Edit for clarity:
I have two matrices, p.valor 2x1000 and p.clase 1x1000. p.valor consists of random numbers spanning from -6 to 6. p.clase contains, in order, 200 1:s, 200 2:s and 600 3:s. What I wan´t to do is
Print p.valor using a diferent color/prompt for each clase determined in p.clase, as in following figure.
I first wrote this, in order to find out which locations in p.valor represented where the 1,2 respective 3 where in p.clase
%identify the locations of all 1,2 respective 3 in p.clase
f1=find(p.clase==1);
f2=find(p.clase==2);
f3=find(p.clase==3);
%define vectors in p.valor representing the locations of 1,2,3 in p.clase
x1=p.valor(f1);
x2=p.valor(f2);
x3=p.valor(f3);
There is 200 ones (1) in p.valor, thus, is x1=(1:200). The problem is that each number one(1) (and, respectively 2 and 3) represents TWO elements in p.valor, since p.valor has 2 rows. So even though p.clase and thus x1 now only have one row, I need to include the elements in the same colums as all locations in f1.
So the different alternatives I have tried have not yet been succesfull. Examples:
plot(x1(:,1), x1(:,2),'ro')
hold on
plot(x2(:,1),x2(:,2),'k.')
hold on
plot(x3(:,1),x3(:,2),'b+')
and
y1=p.valor(201:400);
y2=p.valor(601:800);
y3=p.valor(1401:2000);
scatter(x1,y1,'k+')
hold on
scatter(x2,y1,'b.')
hold on
scatter(x3,y1,'ro')
and
y1=p.valor(201:400);
y2=p.valor(601:800);
y3=p.valor(1401:2000);
plot(x1,y1,'k+')
hold on
plot(x2,y2,'b.')
hold on
plot(x3,y3,'ro')
My figures have the axisies right, but the plotted values does not match the correct figure provided (see top of the question).
Ergo, my question is: how do I include tha values on the second row in p.valor in my plotted figure?
I hope this is clearer!
Values from both rows simultaneously can be accessed using this syntax:
X=p.value(:,findX)
In this case, resulting X matrix will be a matrix having 2 rows and length(findX) columns.
M = magic(5)
M =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
M2 = M(1:2, :)
M2 =
17 24 1 8 15
23 5 7 14 16
Matlab uses column major indexing. So to get to the next row, you actually just have to add 1. Adding 2 to an index on M2 here gets you to the next column, or adding 5 to an index on M
e.g. M2(3) is 24. To get to the next row you just add one i.e. M2(4) returns 5.To get to the next column add the number of rows so M2(2 + 2) gets you 1. If you add the number of columns like you suggested you just get gibberish.
So your method is very wrong. Freude's method is 100% correct, it's much easier to use subscript indexing than linear indexing for this. But I just wanted to explain why what you were trying doesn't work in Matlab. (aside from the fact that X=p.value(findX findX+1000) gives you a syntax error, I assume you meant X=p.value([findX findX+1000]))
Related
In something which made me spent several hours, I have found an inconsistency in how Matlab deals with dimensions. I somebody can explain to me OR indicate me how to report this to Matlab, please enlight me.
For size, ones,zeros,mean, std, and most every other old and common commands existing inside Matlab, the dimension arrangement is like the classical one and like the intended standard (as per the size of every dimension), the first dimension is along the column vector, the second dimension is along the row vector, and the following are the non graphical following indexes.
>x(:,:,1)=[1 2 3 4;5 6 7 8];
>x(:,:,2)=[9 10 11 12;13 14 15 16];
>m=mean(x,1)
m(:,:,1) = 3 4 5 6
m(:,:,2) = 11 12 13 14
m=mean(x,2)
m(:,:,1) =
2.5000
6.5000
m(:,:,2) =
10.5000
14.5000
m=mean(x,3)
m = 5 6 7 8
9 10 11 12
d=size(m)
d = 2 4 2
However, for graphical commands like stream3,streamline and others relying on the meshgrid output format, the dimensions 1 and 2 are swapped!: the first dimension is the row vector and the second dimension is the column vector, and the following (third) is the non graphical index.
> [x,y]=meshgrid(1:2,1:3)
x = 1 2
1 2
1 2
y = 1 1
2 2
3 3
Then, for stream3 to operate with classically arranged matrices, we should use permute(XXX,[2 1 3]) at every 3D argument:
xyz=stream3(permute(x,[2 1 3]),permute(y,[2 1 3]),permute(z,[2 1 3])...
,permute(u,[2 1 3]),permute(v,[2 1 3]),permute(w,[2 1 3])...
,xs,ys,zs);
If anybody can explain why this happens, and can indicate to me why this is not a bug, welcome.
This behavior is not a bug because it is clearly documented as the intended behavior: https://www.mathworks.com/help/matlab/ref/meshgrid.html. Specifically:
[X,Y,Z]= meshgrid(x,y,z) returns 3-D grid coordinates defined by the vectors x, y, and z. The grid represented by X, Y, and Z has size length(y)-by-length(x)-by-length(z).
Without speaking to the original authors, the exact motivation may be a bit obscure, but I suspect it has to do with the fact that the y-axis is generally associated with the rows of an image, while x is the columns.
Columns are either "j" or "x" in the documentation, rows are either "i" or "y".
