I have the matrix below:
a = [1 2 1 4;
3 4 9 16;
0 0 -2 -4;
0 0 -6 -8]
How can I arbitrary remove any given rows or columns? for example second row and third column of the above matrix?
Just assign the column or line to the empty matrix:
a(2,:) = [];
a(:,3) = [];
Note : I compare the other solution to mine, following the link put inside. On a big array (created as rand(1e4)) and on 10 runs where I delete 2 columns and 2 rows, the average times are 0.932ms for the empty-matrix assignment, and 0.905ms for the kept-row (or -column) assignment. So the gap seen there is not as big as 1.5x mentioned in the link. Always perform a little benchmark first :) !
Edit Fastest solution is to create the index mask for rows and columns, and reassign your array with these masks. Ex:
a = rand(10000);
kr = true(size(a,1),1);
kr([72,6144]) = false; % some rows to delete
kc = true(1,size(a,2));
kc([1894,4512]) = false; % some columns to delete
a = a(kr,kc);
On this test, it's clearly twice faster than performing the suppression on rows and columns separately.
A slightly more efficient way (although possibly more complicated to set up) is to reassign all the rows you want to keep (when compared with setting the rows you want to delete to the empty matrix). So for example if you want to delete rows 5 and 7 from a matrix you can either do
A = A([1:4, 6, 8:end],:)
or
A = A(setdiff(1:size(A,1), [5,7] ),:)
but the best method is likely to use logical indexing (which is often a natural step in Matlab workflows anyway):
idx = true(size(A,1),1);
idx([5,7]) = false;
A = A(idx,:)
Related
I have a 2-d matrix that I would like to change the first 6 rows and last 6 rows and their corresponding columns to 0.
I have this right now but it is only doing it by a single row.
at_1(4, :) = zeros(1, 141); % set first 6 rows to zero val
This should make the trick
at_1(1:6, :) = 0;
at_1(end-5:end, :) = 0;
You are basically saying to use the values from 1 to 6 in all columns by using (1:6, :) and the second statement is the same but using the end instead.
I have a Zeros-matrix P of size (16,16); I need to replace first 4 rows in columns 1:4:end by 1, then next 4 rows in columns 2:4:end by one, and so on.
I have done this, but I noticed that in second rows, the ones are repeated in columns 1 and 2, however what I want is to be only in columns 1:4:end.
Here is the code I made:
P = zeros(16,16);
for i = 1 : 4
P(i:i*4,i:4:end)=1;
end
Could you help to solve that issue. ?
Shift the selected row by (ii-1)*4
P = zeros(16,16);
for ii = 1:4
P([1:4]+(ii-1)*4,ii:4:end)=1;
end
BTW: Avoid to use i as a variable since it's the matlab in-built function that return the imaginary unit.
I have this matrix:
A=[2,2,4;1,2,3;4,5,6;4,5,6;4,5,6;7,8,9]
How can I use a for loop to delete a row that has its second column element the same as the previous row second column element in matlab? The objective is to arrive at:
A=[2,2,4;4,5,6;7,8,9]
No loop needed!
What you can do here is create a logical vector with true in the places where there is a difference between the second columns, and false where the value is equal:
This can be achieved using diff like this: diff(A(:,2))~=0. Now, you need to include the first row too, so add a true in the start of this vector: [true; diff(A(:,2))~=0)]. Use this vector to choose which rows you want, and use : to make sure you get all the columns:
A=[2,2,4;1,2,3;4,5,6;4,5,6;4,5,6;7,8,9]
B = A([true; diff(A(:,2))~=0],:)
B =
2 2 4
4 5 6
7 8 9
I think this sample code can do the task:
A=[2,2,4;1,2,3;4,5,6;4,5,6;4,5,6;7,8,9]
% First row will always be the same of the A matrix
res_mat(1,:) = A(1,:);
row = 2;
for i = 2 : size(A,1)
if A(i,2) ~= A(i-1,2)
res_mat(row,:) = A(i,:);
row = row + 1;
end
end
res_mat
HTH ;)
I need to find all possible combinations of numbers 1:8 such that sum of all elements is equal to 8
The combinations need to be arranged in an ascending order.
