How I can know elements of a matrix

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I need know the elements of a matrix if true
I=round(rand(10,10)*1000);
What are the elements of this matrix
Thank you very much
  1 Comment
Jan
Jan on 7 Mar 2013
A small example of the wanted output would avoid, that we waste your and our time with guessing, what you are searching for. Please add an example with a 2x2 matrix by editing the question (not as comment or answer).

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Accepted Answer

Jan
Jan on 7 Mar 2013
Another guess:
A = round(rand(10,10,3) * 1000);
existingElements = unique(A(:));

More Answers (5)

Carlos
Carlos on 7 Mar 2013
When you multiply the matrix by 1000 you have binary values(1 and 0 true false)no more.So I don´t know what you exactly mean by true. However, assuming you have a 0 and 1 valued matrix, you could do the following
>> I=round(rand(10,10));
>> I
I =
0 0 0 0 1 0 1 0 1 1
1 0 1 1 1 0 0 0 0 1
0 0 0 0 0 1 0 1 1 0
1 1 1 0 1 1 0 0 0 0
0 0 1 0 0 0 0 0 1 0
1 1 1 0 0 0 0 0 0 0
0 1 0 1 0 0 1 1 0 1
1 1 0 1 0 1 1 1 1 1
1 0 0 1 0 0 1 1 1 1
1 0 1 0 0 0 0 0 0 1
Now using the find command you can find out the indices that contain a 1.
>> find(I==1)
ans =
2
4
6
8
9
10
14
16
17
18
22
24
25
26
30
32
37
38
39
41
42
44
53
54
58
61
67
68
69
73
77
78
79
81
83
85
88
89
91
92
97
98
99
100
However you have to translate these index values to get what you want, for instance the index 100 corresponds to the matrix position I(10,10), the index 99 to the matrix position I(9,9), the index 88 refers to I(8,8) and so on.
  1 Comment
Sebastian G
Sebastian G on 7 Mar 2013
I have 1000 in the matrix elements as possible and I have to see which ones are.

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Sebastian G
Sebastian G on 7 Mar 2013
I have 1000 in the matrix elements as possible and I have to see which ones are.

Carlos
Carlos on 7 Mar 2013
If you want to find if you have a 1000 in your matrix try the same thing, but change 1 for 1000
find(I==1000)
Will return
ans =
100
If the element I(10,10) is 1000
  1 Comment
Sebastian G
Sebastian G on 7 Mar 2013
I have 1000 potential in the matrix element.
I=round(rand(10,10)*1000);

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Sebastian G
Sebastian G on 7 Mar 2013
This method is a possible but not optimal. There must be something faster.
I=round(rand(10,10,3)*1000);
n=1;
for i=1:1000;
a=sum(find(I==i));
if a>=1;
elm(n)=i;
n=n+1;
end
end

Image Analyst
Image Analyst on 7 Mar 2013
Edited: Image Analyst on 7 Mar 2013
Not sure what you want. Virtually every one of the elements in (the badly-named) I will be considered as true. Only those few which might happen to have a value of 0 will be considered as false. All others, with values from 1 to 1000 will be considered as true in a logical test, like an if statement.
From your answer above I can't tell if you're trying to build a histogram of values, or something else. If you want a histogram, just call hist():
[counts, values] = hist(I, numberOfBins);

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