Here's a small example of what I want. Given the following array:
1 1 2
2 2 1
1 1 1
1 1 6
Sorted (row sum is shown in parenthesis):
1 1 6 (8)
2 2 1 (5)
1 1 2 (4)
1 1 1 (3)
Is there a quick way to achieve this in Matlab?
Since sort returns the indexes in order as well as the sorted matrix, you can use these indices to access the original data -- try this:
% some data
A = [
1 1 2;
2 2 1;
1 1 1;
1 1 6;
];
% compute the row totals
row_totals = sum(A,2);
% sort the row totals (descending order)
[sorted, row_ids] = sort(row_totals, 'descend');
% and display the original data in that order (concatenated with the sums)
disp([A(row_ids,:), row_totals(row_ids)])
>>>
1 1 6 8
2 2 1 5
1 1 2 4
1 1 1 3
The ugliest one-liner I could come up with:
>> subsref( sortrows( [sum(A,2) A], -1 ), struct('type','()','subs',{{':',1+(1:size(A,2))}}) )
ans =
1 1 6
2 2 1
1 1 2
1 1 1
Disclaimer: I don't think anyone should write this kind of code, but it's a nice practice to keep your Matlab's skills sharp.
Just do something very simple like follows
temp = [1 1 2
2 2 1
1 1 1
1 1 6];
rowSums = sum(temp,2);
[~,idx] = sort(rowSums,'descend');
output = [temp(idx,:),rowSums(idx)];
EDIT
Changed the above code to make sure the sum is appended to the last column. I did not notice that this was a requirement when I read the question initially.
I leave it for you to judge if this is uglier than #Shai's:
fliplr(diff([sortrows(fliplr(-cumsum(A,2))) zeros(size(A,1),1) ],1,2))
Let's do some matrix multiplication
>> sortrows([sum(A,2) A], -1)*[zeros(1,size(A,2)); eye(size(A,2))]
returns
ans =
1 1 6
2 2 1
1 1 2
1 1 1
Related
I created a cell array of shape m x 2, each element of which is a matrix of shape d x d.
For example like this:
A = cell(8, 2);
for row = 1:8
for col = 1:2
A{row, col} = rand(3, 3);
end
end
More generally, I can represent A as follows:
where each A_{ij} is a matrix.
Now, I need to randomly pick a matrix from each row of A, because A has m rows in total, so eventually I will pick out m matrices, which we call a combination.
Obviously, since there are only two picks for each row, there are a total of 2^m possible combinations.
My question is, how to get these 2^m combinations quickly?
It can be seen that the above problem is actually finding the Cartesian product of the following sets:
2^m is actually a binary number, so we can use those to create linear indices. You'll get an array containing 1s and 0s, something like [1 1 0 0 1 0 1 0 1], which we can treat as column "indices", using a 0 to indicate the first column and a 1 to indicate the second.
m = size(A, 1);
% Build all binary numbers and create a logical matrix
bin_idx = dec2bin(0:(2^m -1)) == '1';
row = 3; % Loop here over size(bin_idx,1) for all possible permutations
linear_idx = [find(~bin_idx(row,:)) find(bin_idx(row,:))+m];
A{linear_idx} % the combination as specified by the permutation in out(row)
On my R2007b version this runs virtually instant for m = 20.
NB: this will take m * 2^m bytes of memory to store bin_idx. Where that's just 20 MB for m = 20, that's already 30 GB for m = 30, i.e. you'll be running out of memory fairly quickly, and that's for just storing permutations as booleans! If m is large in your case, you can't store all of your possibilities anyway, so I'd just select a random one:
bin_idx = rand(m, 1); % Generate m random numbers
bin_idx(bin_idx > 0.5) = 1; % Set half to 1
bin_idx(bin_idx < 0.5) = 0; % and half to 0
Old, slow answer for large m
perms()1 gives you all possible permutations of a given set. However, it does not take duplicate entries into account, so you'll need to call unique() to get the unique rows.
