I currently have a 3 by 3 matrix "m":
1 2 3
4 5 6
7 8 9
I would like to add a row to matrix 'm' to have a resultant matrix of:
1 2 3
4 5 6
7 8 9
10 11 12
A matrix in q is just a list of lists where inner lists represent rows.
m: ((1 2 3);(4 5 6);(7 8 9))
In order to add one more row all you have to do is add one more inner list to it:
m: m,enlist 10 11 12
enlist is important here, without it you'll end up with this:
q)((1 2 3);(4 5 6);(7 8 9)),10 11 12
1 2 3
4 5 6
7 8 9
10
11
12
I agree; using 0N!x to view the structure is very useful.
To achieve what you want then you can simply do;
q)show m:3 cut 1+til 9 /create matrix
1 2 3
4 5 6
7 8 9
q)show m,:10 11 12 /join new 'row'
1 2 3
4 5 6
7 8 9
10 11 12
q)
Related
Consider the matrix:
1 2 3
4 5 6
7 8 9
I'd like to take the middle column, assign it to a variable, and replace the middle row with it, giving me
1 2 3
2 5 8
7 8 9
I'm extracting the middle column using
a:m[;enlist1]
which returns
2
5
8
How do I replace the middle row with a? Is a flip necessary?
Thanks.
If you want to update the matrix in place you can use
q)show m:(3;3)#1+til 10
1 2 3
4 5 6
7 8 9
q)a:m[;1]
q)m[1]:a
q)show m
1 2 3
2 5 8
7 8 9
q)
cutting out "a" all you need is:
m[1]:m[;1]
You can use dot amend -
q)show m:(3;3)#1+til 10
1 2 3
4 5 6
7 8 9
q)show a:m[;1]
2 5 8
q).[m;(1;::);:;a]
1 2 3
2 5 8
7 8 9
Can see documentation here:
http://code.kx.com/wiki/Reference/DotSymbol
http://code.kx.com/wiki/JB:QforMortals2/functions#Functional_Forms_of_Amend
Making it slightly more generic where you can define the operation, row, and column
q)m:3 cut 1+til 9
1 2 3
4 5 6
7 8 9
Assigning the middle column to middle row :
q){[ m;o;i1;i2] .[m;enlist i1;o; flip[m] i2 ] }[m;:;1;1]
1 2 3
2 5 8
7 8 9
Adding the middle column to middle row by passing o as +
q){[ m;o;i1;i2] .[m;enlist i1;o; flip[m] i2 ] }[m;+;1;1]
1 2 3
6 10 14
7 8 9
Given two parameters:
n %number of repetitions per value
k %max value to repeat
I would like to create a vector of size n*k, which is a concatenation of k vectors of size n, such that the i'th vector contains the value i at each coordinate.
Example:
n = 5;
k = 9;
Desired result:
[1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,9];
Is there an elegant way to achieve this?
Thanks!
quite a few ways to do it:
method 1:
A=1:k
repelem(A',n,1)'
method 2:
A=1:k
kron(A', ones(n,1))'
method 3:
A=1:k
B=repmat(A, n, 1)
B(:)'
method 4:
A=1:k
B=ones(n,1)*A
B(:)'
Here is an alternative method
A = reshape(mtimes((1:k).',ones(1,n)).',1,n*k)
A =
Columns 1 through 22
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5
Columns 23 through 44
5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9
Column 45
9
It multiplies each element by ones n times
>> mtimes((1:k).',ones(1,5)).'
ans =
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
and then reshapes the whole matrix to one vector
I have a matrix A ,and vector x as following (left side)
where S0, H0,...is row number of each block. I want to exchange these blocks such that S0 and S1; H0 and H1 are near together as right side. This is my code
S0=3;
H0=2;
N0=2;
S1=4;
H1=5;
N1=4;
Cols=5;
Rows=S0+H0+N0+S1+H1+N1;
A=randi(10,[ Rows Cols]);
x=randi(10,[Rows 1]);
%% Exchange two block
temp=A(S0+H0+1:S0+H0+N0,1:end);
A(S0+H0+1:S0+H0+H1,1:end)=A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end);
A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end)=temp;
%% How exchange x
The above code is not work. How can I fixed it in MATLAB? Thank in advance.