Eg
1 7
2 2 4
1 3 5
1 2 2 3
1 1 1 1 1 1 1 1
A number can repeat itself. But a combination must not..
i.e 1 2 2 3 and 2 1 2 3
I need the the solution in ascending order So there will be only one possibility of every combination
I tried a few codes online suggested on Find vector elements that sum up to specific number in MATLAB
VEC = [1:8];
NUM = 8;
n = length(VEC);
finans = zeros(2^n-1,NUM);
for i = 1:(2^n - 1)
ndx = dec2bin(i,n) == '1';
if sum(VEC(ndx)) == NUM
l = length(VEC(ndx));
VEC(ndx)
end
end
but they dont include the possibilities where the numbers repeat.
I found a better approach through recursion and it's more elegant (I like elegant) and faster than my previous attempt (0.00399705213 seconds on my computer).
EDIT: You will need my custom function stretchmat.m that stretches a vector to fit the size of another matrix. Kinda like repmat but stretching the first parameter (see help for details). Very useful!
script.m
% Define funciton to prepend a cell x with a variable i
cellprepend = #(x,i) {[i x]};
% Execute and time function
tic;
a = allcomb(cellprepend,1,8); % Solution in a
toc;
allcomb.m
function a = allcomb( cellprepend, m, n )
% Add entire block as a combination
a{1} = n;
% Exit recursion if block size 1
if n == 1
return;
end
% Recurse cutting blocks at different segments
for i = m:n/2
b = allcomb(cellprepend,i,n-i);
a = [a cellfun( cellprepend, b, num2cell( stretchmat( i, b ) ) )];
end
end
So the idea is simple, for solutions that add to 8 is exhaustive. If you look for only valid answers, you can do a depth first search by breaking up the problem into 2 blocks. This can be written recursively as I did above and is kinda similar to Merge Sort. The allcomb call takes the block size (n) and finds all the ways of breaking it up into smaller pieces.
We want non-zero pieces so we loop it from 1:n-1. It then prepends the first block to all the combinations of the second block. By only doing all comb on one of the blocks, we can ensure that all solutions are unique.
As for the sorting, I'm not quite sure what you mean by ascending. From what I see, you appear to be sorting from the last number in ascending order. Can you confirm? Any sort can be appended to the end of script.m.
EDIT 2/3 Notes
For the permutatively unique case, the code can be found here
Thanks to #Simon for helping me QA the code multiple times
EDIT: Look at my second more efficient answer!
The Naive approach! Where the cartprod.m function can be found here.
% Create all permutations
p(1:8) = {0:8};
M = fliplr( cartprod( p{:} ) );
% Check sums
r = sum( M, 2 ) == 8;
M = M(sum( M, 2 ) == 8,:); % Solution here
There are definitely more efficient solutions than this but if you just need a quick and dirty solution for small permutations, this will work. Please note that this made Matlab take 3.5 GB of RAM to temporarily store the permutations.
First save all combinations with repetitions in a cell array. In order to do that, just use nmultichoosek.
v = 1 : 8;
combs = cell(length(v),0);
for i = v
combs{i} = nmultichoosek(v,i);
end
In this way, each element of combs contains a matrix where each row is a combination. For instance, the i-th row of combs{4} is a combination of four numbers.
Now you need to check the sum. In order to do that to all the combinations, use cellfun
sums = cellfun(#(x)sum(x,2),combs,'UniformOutput',false);
sums contains the vectors with the sum of all combinations. For
instance, sums{4} has the sum of the number in combination combs{4}.
The next step is check for the fixed sum.
fixed_sum = 10;
indices = cellfun(#(x)x==fixed_sum,sums,'UniformOutput',false);
indices contains arrays of logical values, telling if the combination satisfies the fixed sum. For instance, indices{4}(1) tells you if the first combination with 4 numbers sums to fixed_sum.
Finally, retrieve all valid combinations in a new cell array, sorting them at the same time.
valid_combs = cell(length(v),0);
for i = v
idx = indices{i};
c = combs{i};
valid_combs{i} = sortrows(c(idx,:));
end
valid_combs is a cell similar to combs, but with only combinations that sum up to your desired value, and sorted by the number of numbers used: valid_combs{1} has all valid combinations with 1 number, valid_combs{2} with 2 numbers, and so on. Also, thanks to sortrows, combinations with the same amount of numbers are also sorted. For instance, if fixed_sum = 10 then valid_combs{8} is
1 1 1 1 1 1 1 3
1 1 1 1 1 1 2 2
This code is quite efficient, on my very old laptop I am able to run it in 0.016947 seconds.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1