Creating a frequency table

Hello,
suppose I have a vector in legnth of 100 that contains a number between 1-4 how can I make a frequncy vector using the data in the vector? I'd tryed the function tabulate but if there is no 4 or 3 for example it will not be in the table and I want it to be there just with the value 0.
Thank's!

3 Kommentare

Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013
This is not clear
googo
googo am 31 Mai 2013
Bearbeitet: googo am 31 Mai 2013
Ok, sorry.
For example : [4 3 2 2 4 3 3] {only numbers between 0-4}
I'm looking for a function that will return a frequency vector :
0
0
2
3
4
Image Analyst
Image Analyst am 31 Mai 2013
Bearbeitet: Image Analyst am 31 Mai 2013
That's simply the histogram - I think - and you can use histc() or hist(). It's "I think" because I'm not sure why you say there are four 4's when I only see 2. The histogram would say there is two 4's in [4 3 2 2 4 3 3]. Perhaps you can explain your thought process around how you came up with four 4's.

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 Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013

0 Stimmen

a=arrayfun(@(z) sum(ismember(x,z)),0:4)'
out=[(0:4)' a 100*a/numel(x)]

Weitere Antworten (2)

Andrei Bobrov
Andrei Bobrov am 31 Mai 2013

1 Stimme

x=[1 1 2 2 1 2 ];
a = histc(x(:),0:4);
out = [(0:4)', a, a/sum(a)*1e2];
Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013

0 Stimmen

x=[1 1 2 2 1 2 ]
y=[x 5]
out=tabulate(y)
out(end,:)=[]

6 Kommentare

googo
googo am 31 Mai 2013
>> x=[1 1 2 2 1 2 ];
y=[x 5];
out=tabulate(y);
out(end,:)=[];
>> out
out =
1.0000 3.0000 42.8571
2.0000 3.0000 42.8571
3.0000 0 0
4.0000 0 0
it was suppose to be
0 0 0%
1 3 50%
2 3 50%
3 0 0%
4 0 0%
didn't understand the purpose of the "5" in your code. Thank's!
Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013
You said between 1 and 4
googo
googo am 31 Mai 2013
Bearbeitet: googo am 31 Mai 2013
ho... sorry.. my mistake but anyway why it's 42.85% an not 50%?
Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013
Bearbeitet: Azzi Abdelmalek am 31 Mai 2013
If there is no 4 in your list, tabulate will not show 4 0 0 like you said, To resolve the problem,I've added 5 and removed its result
x=[1 1 2 2 1 2 ];
[0:4 ;histc(x,0:4)]'
Azzi Abdelmalek
Azzi Abdelmalek am 31 Mai 2013
Iman , make it as an answer

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