finding the frequency of a series of numbers in an array

given the data set

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,6,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1];

I want to find the number of occurrences (frequency) of each series of numbers in the array. For example, I know there are eleven places that 1 occurs.

k=1
location=find(data==k)

location = 1 2 3 22 23 24 25 28 29 30 35

But I want the frequency (number of series that occur in the data set for a given value in the array. Also, I want to exclude series of data in the frequency count if they fall below a threshold value (arbitrarily a value of 2 in this case). The output for the one value should be freq=3

I want the output to be a 10x2 array with the frequency of each number value 1 through 10 to be sorted in order

seriesfreq=

1     3
2     2
3     1
4     1
5     1
6     0
7     0
8     0
9     1
10    1

Please help!

Answer 1

[Guillaume]

A method with no loop:

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1];
threshold = 2;
transpos = find(diff(data)); %gives the position of the end of each series of identical number
transvals = data(transpos);  %gives the corresponding value of each series
transvals(diff([0 transpos numel(data)+1]) < threshold) = []; %remove values for which the run is shorter than threshold
bins = min(data):max(data)';
[bins; histcounts(transvals, [bins Inf])]' %get the histogram of the runs

[Joe]

I've combined the answeres into one use with a threshold

data=[1,1,1,2,2,2,3,6,3,3,10,3,10];
threshold=1;
transpos = find(diff(data));
transpos1=[transpos, transpos(end)+1];
transvals = data(transpos1);  
transvals(diff([0 transpos1 numel(data)+1])<threshold)=[0];

for k=1:1:10
frequency(k)=length(find(transvals==k));
end

Thanks for all the help!

Answer 2

[Jan]

With FEX: RunLength:

[value, len] = RunLength(data);
bin          = min(value):max(value);
h            = histc(value(len > 2), bin);
result       = [bin.', h.']

Answer 3

[Azzi Abdelmalek]

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,6,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1];
ii=unique(data);
for k=1:numel(ii)
  a=data==ii(k);
  b=[0 a 0];
  out(k)=sum(strfind(b,[1 0])-strfind(b,[0 1])>=2);
end
out

[Joe]

If there are a fixed number of places in this array (for example k=1:1:10) and the data array is missing the number 6

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1]

is there a way to output a zero in its place to make it a 10x1 array?

[Azzi Abdelmalek]

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,6,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1];
ii=min(data):max(data);
for k=1:numel(ii)
a=data==ii(k);
b=[0 a 0];
out(k)=sum(strfind(b,[1 0])-strfind(b,[0 1])>=2);
end
out

Answer 4

[Suvidha Tripathi]

data=[1,1,1,2,2,3,3,3,4,4,3,3,5,5,5,7,8,9,9,9,1,1,1,1,10,10,1,1,1,2,2,2,2,1]
unique_data=unique(data);
disp("series_no "+"frequency")
for i=1:length(unique_data)
    [~,indices]=find(data==unique_data(i));
    freq(i)=length(indices);
    disp(unique_data(i)+" "+freq(i))
end

Answer 5

[Manoel Rodrigues Trigueiro]

% Frequency order function %

function f = freq(X)

a = unique(X); len_X = length(a);

for i=1:len_X

f(i) = mode(X);

ind = find(mode(X)==X);

for j=1:length(ind)

X(ind(j)) = NaN;

end

end

end