I have an array in MATLAB
For example
a = 1:100;
I want to select the first 4 element in every successive 10 elements.
In this example I want to b will be
b = [1,2,3,4,11,12,13,14, ...]
can I do it without for loop?
I read in the internet that i can select the element for each step:
b = a(1:10:end);
but this is not working for me.
Can you help me?
With reshape
%// reshaping your matrix to nx10 so that it has successive 10 elements in each row
temp = reshape(a,10,[]).'; %//'
%// taking first 4 columns and reshaping them back to a row vector
b = reshape(temp(:,1:4).',1,[]); %//'
Sample Run for smaller size (although this works for your actual dimensions)
a = 1:20;
>> b
b =
1 2 3 4 11 12 13 14
To vectorize the operation you must generate the indices you wish to extract:
a = 1:100;
b = a(reshape(bsxfun(#plus,(1:4)',0:10:length(a)-1),[],1));
Let's break down how this works. First, the bsxfun function. This performs a function, here it is addition (#plus) on each element of a vector. Since you want elements 1:4 we make this one dimension and the other dimension increases by tens. this will lead a Nx4 matrix where N is the number of groups of 4 we wish to extract.
The reshape function simply vectorizes this matrix so that we can use it to index the vector a. To better understand this line, try taking a look at the output of each function.
Sample Output:
>> b = a(reshape(bsxfun(#plus,(1:4)',0:10:length(a)-1),[],1))
b =
Columns 1 through 19
1 2 3 4 11 12 13 14 21 22 23 24 31 32 33 34 41 42 43
Columns 20 through 38
44 51 52 53 54 61 62 63 64 71 72 73 74 81 82 83 84 91 92
Columns 39 through 40
93 94
Related
Say I have an Mx4 array A where the values in the first column are a number 1 to 12. Now I want to gather the rest of the columns in 12 separate Mx3 arrays depending on which number is in column 1.
How would I go about doing that?
You can use unique and splitapply as follows. The result is a cell array of arrays.
M = [2 11 41 51;
1 10 20 30;
1 62 83 22;
4 73 53 53;
2 84 94 14]; % example data
L = 5; % Group labels are 1:L (L=12 in your case)
[u,~,w] = unique(M(:,1));
result = cell(L,1);
result(u) = splitapply(#(x){x}, M(:,2:end), w);
This gives
>> celldisp(result)
result{1} =
10 20 30
62 83 22
result{2} =
11 41 51
84 94 14
result{3} =
[]
result{4} =
73 53 53
result{5} =
[]
I have an array with three columns like this:
A B C
10 75 20
30 67 50
85 12 30
98 49 70
I have A and B values, and I want to get the corresponding C value.
For example if I enter (30,67) it should display 50.
Does Matlab have any trick for getting C value?
(my dataset is very large, and I need a fast way)
you can use ismember:
ABC = [10 75 20
30 67 50
85 12 30
98 49 70];
q = [30 67
85 12];
[~, locb] = ismember( q, ABC(:,1:2), 'rows' );
C = ABC(locb,3);
The result you get is
C =
50
30
Note that the code assume all pairs in q can be found in ABC.
Let your input data be defined as
data = [ 10 75 20
30 67 50
85 12 30
98 49 70];
values = [ 30 67];
This should be pretty fast:
index = data(:,1)==values(1) & data(:,2)==values(2); %// logical index to matching rows
result = data(index,3); %// third-column value for those rows
This gives all third-column values that match, should there be more than one.
If you want to specify several pairs of values at once, and obtain all matching results:
index = any(bsxfun(#eq, data(:,1).', values(:,1)), 1) & ...
any(bsxfun(#eq, data(:,2).', values(:,2)), 1);
result = data(index,3);
For example, given
data = [ 10 75 20
30 67 50
85 12 30
98 49 70
30 67 80 ];
values = [ 30 67
98 49];
the result would be
result =
50
70
80
You can create a sparse matrix. This solution only works if C does not contain any zeros and A and B are integers larger 0
A = [10 30 85 98]';
B = [75 67 12 49]';
C = [20 50 30 70]';
S = sparse(A,B,C);
S(10,75) % returns corresponding C-Value if found, 0 otherwise.
