how to find mode without built-in mode function?

any suggestions how to write a function that finds a vectors mode WITHOUT using the built-in mode function?

Answer 1

[the cyclist]

You could use the hist() function, and use the value that has the highest count.

[izabela kudela]

How can I precisely read which value has the highest count?

[Walter Roberson]

[counts, centers] = hist(YourVector);
[~, maxidx] = max(counts);
mode_value = centers(maxidx);

[Louise Wade]

This gives me non-integer values when all values in my array are integers.

[the cyclist]

I think this is because by default, hist() chooses the bin centers by an algorithm that does not guarantee they are in the original vector. Try this instead:

[counts, centers] = histcounts(YourVector,[unique(YourVector) Inf]);
[~, maxidx] = max(counts);
mode_value = centers(maxidx);

[the cyclist]

I realized that that code will not find all the modes, if there are multiple. This code should:

[counts, centers] = histcounts(YourVector,[unique(YourVector) Inf]);
max_value = max(counts);
max_idx = (counts==max_value);

mode_values = centers(max_idx)

Answer 2

[Walter Roberson]

unique(), take the third output, put it through accumarray(), find the max()

Answer 3

[Roger Stafford]

If you are not allowed to use the 'mode' function, it sounds as though you must use only more primitive functions. Are you allowed to do the following with v as the given column vector?

 u = sort(v);
 f = find([true;diff(u)~=0;true]);
 [~,ix] = max(diff(f));
 m = u(f(ix)); % m is most frequent value occurring in v

(Corrected)

[dana]

thanks. i am indeed not allowed to use the mode function and need to do it in a more primitive way. i am not sure, though, that i completely understand what you did here. could you please explain?

[Roger Stafford]

There's nothing magic about it. I'll give an example. Let v be:

 v = [9;4;2;4;9;2;4]

Then you get

 u = sort(v) = [2;2;4;4;4;9;9]
 diff(u) = [0;2;0;0;5;0]
 diff(u)~=0 = [F;T;F;F;T;F]   Where 'T' & 'F' stand for true & false
 [true;diff(u)~=0;true] = [T;F;T;F;F;T;F;T]
 f = find([true;diff(u)~=0;true]) = [1;3;6;8]
 diff(f) = [2;3;2]
 max(diff(f)) is max at second value, hence ix = 2
 f(ix) = f(2) = 3
 m = u(f(ix)) = u(3) = 4  Therefore the mode value is 4

Does that show you the principles involved?

[Louise Wade]

I am trying this and I am getting the below error. Sorry to bring this up on a 7.5 year comment, it's just relevant to something I am doing right now.

Error using vertcat

Dimensions of arrays being concatenated are not consistent.

Error in Question_1 (line 29)

f = find([true;diff(u)~=0;true]);

[Rik]

That depends on the dimensions of v. If it is a column vector this runs without error.

v = [9;4;2;4;9;2;4];
u = sort(v);
[true;diff(u)~=0;true]

ans = 8x1 logical array
   1
   0
   1
   0
   0
   1
   0
   1

v = [9;4;2;4;9;2;4];
v = v.';
u = sort(v);
[true;diff(u)~=0;true]

Error using vertcat
Dimensions of arrays being concatenated are not consistent.

[Louise Wade]

I'll try and transpose my row vector into a column vector and try again. If there are multiple values as the mode, would this cause an issue?

EDIT: It seems to be working for me. Thank you so much for helping. All I needed to do was transpose the array.