Some functions deal with spatial coordinates. The documentation will refer to "x, y, z". These functions will thus take column values before row values as input arguments.
Some functions deal with array indices. The documentation will refer to "i, j" (or sometimes "i1, i2, i3, ..., in", or using specific names instead of "i" before the dimension number). These functions will thus take row values before column values as input arguments.
Yes, this can be confusing. But if you pay attention to the names of the variables in the documentation, you will quickly figure out the right order.
With meshgrid in particular, if the "x, y, ..." order of arguments is confusing, use ndgrid instead, which takes arguments in array indexing order.
I have a matrix A and a vector b. I don't know their sizes, the size varies because it is the output of another function. What I want to do is filter A by a column (let's say jth column) which has at least one value that is in b.
How do I do this without measuring the size of b and concatenating every filtered result. Right now, the code is like this (assume j is a given value)
bsize=size(b,1);
for i=1:bsize
if i==1
a=A(A(:,j)==b(i),:);
else
a=[a; A(A(:,j)==b(i),:)];
end
end
I want to code a faster solution.
I am adding a numerical example just to make it clear. Let's say
A=[2 4
7 14
11 13
15 14]
and b=[4 14]
What I'm trying to do is filter to obtain the A matrix whose values are 4 and 14 in the second column, the elements of b to obtain the following output.
A=[2 4
7 14
15 14]
In my data A has more than 12000 rows and b has more than 100 elements. It doesn't always have to be the second column, sometimes the column index changes but that's not the problem now.
Use the ismember function to create a logical index based on column j=2 of A and vector b, and use that index into the rows of A:
output = A(ismember(A(:,j), b), :);
I am trying to achieve multiple matrices that will cover the full set of numbers. For example say I want to generate 5 matrices of length 10 that cover all the numbers from 1-20.
So matrix one will contain half the numbers say
m1 = [1 2 3 4 5 6 7 8 9 10];
while matrix two contains
m2 = [11 12 13 14 15 16 17 18 19 20];
Although this satisfies my condition with only two matrices not 5, I preferably need to generate all matrices randomly. Other than randomly generating the matrices and checking all values are generated is there a more efficient way to do this?
You can do it like that:
>> l=[1:20,randi(20,1,30)];
>> vec=l(randperm(length(l)));
>> v=reshape(vec,5,10);
The first line generates an array of 50 numbers from 1 to 20. It guarantees that each such number appears at least once. The second line randomizes the order of the numbers. The third line reshapes the vector into an array of arrays (that is, a 2D matrix, where each row is one of the arrays).
all.
I have a 15 element array = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15];.
I was wondering if there was a command such that it would step through iterations of the array without repeating itself. In other words, since there is a chance that randperm() will create the same matrix twice, I want to step through each permutation only once and perform a calculation.
I concede that there are factorial(15) permutations, but for my purposes, these two vectors (and similar) are identical and don't need to be counted twice:
[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15]
[15 14 13 12 11 10 9 8 7 6 5 4 3 2 1]
Thus, is there any way to step through this?
Thanks.
I think what you are looking for is perms. randperm returns a single random permutation, you want all the permutations.
So use
my_permuations = perms([1:15]);
If forward-backward is the same as backward-foward then you can use the top half of the list only...
my_permutation_to_use = my_permutations(1:length(my_permutations)/2, :);
You may compare all permutations, but this would require to store all past permutations. Instead a local decision is better. I recommend this simple rule:
A permutation is valid, if the first element is smaller than the last element.
A permutation is redundant, if the first element is larger than the last element.
For small sizes, this could simply be done with this code:
%generate all permutations
x=perms(1:10)
%select only the valid lines, remove all redundant lines
x(x(:,1)<x(:,end),:)
Remains the problem, that generating x for 1:15 breaks all memory limits and would require about 100h.
I have two datasets, the original have all the labels and description of each variable, but the second is a reduced version of this dataset, used for specifics experiments, but don't have any of the information about the variables, contained in the original. So, I'm trying to match both datasets.
My question here is how can I find if a row from the original dataset is present in the new dataset, if a slight data reduction have been performed in both matrix dimensions?
Being more specific, the original dataset is a 24481 x 117 matrix and the new one is a 24188 x 97 matrix. However, the problem here is that I have no information of which rows or columns were or were not included in the new dataset
what you can do is zero pad the matrix with less number of elements so that it matches the size of the original data. then use
find(A==B)
A and B are the matrices
Using intersect function worked for me. Since a data reduction have been performed in both dimensions, first I look for the intersection of the first two columns vectors in the matrices (assuming that at least the columns order have been preserved in the reduction).
>> M = magic(5)
M =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
>> X = M([2,3,5], [1,2,4,5])
X =
23 5 14 16
4 6 20 22
11 18 2 9
>> [c,xi, mi]=intersect(X(:,1),M(:,1))
mi is the column index vector of all rows from the original matrix M present in the reduced matrix X.
Doing the same for the two first rows in the matrices gave me a row index vector for all columns selected from the original matrix M.
>> [c,xi, mi]=intersect(X(1,:),M(1,:))
This solution has a drawback is that when the first row or column of the original matrix was not selected in the new set, then there you go moving the index of the compared vector from the original matrix, luckily not too much ;).
>> [c,xi, mi]=intersect(X(1,:),M(2,:))