One approach with mat2cell and cell2mat -
grps = [S0,H0,N0,S1,H1,N1]
new_pattern = [1 4 2 5 3 6]
celldata_roworder = mat2cell((1:size(A,1))',grps); %//'
newx = cell2mat(celldata_roworder(new_pattern)).'; %//'
newA = A(newx,:)
Sample run -
Input :
A =
6 8 9 8 7
4 8 8 3 4
3 8 2 1 10
5 2 6 8 3
5 7 4 7 7
4 5 6 8 7
6 3 4 7 4
8 1 5 5 2
5 9 2 4 1
5 2 3 9 5
2 2 1 4 2
1 7 10 9 8
3 9 7 8 4
4 6 10 9 9
7 8 2 6 8
10 2 10 7 6
10 10 8 10 2
5 6 6 5 10
3 7 5 1 3
8 1 3 9 10
grps =
3 2 2 4 5 4
new_pattern =
1 4 2 5 3 6
Output:
newx =
1 2 3 8 9 10 11 4 5 12 ...
13 14 15 16 6 7 17 18 19 20
newA =
3 3 2 5 8
4 3 3 7 7
1 5 2 8 1
4 6 4 1 4
7 1 5 8 8
4 9 10 10 8
7 10 10 4 3
7 3 1 6 9
2 9 2 6 10
1 1 7 10 3
10 10 10 4 7
9 1 8 9 5
8 7 4 5 7
9 8 7 5 3
1 10 7 6 8
8 1 10 6 1
4 6 3 3 2
7 9 3 2 9
6 9 7 4 8
6 7 6 8 10
I assume you are using a 2-dimensional matrix with Row rows and Cols columns.
You can use the colon : as a second index to address a full row, e.g. for the third row:
A(3, :)
(equal to A(3, 1:end) but little bit clearer).
So you could split your matrix into lines and re-arrange them like this (putting back together the lines to a two-dimensional matrix):
A = [ A(3:4, :); A(1:2, :); A(5:end, :) ]
This moves rows 3 and 4 at the beginning, then old lines 1 and 2 and then all the rest. Does this help you?
Hint: you can use eye for experimenting.
I have a vector V periodic an I want to write a program which associates to each period the set of different elements in that period and gives its cardinal.
For example:
For the vector v=(2 3 7 2 7 3 2 3 7 2 7 3) and cardinal 6, give me only the vector P=(2 3 7 2 7 3).
For The vector v=(2 3 7 5 8 6 10 11 10 6 8 5 7 3 2 3 7 5 8 6) and the cardinal 14, give me P=(2 3 7 5 8 6 10 11 10 6 8 5 7 3).
If I understood you correctly you have to use build-in function seqperiod(v).
In your case, for example:
v=[2 3 7 5 8 6 10 11 10 6 8 5 7 3 2 3 7 5 8 6];
>> seqperiod(v)
ans =
14
Interesting moment: in your second example there aren't full repetition. So we can't really say is it periodic... But seqperiod still works and returns 14 as you wish.
go further you can use it in this way:
[p, num] = seqperiod(v);
p = 14
num = 1.4286
num - is a number of repetitions.
Ok. Now you say you need not only the cardinal, but the vector. So you can do it easily:
result = v(1:p);
Hope it helps!
Anybody know a fast way to produce a matrix consisting of a linspace for each row? For example, the sort of pattern I'm looking for in this matrix is:
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
...
1 2 3 4 5 6 7 8 9 10
Anyone know any fast tricks to produce this WITHOUT using a for loop?
I just figured this out, so just in case anyone else was troubled by this, we can achieve this exact pattern by:
a=linspace(1,10,10);
b=ones(3,1)*a;
This will give:
>> a = 1 2 3 4 5 6 7 8 9 10
>> b = 1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You need to use repmat.
Example:
>> B = repmat(1:10,[3 1])
B =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You can vary the value of 3 there. You can change it accordingly.
Another shortcut I can recommend is similar to repmat, but you specify a base array first of a = 1:10;. Once you do this, you specify a series of 1s in the first dimension when indexing which should produce a matrix of the same vectors with many rows as you want, where each row consists of the base array a. As such:
%// number of times to replicate
n = 4;
a = 1:10;
a = a(ones(1,n),:);
Result:
a =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Insert this command: transpose(ndgrid(1:10,1:n));, where n is the number of rows desired in the result.
You can consider these solutions:
With basic matrix indexing (taken from here)
b=a([1:size(a,1)]' * ones(1,NumToReplicate), :) %row-wise replication
b=a(:, ones(NumToReplicate, 1)) %column-wise replication
With bsxfun:
bsxfun(#times,a,(ones(1,NumToReplicate))') %row-wise replication
bsxfun(#times,a',(ones(1,NumToReplicate))) %column-wise replication
You are welcome to benchmark above two solutions with repmat.