A = [5 10 16 22 28 32 36 44 49 56]
B = [2 1 1 2 1 2 1 2 2 2]
How to get this?
C1 = [10 16 28 36]
C2 = [5 22 32 44 49 56]
C1 needs to get the values from A, only in the positions in which B is 1
C2 needs to get the values from A, only in the positions in which B is 2
You can do this this way :
C1 = A(B==1);
C2 = A(B==2);
B==1 gives a logical array : [ 0 1 1 0 1 0 1 0 0 0 ].
A(logicalArray) returns elements for which the value of logicalArray is true (it is termed logical indexing).
A and logicalArray must of course have the same size.
It is probably the fastest way of doing this operation in matlab.
For more information on indexing, see matlab documentation.
To achieve this with an arbitrary number of groups (not just two as in your example), use accumarray with an a anoynmous function to collect the values in each group into a cell. To preserve order, B needs to be sorted first (and the same order needs to be applied to A):
[B_sort, ind_sort] = sort(B);
C = accumarray(B_sort.', A(ind_sort).', [], #(x){x.'});
This gives the result in a cell array:
>> C{1}
ans =
10 16 28 36
>> C{2}
ans =
5 22 32 44 49 56
Related
Suppose I have a matrix of dimension [4x4], and a vector of [16x1], I need to multiply every 4 element in the vector in one element in the matrix, (instead of multiplying element in row by element in vector), how can I do that using loop ?
For example here below, the results of the first four elements in the resulted vector as shown in the below example, then the same thing for the second, third and fourth rows in the matrix. :
So the results in that example is supposed to be [16x1]
Thank you
Using kron you can use this one-liner:
%A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
%v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
sum(kron(A,ones(4,4)).'.*v).'/4
I use the kronecker tensor product to "replicate" 4x4 time the A matrice. After that it's pure algebra.
This is just matrix multiplication in disguise... If your tall vector was a matrix of the same size as the matrix shown, where each highlighted block is a row, it's matrix multiplication. We can set this up, then reshape back into a vector.
You can use indexing to turn this into simple matrix multiplication. A question I answered already today details how the below indexing works using bsxfun, then we just reshape at the end:
% Setup
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
% Matrix mutliplication
r = numel(v)/size(A,1);
b = A * v( bsxfun( #plus, (1:r:numel(v)).', 0:r-1 ) );
% Make result a column vector
b = reshape( b.', [], 1 );
See if this is what you want:
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
r = reshape(sum(bsxfun(#times, permute(A, [3 2 1]), permute(reshape(v, 1, [], size(A,2)), [2 3 1])), 2), [], 1);
which gives
r =
17
17
17
17
41
41
41
41
65
65
65
65
89
89
89
89
There are details that I assumed, but this shoudl do the trick:
A=reshape(1:16,4,4).';
b=repelem([2,0,1,3],1,4).';
c=[];
for row=1:size(A,1)
c=[ c; sum(reshape(repelem(A(row,:),4).*b.',4,[]),2)];
end
I am assuming here that your demo for the vector is just a bad example and that you wont have repeated values, otherwise an easier version can be achieved by just not doing 3/4ths of the multiplications.
If you do not have access to repelem, have a look at alterative codes that do the same thing:Element-wise array replication in Matlab
I need to loop through coloumn 1 of a matrix and return (i) when I have come across ALL of the elements of another vector which i can predefine.
check_vector = [1:43] %% I dont actually need to predefine this - i know I am looking for the numbers 1 to 43.
matrix_a coloumn 1 (which is the only coloumn i am interested in looks like this for example
1
4
3
5
6
7
8
9
10
11
12
13
14
16
15
18
17
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
1
3
4
2
6
7
8
We want to loop through matrix_a and return the value of (i) when we have hit all of the numbers in the range 1 to 43.
In the above example we are looking for all the numbers from 1 to 43 and the iteration will end round about position 47 in matrix_a because it is at this point that we hit number '2' which is the last number to complete all numbers in the sequence 1 to 43.
It doesnt matter if we hit several of one number on the way, we count all those - we just want to know when we have reached all the numbers from the check vector or in this example in the sequence 1 to 43.
Ive tried something like:
completed = []
for i = 1:43
complete(i) = find(matrix_a(:,1) == i,1,'first')
end
but not working.
Assuming A as the input column vector, two approaches could be suggested here.