unique(perms([1,1,2,2]), 'rows')
ans =
1 1 2 2
1 2 1 2
1 2 2 1
2 1 1 2
2 1 2 1
2 2 1 1
The only thing left now is to somehow do this over all possible amounts of 1s and 2s. I suggest using a simple loop:
m = 5;
out = [];
for ii = 1:m
my_tmp = ones(m,1);
my_tmp(ii:end) = 2;
out = [out; unique(perms(my_tmp),'rows')];
end
out = [out; ones(1,m)]; % Tack on the missing all-ones row
out =
2 2 2 2 2
1 2 2 2 2
2 1 2 2 2
2 2 1 2 2
2 2 2 1 2
2 2 2 2 1
1 1 2 2 2
1 2 1 2 2
1 2 2 1 2
1 2 2 2 1
2 1 1 2 2
2 1 2 1 2
2 1 2 2 1
2 2 1 1 2
2 2 1 2 1
2 2 2 1 1
1 1 1 2 2
1 1 2 1 2
1 1 2 2 1
1 2 1 1 2
1 2 1 2 1
1 2 2 1 1
2 1 1 1 2
2 1 1 2 1
2 1 2 1 1
2 2 1 1 1
1 1 1 1 2
1 1 1 2 1
1 1 2 1 1
1 2 1 1 1
2 1 1 1 1
1 1 1 1 1
NB: I've not initialised out, which will be slow especially for large m. Of course out = zeros(2^m, m) will be its final size, but you'll need to juggle the indices within the for loop to account for the changing sizes of the unique permutations.
You can create linear indices from out using find()
linear_idx = [find(out(row,:)==1);find(out(row,:)==2)+size(A,1)];
A{linear_idx} % the combination as specified by the permutation in out(row)
Linear indices are row-major in MATLAB, thus whenever you need the matrix in column 1, simply use its row number and whenever you need the second column, use the row number + size(A,1), i.e. the total number of rows.
Combining everything together:
A = cell(8, 2);
for row = 1:8
for col = 1:2
A{row, col} = rand(3, 3);
end
end
m = size(A,1);
out = [];
for ii = 1:m
my_tmp = ones(m,1);
my_tmp(ii:end) = 2;
out = [out; unique(perms(my_tmp),'rows')];
end
out = [out; ones(1,m)];
row = 3; % Loop here over size(out,1) for all possible permutations
linear_idx = [find(out(row,:)==1).';find(out(row,:)==2).'+m];
A{linear_idx} % the combination as specified by the permutation in out(row)
1 There's a note in the documentation:
perms(v) is practical when length(v) is less than about 10.
I have been trying to write a code where the users enters two numbers in order to get 2 columns. It is very hard to explain by words what I am trying to achieve so here is an example:
If the user inputs a = 1 and b = 1, the following table should be created:
ans =
1 1
If the user inputs a = 2 and b = 2:
ans =
1 1
1 2
2 1
2 2
If the user inputs a = 2 and b = 5:
ans =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
For other values of a and b, the matrix should be constructed according to the above shown sequence.
This can be achieved straight-forward by the use of repelem and repmat:
[repelem((1:a).',b),repmat((1:b).',a,1)]
A more elegant way is using meshgrid and reshape it after:
[A,B] = meshgrid(1:a,1:b);
[A(:),B(:)]
Let's create an anonymous function and test the first approach:
>> fun = #(a,b) [repelem((1:a).',b),repmat((1:b).',a,1)];
>> fun(1,1)
ans =
1 1
>> fun(2,2)
ans =
1 1
1 2
2 1
2 2
>> fun(2,5)
ans =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
Here is an alternative way
for example a = 2 and b = 5
A(1:b*a,1) = reshape(mtimes((1:a).',ones(1,b)).',1,b*a)
A(1:b*a,2) = reshape(mtimes((1:b).',ones(1,a)),1,b*a)
A =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
There is just one logic, in the code below you define a matrix of row size and a and column size b
>> mtimes((1:a).',ones(1,b))
ans =
1 1 1 1 1
2 2 2 2 2
and the next step simply reshapes the matrix column wise for a and row wise for b by taking a transpose
A(1:b*a,1) = reshape(mtimes((1:a).',ones(1,b)).',1,b*a)
A(1:b*a,2) = reshape(mtimes((1:b).',ones(1,a)),1,b*a)
I have a matrix with some zero values I want to erase.
a=[ 1 2 3 0 0; 1 0 1 3 2; 0 1 2 5 0]
>>a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
However, I want to erase only the ones after the last non-zero value of each line.
This means that I want to retain 1 2 3 from the first line, 1 0 1 3 2 from the second and 0 1 2 5 from the third.
I want to then store the remaining values in a vector. In the case of the example this would result in the vector
b=[1 2 3 1 0 1 3 2 0 1 2 5]
The only way I figured out involves a for loop that I would like to avoid:
b=[];
for ii=1:size(a,1)
l=max(find(a(ii,:)));
b=[b a(ii,1:l)];
end
Is there a way to vectorize this code?
There are many possible ways to do this, here is my approach:
arotate = a' %//rotate the matrix a by 90 degrees
b=flipud(arotate) %//flips the matrix up and down
c= flipud(cumsum(b,1)) %//cumulative sum the matrix rows -and then flip it back.
arotate(c==0)=[]
arotate =
1 2 3 1 0 1 3 2 0 1 2 5
=========================EDIT=====================
just realized cumsum can have direction parameter so this should do:
arotate = a'
b = cumsum(arotate,1,'reverse')
arotate(b==0)=[]
This direction parameter was not available on my 2010b version, but should be there for you if you are using 2013a or above.