Try accumarray:
YourMatrix = accumarray([A B],C,[],#mean,true);
This way YourMatrix will be a matrix of size [max(A) max(B)], with the values of C at YourMatrix(A(ind),B(ind)), with ind the desired index of A and B:
A = [10 30 85 98]';
B = [75 67 12 49]';
C = [20 50 30 70]';
YourMatrix = accumarray([A B],C,[],#mean,true);
ind = 2;
YourMatrix(A(ind),B(ind))
ans =
50
This way, when there is a repetition in A B, it will return the corresponding C value, provided each unique pair of A B has the same C value. The true flag makes accumarray output a sparse matrix as opposed to a full matrix.
I have a 3x3x2000 array of rotation matrices that I need to transform into a 2000x9 array.
I think I have to use a combination of permute() and reshape(), but I don't get the correct output order.
This is what I need:
First row of 3x3 array needs to be columns 1:3 in the output
Second row of 3x3 array needs to be columns 4:6 in the output
Third row of 3x3 array needs to be columns 7:9 in the output
I have tried all possible combinations of numbers 1 2 3 in the following code:
out1 = permute(input, [2 3 1]);
out2 = reshape(out1, [2000 9]);
But I always end up with the wrong order. Any tips for a Matlab newbie?
How about a simple for-loop?
for i=1:size(myinput,3)
myoutput(i,:)=[myinput(1,:,i) myinput(2,:,i) myinput(3,:,i)];
% or
% myoutput(i,:)=reshape(myinput(:,:,i),[],9);
end
It's not simple as using permute and reshape, but it is transparent and easier for debugging. Once everything in your program runs perfectly, you can consider to rewrite such for-loops in your code...
You had a mix-up in your permute
a = reshape(1:9*6, 3, 3, []);
a is a 3x3x6 matrix, each
a(:,:,i) = 9*(i-1) + [1 4 7
2 5 8
3 6 9];
So
out1 = permute(a, [3,1,2]);
out2 = reshape(out1, [], 9);
Or in one line
out3 = reshape(permute(a, [3,1,2]), [], 9);
So
out2 = out3 =
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
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 44 45
46 47 48 49 50 51 52 53 54
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.
I would like to create an array or vector of musical notes using a for loop. Every musical note, A, A#, B, C...etc is a 2^(1/12) ratio of the previous/next. E.G the note A is 440Hz, and A# is 440 * 2^(1/12) Hz = 446.16Hz.
Starting from 27.5Hz (A0), I want a loop that iterates 88 times to create an array of each notes frequency up to 4186Hz, so that will look like
f= [27.5 29.14 30.87 ... 4186.01]
So far, I've understood this much:
f = [];
for i=1:87,
%what goes here
% f = [27.5 * 2^(i/12)]; ?
end
return;
There is no need to do a loop for this in matlab, you can simply do:
f = 27.5 * 2.^((0:87)/12)
The answer:
f =
Columns 1 through 13
27.5 29.135 30.868 32.703 34.648 36.708 38.891 41.203 43.654 46.249 48.999 51.913 55
Columns 14 through 26
58.27 61.735 65.406 69.296 73.416 77.782 82.407 87.307 92.499 97.999 103.83 110 116.54
Columns 27 through 39
123.47 130.81 138.59 146.83 155.56 164.81 174.61 185 196 207.65 220 233.08 246.94
Columns 40 through 52
261.63 277.18 293.66 311.13 329.63 349.23 369.99 392 415.3 440 466.16 493.88 523.25
Columns 53 through 65
554.37 587.33 622.25 659.26 698.46 739.99 783.99 830.61 880 932.33 987.77 1046.5 1108.7
Columns 66 through 78
1174.7 1244.5 1318.5 1396.9 1480 1568 1661.2 1760 1864.7 1975.5 2093 2217.5 2349.3
Columns 79 through 88
2489 2637 2793.8 2960 3136 3322.4 3520 3729.3 3951.1 4186
maxind = 87;
f = zeros(1, maxind); % preallocate, better performance and avoids mlint warnings
for ii=1:maxind
f(ii) = 27.5 * 2^(ii/12);
end
The reason I named the loop variable ii is because i is the name of a builtin function. So it's considered bad practice to use that as a variable name.
Also, in your description you said you want to iterate 88 times, but the above loop only iterates 1 through 87 (both inclusive). If you want to iterate 88 times change maxind to 88.