Approach #1
With arrayfun -
check_vector = [1:43]
idx = find(arrayfun(#(n) all(ismember(check_vector,A(1:n))),1:numel(A)),1)+1
gives -
idx =
47
Approach #2
With customary bsxfun -
check_vector = [1:43]
idx = find(all(cumsum(bsxfun(#eq,A(:),check_vector),1)~=0,2),1)+1
To find the first entry at which all unique values of matrix_a have already appeared (that is, if check_vector consists of all unique values of matrix_a): the unique function almost gives the answer:
[~, ind] = unique(matrix_a, 'first');
result = max(ind);
Someone might have a more compact answer but is this what your after?
maxIndex = 0;
for ii=1:length(a)
[f,index] = ismember(ii,a);
maxIndex=max(maxIndex,max(index));
end
maxIndex
Here is one solution without a loop and without any conditions on the vectors to be compared. Given two vectors a and b, this code will find the smallest index idx where a(1:idx) contains all elements of b. idx will be 0 when b is not contained in a.
a = [ 1 4 3 5 6 7 8 9 10 11 12 13 14 16 15 18 17 19 20 21 22 23 24 25 26 ...
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 1 3 4 2 6 7 8 50];
b = 1:43;
[~, Loca] = ismember(b,a);
idx = max(Loca) * all(Loca);
Some details:
ismember(b,a) checks if all elements of b can be found in a and the output Loca lists the indices of these elements within a. The index will be 0, if the element cannot be found in a.
idx = max(Loca) then is the highest index in this list of indices, so the smallest one where all elements of b are found within a(1:idx).
all(Loca) finally checks if all indices in Loca are nonzero, i.e. if all elements of b have been found in a.
In trying to port an algorithm from C# to Matlab I found that Matlab is inefficient at running for loops. As such I want to vectorize the algorithm.
I have following inputs:
lowrange:
[ 00 10 20 30 40 50 ... ]
highrange:
[ 10 20 30 40 50 60 ... ]
These arrays are equal in length.
I now have a third array Values (which could be any length) and for this array I want to count the occurrences of Values elements between lowerange(i) and highrange(i) (You can see I'm coming from a for loop).
The output should be an array of length lowrange/highrange.
So with the above arrays and input LineData:
[ 1 2 3 4 6 11 12 16 31 34 45 ]
I expect to get:
[ 05 03 00 02 01 00 ... ]
I tried the (for me) obvious thing:
LineData(LineData < PixelEnd & LineData > PixelStart)
But that doesn't work because it just checks LineData on an element by element way. It does not try to apply the comparison over all values in LineData.
Unfortunately, I cannot come up with anything else since I'm not yet used to think in a Matlab 'vector' way, let alone knowing all applicable instructions from memory.
As you are looking to do a basic histogram with given edges, you can use Matlabs built-in function histc:
values = [ 1 2 3 4 6 11 12 16 31 34 45 ];
edges = 0:10:60;
histc(values, edges)
ans =
5 3 0 2 1 0 0
For ranges with identical intervals and starting from 0, here's a bsxfun based counting approach -
LineData = [ 1 2 3 4 6 11 12 16 31 34 45 ] %// Input
interval = 10; %// interval width
num_itervals = 6; %// number of intervals
%// Get matches for each interval and sum them within each interval for the counts
out = sum(bsxfun(#eq,ceil(LineData(:)/interval),1:num_itervals))
Output -
LineData =
1 2 3 4 6 11 12 16 31 34 45
out =
5 3 0 2 1 0
Assuming that the last interval would be the one holding the max of input data, you can try out a diff + indexing based approach too -
LineData = [ 1 2 3 4 6 11 12 16 31 34 45 ] %// Input
interval = 10; %// interval width
labels = ceil(LineData(:)/interval); %// set labels to each input entry
df_labels = diff(labels)~=0; %// mark the change of labels
df_labels_pos = find([df_labels; 1]); %// get the positions of label change
intv_pos= labels([true;df_labels]);%// position of each interval with nonzero counts
%// get counts from interval between label position change and put at right places
out(intv_pos) = [ df_labels_pos(1) ; diff(df_labels_pos)];
Assume that we have a 100*4 array.
We also have a 100*1 array of 1 and 0. Assume there are n 1's.
We want to create a n*4 array from the 100*4 array, where we only include the columns for which the second array is a 1.
One way to do it is through a double for loop. Is there a simpler method?
So, We have
A = [ [ 332 44 33 22 33 55 33 211 .....
[ 823 44 12 98 19 23 32 911 .....
....
....
]
and
B = [1 0 0 1 0 0 0 ....]
and we want
C = [ [ 332 22 ...
[ 823 98 ...
....
....
]
You should use logical indexing:
C = A(:, B==1 );
First you repmat the logical vector so that it has the exact same size as the matrix A.
idx2keep = repmat(b, [1 4]); % Or [4 1] depending on if it's a col or row vector
Then you can simply index them with
B = A( idx2keep )
you can then make it into a column vector:
B = B(:)
That should do the job. Next time please always post some code or notation so it's easier and clearer to answer this.
I have two arrays in Matlab
say
A = [1 4 89 2 67 247 2]
B = [0 1 1 1 0 0 1]
I want an array C, which contains elements from array A, if there is 1 in B at the corresponding index. In this case, C = [4 89 2 2].
How to do this?
Use logical indexing:
>> C = A(logical(B))
C =
4 89 2 2