Here's an approach using bsxfun's masking capability -
M = size(a,2); %// Save size parameter
at = a.'; %// Transpose input array, to be used for masked extraction
%// Index IDs of last non-zero for each row when looking from right side
[~,idx] = max(fliplr(a~=0),[],2);
%// Create a mask of elements that are to be picked up in a
%// transposed version of the input array using BSXFUN's broadcasting
out = at(bsxfun(#le,(1:M)',M+1-idx'))
Sample run (to showcase mask usage) -
>> a
a =
1 2 3 0 0
1 0 1 3 2
0 1 2 5 0
>> M = size(a,2);
>> at = a.';
>> [~,idx] = max(fliplr(a~=0),[],2);
>> bsxfun(#le,(1:M)',M+1-idx') %// mask to be used on transposed version
ans =
1 1 1
1 1 1
1 1 1
0 1 1
0 1 0
>> at(bsxfun(#le,(1:M)',M+1-idx')).'
ans =
1 2 3 1 0 1 3 2 0 1 2 5
I am trying to find the number of the lines where the values of two matrices are not the same
I found only a way to know the indexs on the not same items by:
find(a~=b)
where a is N*N and b is N*N
How can I know the rows numbers of the not same items
ps
looking for nicer way then
dint the find and then having some vector in a loop filling with
ind2sub(size(A), 6)
You can use max on the logical array of such matches or mis-mistaches in this case along a certain dimension, alongwith find.
If you are looking to find unique row IDs for mismatches, do this -
find(max(a~=b,[],2))
For unique column IDs, just change the dimension specifier -
find(max(a~=b,[],1))
Sample run -
>> a
a =
1 2 2 2 1
1 2 1 1 1
2 2 2 2 1
1 1 2 1 1
>> b
b =
1 2 1 1 2
1 2 1 2 1
2 2 2 2 1
1 1 2 2 2
>> a~=b
ans =
0 0 1 1 1
0 0 0 1 0
0 0 0 0 0
0 0 0 1 1
>> find(max(a~=b,[],2)) %// unique row IDs
ans =
1
2
4
>> find(max(a~=b,[],1)) %// unique col IDs
ans =
3 4 5
here I found an easy way if any one will need it
indexs=find(a~=b)
[~,rows]=ind2sub(size(a),indexs)
rows=unique( sort( rows ) )
now rows are only the different rows
NotSame = 0;
for ii = 1:size(a,1)
if a(ii,:) ~= b(ii,:)
NotSame = NotSame+1;
end
end
This checks it row by row and when a row in a is not the same as the row in b this will increase the count of NotSame. Not the fastest way, I'm sure someone can produce a solution using bsxfun, but I'm not an expert in that.
You can also use the double output of find
[row, col] = find(a~=b)
myrows = unique(row);
You can also have the columns where a & b have different values
mycols = unique(col);
So here's my problem i want to count the numbers of same values in a column, this my data:
a b d
2 1 5
1 3 10
1 -2 5
0 5 25
5 0 25
1 1 2
-1 -1 2
i want to count the same values of d (where d = a^2 + b^2), so this is the output i want:
a b d count
2 1 5 2
1 3 10 1
0 5 25 2
1 1 2 2
so as you can see, only positive combinations of a and b displayed. so how can i do that? thanks.
Assuming your data is a matrix, here's an accumarray-based approach. Note this doesn't address the requirement "only positive combinations of a and b displayed".
M = [2 1 5
1 3 10
1 -2 5
0 5 25
5 0 25
1 1 2
-1 -1 2]; %// data
[~, jj, kk] = unique(M(:,3),'stable');
s = accumarray(kk,1);
result = [M(jj,:) s];
Assuming your input data to be stored in a 2D array, this could be one approach -
%// Input
A =[
2 1 5
1 3 10
1 -2 5
0 5 25
5 0 25
1 1 2
-1 -1 2]
[unqcol3,unqidx,allidx] = unique(A(:,3),'stable')
counts = sum(bsxfun(#eq,A(:,3),unqcol3.'),1) %//'
out =[A(unqidx,:) counts(:)]
You can also get the counts with histc -
counts = histc(allidx,1:max(allidx))
Note on positive combinations of a and b: If you are looking to have positive combinations of A and B, you can select only those rows from A that fulfill this requirement and save back into A as a pre-processing step -
A = A(all(A(:,1:2)>=0,2),:)