I have 2 questions.
Question 1:
I have a column vector, v, with (say) 1000 rows. I want to delete first (say) n rows (for example n=300 to remove the first 300 rows of v).
Question 2:
I have an array of indices. How can it be used to get rows of another array whose index values are in first array?
I = [ 1 2 4 5] %// array of indices
Q = [ 45 22 66 87 99 10 ] %// input array
Desired Output:
O = [45 22 87 99] %// output array
For you first question:
N=300;
v=rand(1000,1); %// sample data
v(1:N)=[];
And the second:
O=Q(I)
Related
Hi i have a 289x2 array that i want to sort in MatLab. I want to sort the first column into numerical ascending order. However I want to keep the second column entry that is associated with it. Best way to explain is through an example.
x = 76 1
36 2
45 3
Now I want to sort x so that it returns an array that looks like:
x = 36 2
45 3
76 1
So the first column has been sorted into numerical order but has retained its second column value. So far I have tried sort(x,1). This sorts the first column as i want but does not keep the pairing. This returns x as:
x = 36 1
45 2
76 3
Any help would be great. Cheers!!
This is exactly what sortrows does.
x=sortrows(x); % or x=sortrows(x,1);
or if you want to use sort then get the sorted indexes first and then arrange the rows accordingly like this:
[~, idx] = sort(x); %Finding the sorted indexes
x = x(idx(:,1),:) ; %Arranging according to the indexes of the first column
Output for both approaches:
x =
36 2
45 3
76 1
I have a vector from 1 to 40 and want to shuffle it in such a way that each block of four integers (ten blocks in total) are shuffled only with themselves.
For example: 3 4 2 1 | 7 6 5 8 | 9 11 10 12 | ...
My original idea was to append ten permutation vectors to eachother and then add a 1 to 40 vector to the big permutation vector, but it didn't work at all as expected and was logically wrong.
Has anyone an idea how to solve this?
data = 10:10:120; % input: values to be permuted
group_size = 4; % input: group size
D = reshape(data, group_size, []); % step 1
[~, ind] = sort(rand(size(D)), 1); % step 2
result = D(bsxfun(#plus, ind, (0:size(D,2)-1)*group_size)); % step 3
result = result(:).'; % step 4
Example result:
result =
20 10 30 40 60 50 70 80 110 100 120 90
How it works
Reshape the data vector into a matrix D, such that each group is a column. This is done with reshape.
Generate a matrix, ind, where each column contains the indices of a permutation of the corresponding column of D. This is done generating independent, uniform random values (rand), sorting each column, and getting the indices of the sorting (second output of sort).
Apply ind as column indices into D. This requires converting to linear indices, which can be done with bsxfun (or with sub2ind, but that's usually slower).
Reshape back into a vector.
You can use A = A(randperm(length(A))) to shuffle an array.
Example in Octave:
for i = 1:4:40
v(i:i+3) = v(i:i+3)(randperm(4));
end
I have an array :
Z = [1 24 3 4 52 66 77 8 21 100 101 120 155];
I have another array:
deletevaluesatindex=[1 3; 6 7;10 12]
I want to delete the values in array Z at indices (1 to 3, 6 to 7, 10 to 12) represented in the array deletevaluesatindex
So the result of Z is:
Z=[4 52 8 21 155];
I tried to use the expression below, but it does not work:
X([deletevaluesatindex])=[]
Another solution using bsxfun and cumsum:
%// create index matrix
idx = bsxfun(#plus , deletevaluesatindex.', [0; 1])
%// create mask
mask = zeros(numel(Z),1);
mask(idx(:)) = (-1).^(0:numel(idx)-1)
%// extract unmasked elements
out = Z(~cumsum(mask))
out = 4 52 8 21 155
This will do it:
rdvi= size(deletevaluesatindex,1); %finding rows of 'deletevaluesatindex'
temp = cell(1,rdvi); %Pre-allocation
for i=1:rdvi
%making a cell array of elements to be removed
temp(i)={deletevaluesatindex(i,1):deletevaluesatindex(i,2)};
end
temp = cell2mat(temp); %Now temp array contains the elements to be removed
Z(temp)=[] % Removing the elements
If you control how deletevaluesatindex is generated, you can instead directly generate the ranges using MATLAB's colon operator and concatenate them together using
deletevaluesatindex=[1:3 6:7 10:12]
then use the expression you suggested
Z([deletevaluesatindex])=[]
If you have to use deletevaluesatindex as it is given, you can generate the concatenated range using a loop or something like this
lo = deletevaluseatindex(:,1)
up = deletevaluseatindex(:,2)
x = cumsum(accumarray(cumsum([1;up(:)-lo(:)+1]),[lo(:);0]-[0;up(:)]-1)+1);
deleteat = x(1:end-1)
Edit: as in comments noted this solution only works in GNU Octave
with bsxfun this is possible:
Z=[1 24 3 4 52 66 77 8 21 100 101 120 155];
deletevaluesatindex = [1 3; 6 7;10 12];
idx = 1:size(deletevaluesatindex ,1);
idx_rm=bsxfun(#(A,B) (A(B):deletevaluesatindex (B,2))',deletevaluesatindex (:,1),idx);
Z(idx_rm(idx_rm ~= 0))=[]
I have 50 4x4 matrices. I want to delete a matrix at specific index e.g. index 2, 12 and 34. I have tried this but I am not getting the desired result:
for i = 1:50
index = true(length(AB));
index([2,12,34]) = false;
AB(:,:,i) = AB(:,:,index);
end
You are apparently using a 3D matrix AB of size 4 x 4 x 50. Removing elements can be done like this:
index = [2 12 34];
AB(:,:,index) = [];
note that this will change your indices, since indices 2, 12 and 34 have been removed.
To set the element to 0 however:
index = [2 12 34];
AB(:,:,index) = 0;
I have a cell array like this: [...
0
129
8...2...3...4
6...4
0
I just want to find and replace specific values, but I can't use the ordinary function because the cells are different lengths. I need to replace many specific values at the same time and there is no general function about how values are replaced. However, sometimes several input values should be replaced by the same output.
so I want to say
for values 1:129
'if 0, then 9'
'elseif 1 then 50'
'elseif 2 or 3 or 4 then 61'
etc...up to 129
where these rules are applied to the entire array.
I've tried to work it out myself, but still getting nowhere. Please help!
Since your values appear to span the range 0 to 129, one solution is to add one to these values (so they span the range 1 to 130) and use them as indices into a vector of replacement values. Then you can apply this operation to each cell using the function CELLFUN. For example:
>> C = {0, 129, [8 2 3 4], [6 4], 0}; %# The sample cell array you give above
>> replacement = [9 50 61 61 61 100.*ones(1,125)]; %# A 1-by-130 array of
%# replacement values (I
%# added 125 dummy values)
>> C = cellfun(#(v) {replacement(v+1)},C); %# Perform the replacement
>> C{:} %# Display the contents of C
ans =
9
ans =
100
ans =
100 61 61 61
ans =
100 61
ans =
9