts(C);
N        = sN(q)
N = 4×1
     3
     2
     4
     1
EVENTS   = sEVENTS(q)
EVENTS = 4×1 cell array
    {'C'}
    {'A'}
    {'E'}
    {'B'}

If the equal keys are guaranteed to be neighboring:

% Call this instead of GROUPCOUNTS to keep the order
function [n, b] = RunLength_CStr(x)
x = x(:);
nx = numel(x);
d = [true; ~strcmp(x(1:nx-1), x(2:nx))];   % TRUE if values change
b = x(d);                                  % Elements without repetitions
n = diff(find([d.', true]));               % Number of repetitions
end

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Gianluca Regina 2022-6-8

I apologize, as I was writing that it didn't work (the old code gave the same results of groupcounts, so still no alphabetical order) you edited the answer and added the function, which at first glance is not immediate to understand (at least for me). Nonetheless, your code works perfectly now, thank you very much!

Jan 2022-6-9

There is no need for apologies. You are the only person who needs this function and I spend my time voluntarily to find solutions. Questions for clarifications are a standard part of solving problems.

I'm happy, if it is working now :-)

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