how to count a sequence(pattern) of numbers - tsql

I have table like this
ID
MW1
1
0
2
0
3
0
4
1
5
1
6
1
7
0
9
1
10
0
11
0
My output should look like this
ID
MW1
CountSequenz
1
0
1
2
0
1
3
0
1
4
1
2
5
1
2
6
1
2
7
0
3
9
1
4
10
0
5
11
0
5


The window functions would be a nice fit here.
Select ID
,MW1
,Seq = sum(Flg) over (order by ID)
From (
Select *
,Flg =case when lag(MW1,1) over (order by ID)=MW1 then 0 else 1 end
From YourTable
) A
Order by ID
Results

Related

Finding the non-intersecting rows in a matrix

I want to find the non-intersecting rows in a large matrix. As an example:
A=[1 5 3; 3 4 5; 7 9 10;4 5 6;11 2 8; 3 5 10]
In this matrix, the non-intersecting rows are: [1 5 3], [11 2 8] and [7 9 10]. How can I program this in Matlab in a fast way?

If I may bsxfun -
M = squeeze(any(bsxfun(#eq,A,permute(unique(A),[3 2 1])),2))
[~,row_idx] = max(M,[],1)
out = A(sum(M,2).' == histc(row_idx,1:size(A,1)),:)
Sample step-by-step run -
A =
1 5 3
3 4 5
7 9 10
4 5 6
11 2 8
3 5 10
M =
1 0 1 0 1 0 0 0 0 0 0
0 0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 1 0 1 1 0
0 0 0 1 1 1 0 0 0 0 0
0 1 0 0 0 0 0 1 0 0 1
0 0 1 0 1 0 0 0 0 1 0
row_idx =
1 5 1 2 1 4 3 5 3 3 5
out =
1 5 3
7 9 10
11 2 8

You can look for rows that adding them to union of previous rows increases the number of elements in the union by the number of columns (i.e. all elements in that row are new):
B = [];
C = zeros(1,size(A,1));
for k=1:size(A,1),
B1 = union(B, A(k,:));
C(k) = numel(B1)-numel(B);
B=B1;
end
result = A(C==size(A,2),:);

Every possible sum combination in a vector

Assuming I'm having a vectors of numbers A, for example: A=[1 3 5 3 9 6](A's length >= 2) and an Integer X=6. Need to find how many pairs (A[i],A[j]) where i<j exist in the vector which answer this condition: A[i]+A[j]=X. The number of pairs is printed.
Not allowed to use for/while. Allowed only ceil,floor,mod,repmat,reshape,size,length,transpose,sort,isempty,all,any,find ,sum,max,min.

With repmat, length and sum -
integer1 = 6; %// One of the paramters
A_ind = 1:length(A) %// Get the indices array
%// Expand A_ind into rows and A_ind' into columns, to form a meshgrid structure
A_ind_mat1 = repmat(A_ind,[length(A) 1])
A_ind_mat2 = repmat(A_ind',[1 length(A)]) %//'
%// Expand A into rows and A' into columns, to form a meshgrid structure
A_mat1 = repmat(A,[length(A) 1])
A_mat2 = repmat(A',[1 length(A)]) %//'
%// Form the binary matrix of -> (A[i],A[j]) where i<j
cond1 = A_ind_mat1 < A_ind_mat2
%// Use the binary matrix as a logical mask to select elements from the two
%// matrices and see which element pairs satisfy -> A[i]+A[j]=X and get a
%// count of those pairs with SUM
pairs_count = sum((A_mat1(cond1) + A_mat2(cond1))==integer1)
Outputs from code run to make it clearer -
A =
1 3 5 3 9 6
A_ind =
1 2 3 4 5 6
A_ind_mat1 =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
A_ind_mat2 =
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
A_mat1 =
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
A_mat2 =
1 1 1 1 1 1
3 3 3 3 3 3
5 5 5 5 5 5
3 3 3 3 3 3
9 9 9 9 9 9
6 6 6 6 6 6
cond1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 1 1 1 1 0
pairs_count =
2
A bit more explanation -
Taking few more steps to clarify why pairs_count must be 2 here -
Set all values in A_mat1 and A_mat2 to be zeros that do not satisfy the less than criteria
>> A_mat1(~cond1)=0
A_mat1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 3 0 0 0 0
1 3 5 0 0 0
1 3 5 3 0 0
1 3 5 3 9 0
>> A_mat2(~cond1)=0
A_mat2 =
0 0 0 0 0 0
3 0 0 0 0 0
5 5 0 0 0 0
3 3 3 0 0 0
9 9 9 9 0 0
6 6 6 6 6 0
Now, add A_mat1 and A_mat2 and see how many 6's you got -
>> A_mat1 + A_mat2
ans =
0 0 0 0 0 0
4 0 0 0 0 0
6 8 0 0 0 0
4 6 8 0 0 0
10 12 14 12 0 0
7 9 11 9 15 0
As you can see there are two 6's and thus our result is verified.

