I am trying with following code, but get indice of last value only
A=[ 3 4 1 2 4 4 4]
B=unique(A)
[b1 b2]=max(B)
while i<=numel(A)
if A(i)==A(b2)
ID=A(i)
end
end
Is there any way other in matlab to get all indices of value 4 (max value).
If you want to find the indices of the largest value in your matrix, there is no need for unique at all. It's superfluous. Just use find and max simultaneously:
ID = find(A == max(A));
max(A) returns the largest value in A. A == max(A) returns a logical vector where 1 corresponds to a value in A matching the largest value in A and 0 otherwise. Finally, find determines the locations in the input that are non-zero, so in effect we are finding the locations that match the largest value in A.
Related
I have a m-by-1 matrix, A. I want to find out which elements are duplicates and get their row values. (Just the row values because the matrix is m-by-1.)
I've tried
k = find(~unique(A));
but k contains the wrong values.
Here's an example of what I'm trying to do. Consider the array A;
A = [4
5
5
5
7
8
4];
Since 4 and 5 are the repeated elements here, I would like to get the row values of these elements and put them in a new array. The resulting array would be
RowValues= [1
2
3
4
7];
Note: The above is just an example and the actual array I am dealing with contains rational numbers of the type -0.0038, 1.3438 and so on in the array A.
Here is a solution using intersect:
s = sort(A);
c = intersect(s(1:2:end),s(2:2:end));
RowValues = find(ismember(A,c));
I have compared this method with the method proposed by #SardarUsama with a large [1 x 10000000] input in Octave. Here is the result:
=======INTERSECT==========
Elapsed time is 1.94979 seconds.
=======ACCUMARRAY==========
Elapsed time is 2.5205 seconds.
Find the count of each element of A using unique and accumarray, filter out the non-repeating values, use ismember to get the logical indices of repeating values and then use find to convert them to linear indices.
[C, ~, ic] = unique(A);
RowValues = find(ismember(A, C(accumarray(ic,1)>1)));
Why you get the wrong indices with your code?
Applying logical NOT on the vector of unique values would convert them to a logical vector containing true at the index where unique value is zero and false where it is non-zero and hence finding the non-zero (false in this case) elements of such a vector would lead nowhere.
This doesn't make any sense to me and I'm not sure what to even search for.
Matlab code:
[a b] = max(.9);
Output:
a =
0.9
b =
1
Why is it increasing by 1/10? What does [a b] do when on the left side of equal sign?
max is used to find the maximum value of an array. The second output (if requested), returns the index that corresponds to the first value in the array that is equal to the maximum value.
[max_value, max_index] = max([1 3 3 2]);
% max_value = 3
% max_index = 2
In your case, you are passing a scalar (a 1 x 1 array) to max, therefore the scalar is the maximum and the maximum appears at index 1.
[M,I] = max(A) finds the indices of the maximum values of A and returns them in output vector I, using any of the input arguments in the previous syntaxes. If the maximum value occurs more than once, then max returns the index corresponding to the first occurrence.
Reference: https://www.mathworks.com/help/matlab/ref/max.html?requestedDomain=www.mathworks.com
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
I have a structure A with fields B and C.
Values of B are double and can take on the value of 1 or 2.
Values of C are double and range in value from 1 to about 50.
Both B and C have the same number of elements (when a value is assigned to C then a 1 or 2 is assigned to B).
The following code is supposed to find all index values where B equals 1 and then use those index values to find the minimum value of C.
>> a=find(A(1).B(:)==1);
>> [value,index]=min(A(1).C(a))
value = 5.020000000000000
index = 630
As you can see below, when I put the minimum value index back into B it returns a 2.
>> A(1).B(630)
ans = 2
The problem stems from the fact that vector a is not the same size as original vector A(1).B(:). So the value you get in index refers to the corresponding location in a not in A(1).C(:) to get the correct index use:
A(1).B(a(index))
In Matlab, by the function min（）， I can only get one single minimum element of a vector, even if there can be several equal minimum elements. I was wondering how to get the indices of all minimum elements in a vector?
For example,
v=[1,1];
I would like to get the indices 1 and 2, both of which index the smallest elements 1.
Thanks and regards!
You can use find to find the min values:
find(v == min(v))
v = [1 2 3 1 5];
find( v == min(v) )
ans = 1 4
At least in Octave (don't have matlab), this returns the indexes of all minimums in v