I have a vector:
v1 = [13 13 13 13 5 5 5 2 2 5 5];
and would like to create another vector, v2, such that if an element in v1 is the same as the previous element, it gets the same ordered (sequential) integer, eg:
v2 = [1 1 1 1 2 2 2 3 3 4 4];
Formalising Luis comment as a (community wiki) answer, you could do
cumsum([1, diff(v1)~=0]);
Which uses cumsum to do the running sum of a true/false array identifying where one element is different to the previous one, i.e. 1 where the value changed, 0 otherwise.
Related
Hi I need to sort a vector and assign a ranking for the corresponding sorting order. I'm using sort function [sortedValue_X , X_Ranked] = sort(X,'descend');
but the problem is it assigns different ranks for the same values (zeros).
i.e. x = [ 13 15 5 5 0 0 0 1 0 3] and I want zeros to take the same last rank which is 6 and fives needs to share the 3rd rank etc..
any suggestions?
The syntax [sortedValues, sortedIndexes] = sort(x, 'descend') does not return rank as you describe it. It returns the indexes of the sorted values. This is really useful if you want to use the sort order from one array to rearrange another array.
As suggested by #user1860611, unique seems to do what you want, using the third output as follows:
x = [ 13 15 5 5 0 0 0 1 0 3];
[~, ~, forwardRank] = unique(x);
%Returns
%forwardRank =
% 5 6 4 4 1 1 1 2 1 3
To get the order you want (decending) you'll need to reverse the order, like this:
reverseRank = max(forwardRank) - forwardRank + 1
%Returns
%reverseRank =
% 2 1 3 3 6 6 6 5 6 4
You may be done at this point. But you may want to sort these into the into an acsending order. This is a reorder of the reverseRank vector which keeping it in sync with the original x vector, which is exactly what the 2nd argument of sort is desined to help with. So we can do something like this:
[xSorted, ixsSort] = sort(x, 'descend'); %Perform a sort on x
reverseRankSorted = reverseRank(ixsSort); %Apply that sort to reverseRank
Which generates:
xSorted = 15 13 5 5 3 1 0 0 0 0
reverseRankSorted = 1 2 3 3 4 5 6 6 6 6
tiedrank.m might be the thing you are looking for.
>> x = round(rand(1,5)*10)
x =
8 7 3 10 0
>> tiedrank(x)
ans =
4 3 2 5 1
I would like to get the cumsum of an vector, but need the first element of the vector to count itself as well. An example:
a = [1 2 3 4 5]
and the result needs to look as follow:
2 3 6 10 15
Presumably you already know about the cumsum function so, have you not tried?:
a = [1 2 3 4 5];
s = cumsum(a);
s(1) = s(1)+a(1)
which returns
s =
2 3 6 10 15
Hi I need to sort a vector and assign a ranking for the corresponding sorting order. I'm using sort function [sortedValue_X , X_Ranked] = sort(X,'descend');
but the problem is it assigns different ranks for the same values (zeros).
i.e. x = [ 13 15 5 5 0 0 0 1 0 3] and I want zeros to take the same last rank which is 6 and fives needs to share the 3rd rank etc..
any suggestions?
The syntax [sortedValues, sortedIndexes] = sort(x, 'descend') does not return rank as you describe it. It returns the indexes of the sorted values. This is really useful if you want to use the sort order from one array to rearrange another array.
As suggested by #user1860611, unique seems to do what you want, using the third output as follows:
x = [ 13 15 5 5 0 0 0 1 0 3];
[~, ~, forwardRank] = unique(x);
%Returns
%forwardRank =
% 5 6 4 4 1 1 1 2 1 3
To get the order you want (decending) you'll need to reverse the order, like this:
reverseRank = max(forwardRank) - forwardRank + 1
%Returns
%reverseRank =
% 2 1 3 3 6 6 6 5 6 4
You may be done at this point. But you may want to sort these into the into an acsending order. This is a reorder of the reverseRank vector which keeping it in sync with the original x vector, which is exactly what the 2nd argument of sort is desined to help with. So we can do something like this:
[xSorted, ixsSort] = sort(x, 'descend'); %Perform a sort on x
reverseRankSorted = reverseRank(ixsSort); %Apply that sort to reverseRank
Which generates:
xSorted = 15 13 5 5 3 1 0 0 0 0
reverseRankSorted = 1 2 3 3 4 5 6 6 6 6
tiedrank.m might be the thing you are looking for.
>> x = round(rand(1,5)*10)
x =
8 7 3 10 0
>> tiedrank(x)
ans =
4 3 2 5 1
I have a matrix in Matlab, A =
1 2 3
4 5 6
7 8 9
10 11 12
Now I want to create a new matrix B derived from A. The new matrix should look like: B =
1 0
0 4
2 0
0 5
3 0
0 6
Is it possible to solve that without using FOR loop?
Easy with some indexing:
A=[1 2 3 ;
4 5 6 ;
7 8 9 ;
10 11 12 ];
B = zeros(2*size(A,2),2);
B(1:2:end,1)=A(1,:); % put first row values in first column of c
B(2:2:end,2)=A(2,:); % put énd row values in 2nd column of c
If you're only working with 4x4 matrices then yes it is.
You want to convert a 4x4 and using row x column convention you can access A's elements one at at time like so A[row][column]
Then you want a 6x2 matrix then you just call it with zeros B = zeros(6,2)
Then alternate down B[row][column] = A[row][column] and you should be able to build it out easily.
I have data in two columns that looks as follows:
A B
1,265848208 3
-0,608043611 0
-0,285735893 0
0,006895134 7
0 7
-0,004526196 7
0,176326617 10
-0,159688071 2
0,22439945 2
-0,991045044 1
0,178022324 1
-0,270967397 4
0,285849994 4
1,881705539 23
1,057184204 10
NaN 10
For all unique values in B I want to extract the corresponding value in column A and move it to a new matrix. I'm looking to then compute the mean of all the corresponding values in A and use as a dependent variable (weighted by no of observations per value in B) in a regression with the common value of B being the independent variable to reduce noise. Any help would on how to do this in Matlab (except running the regression) would be great!
Thanks
Oscar
Here is an efficient solution:
X = [
1.265848208 3
-0.608043611 0
-0.285735893 0
0.006895134 7
0 7
-0.004526196 7
0.176326617 10
-0.159688071 2
0.22439945 2
-0.991045044 1
0.178022324 1
-0.270967397 4
0.285849994 4
1.881705539 23
1.057184204 10
NaN 10
];
%# unique values in B, and their indices
[valB,~,subs] = unique(X(:,2));
%# values of A for each unique number in B (cellarray)
valA = accumarray(subs, X(:,1), [], #(x) {x});
%# mean of each group
meanValA = cellfun(#nanmean, valA)
%# perform regression here...
The result:
%# B values, mean of corresponding values in A, number of A values
>> [valB meanValA cellfun(#numel,valA)]
ans =
0 -0.44689 2
1 -0.40651 2
2 0.032356 2
3 1.2658 1
4 0.0074413 2
7 0.00078965 3
10 0.61676 3
23 1.8817 1