Function to find the more recurrent number

Hye, I have a matrix 1 by (a big number, like 1000), and I would like to extract the element of the matrix that appears most of the time.

Can you help me?

Answer 1

[Friedrich]

Hi,

i think the hist function can help here:

%create some random numbers between 1 and 20
a = ceil(rand(100,1)*20);
%put them in as much bin as different numbers exists
n = hist(a,numel(unique(a)));
%show the histogram
hist(a,numel(unique(a)));
%number appears most
element = find(n == max(n))
%numel(find(a==element)) should return the same as max(n)

[Liber-T]

Let say we have this vector: Y= 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 5.9139 5.9432 5.9140 5.9432 5.9434 5.9139

The element that appear the most is 0.0084, and that is the answer I want to extract in this case. I want the program to do it by itself. I used the line x=find(Y==max(Y)), but it give me 15. What have I done wrong?

[Friedrich]

You have to do:

>> Y_un = unique(Y);

>> n = hist(Y,numel(Y_un));

>> element = find(n == max(n));

>> most_occ = Y_un(element)

Answer 2

[Walter Roberson]

Much more simple:

x = mode(Y);

[Liber-T]

I tried it, but I received the message:

??? Undefined command/function 'mode'.

[Walter Roberson]

You must have an earlier MATLAB. mode() has been part of MATLAB for several versions, but I would have to research to find out when it was introduced.

[Paulo Silva]

the mode function was introduced in the Service Pack 3 of Release 14 back in Setember 2005, I think it's the first time I see it, thanks for bring it up Walter :)

[Liber-T]

function X=modi(Y)

%Y is a vector, it cannot be a matrix, reshape it before entering it in

%modi.

%First, FOR MY NEED, I adjust the value of Y

Y=round(10000*Y)/10000;

%Parameter used in the boucle.

n=length(Y);

%Preallocation of the memories for the matrix T

T(2,n)=0;

%For the first iteration, the first row of T is the different value of Y,

%the second is number of time the value appears.

T(1,1)=Y(1);

T(2,1)=1;

%This is a parameter to know where will go the next new value in the matrix

%T.

s=1;

for j=2:n %First boucle to pass every element of Y.

for i=1:s %Second boucle to compare the value of Y(j) with every value of T.

k=1; %Parameter of control to know if the value exist in T or not

if Y(j)==T(1,i) %Condition to adjust the count of the value Y(j), if not already in T, the value will be added after.

T(2,i)=T(2,i)+1;

k=0;

break

end

end

if k==1 %If the value didn't exist in T, it is now added.

s=s+1;

T(1,s)=Y(j);

T(2,s)=1;

end

end

%Now looking for the most recurrent value by comparing the value of the

%second row.

H=T(2,1);

d=1;

for j=2:s

if H<T(2,j)

H=T(2,j);

d=j;

end

end

%Time to give the answer.

X=T(1,d);

[Liber-T]

I've programmed one that dos the job for me, if someone need it. Just make sure it's good for you, cause at the line 5, I added a line that might create some problem, you can just erase it.

Answer 3

[Paulo Silva]

Another possible way

a=randi([1 20],1,1000);
u=unique(a);
[C,I]=max(arrayfun(@(x)sum(a==u(x)),1:numel(u)));
disp('The value that appears most times is:')
u(I)
disp('Number of times it appears:')
C

In case of having two values that appear the same number of times it will choose just one of them.

Answer 4

[huda nawaf]

hi, see it please,

X=[1 1 1 2 3 4 4 4 0]; >> v=mode(X) ??? No appropriate methods for function mode. . I need the mode which compute frequencies of appearing of number. thanks

[Liber-T]

function T=modi(Y)

%Y is a vector, it cannot be a matrix, reshape it before entering it in

%modi.

%Parameter used in the boucle.

n=length(Y);

%Preallocation of the memories for the matrix T

T(2,n)=0;

%For the first iteration, the first row of T is the different value of Y,

%the second is number of time the value appears.

T(1,1)=Y(1);

T(2,1)=1;

%This is a parameter to know where will go the next new value in the matrix

%T.

s=1;

for j=2:n %First boucle to pass every element of Y.

for i=1:s %Second boucle to compare the value of Y(j) with every value of T.

k=1; %Parameter of control to know if the value exist in T or not

if Y(j)==T(1,i) %Condition to adjust the count of the value Y(j), if not already in T, the value will be added after.

T(2,i)=T(2,i)+1;

k=0;

break

end

end

if k==1 %If the value didn't exist in T, it is now added.

s=s+1;

T(1,s)=Y(j);

T(2,s)=1;

end

end

for j=n:-1:1

if T(2,j)==0

T(:,j)=[];

else

break

end

end

I change my code, and this one should return you a matrix of two row. The first row gives you the different number, and the second row return you the frequencies of their appearence. I didn't test it.