MATLAB: How do you add all the first n rows of a matrix within the same column, and save the result to the last row?

It was hard to phrase the question, but here's an example of what I'm looking for:
1 2 3 4
2 1 1 1
2 2 3 1
0 0 0 0
and in column one, I add all the value of all of the first three rows and save it to the third and so on, so that it becomes:
1 2 3 4
2 1 1 1
2 2 3 1
5 5 7 6

I think you can use sum:
octave:23> m = [1 2 3 4; 2 1 1 1; 2 2 3 1; 0 0 0 0]
m =
1 2 3 4
2 1 1 1
2 2 3 1
0 0 0 0
octave:24> m(length(m), :) = sum(m)
m =
1 2 3 4
2 1 1 1
2 2 3 1
5 5 7 6

(matlab) qtdecomp works with uint8 matrix?

I haven't fully understood how qtdecomp works...
I = [1 1 1 1 2 3 6 6
1 1 2 1 4 5 6 8
1 1 1 1 10 15 7 7
1 1 1 1 20 25 7 7
20 22 20 22 1 2 3 4
20 22 22 20 5 6 7 8
20 22 20 20 9 10 11 12
22 22 20 20 13 14 15 16];
S = qtdecomp(I,2);
disp(full(S));
The results of this are:
4 0 0 0 1 1 2 0
0 0 0 0 1 1 0 0
0 0 0 0 1 1 2 0
0 0 0 0 1 1 0 0
4 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
in the left bottom 4*4 matrix, maximum value (22) of the block elements minus the minimum value (20) is 2, so when decomposing this part, it will left as is.
When I do this on a uint8 matrix:
I = uint8([...
1 1 1 1 2 3 6 6
1 1 2 1 4 5 6 8
1 1 1 1 10 15 7 7
1 1 1 1 20 25 7 7
20 22 20 22 1 2 3 4
20 22 22 20 5 6 7 8
20 22 20 20 9 10 11 12
22 22 20 20 13 14 15 16]);
S = qtdecomp(I,2/255);
disp(full(S));
the answer is just like before. But when I change S to this:
S = qtdecomp(I,1.9/255);
The answer is
4 0 0 0 1 1 2 0
0 0 0 0 1 1 0 0
0 0 0 0 1 1 2 0
0 0 0 0 1 1 0 0
4 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
I suppose the left bottom 4*4 matrix should decompose, but why doesn't it?

What matlab does here is when I is uint8 it multiples the threshold by 255 and rounds it, so 1.9/255 is evaluated to 2.
You can see this by opening the source code for qtdecomp (by pressing ctrl+D) or here. There's an if/elseif near the end of the file (params{1} = round(255 * params{1});).
You should be able to use S = qtdecomp(I,1/255); to get the result you are looking for.

Matlab: Unique values between two matrices

I have 2 matrices:
T3(:,:,1) =
0 0 0 0 1 0 0 0 0
0 0 0 0 2 0 0 0 0
0 0 0 0 3 0 0 0 0
0 1 0 1 4 2 0 4 0
0 3 0 2 6 3 0 5 0
2 4 2 5 7 5 4 6 5
4 5 5 7 8 8 5 7 6
5 6 6 8 9 9 8 9 8
T3(:,:,2) =
2 1 1 1 1 1 1 1 1
3 3 2 2 2 2 2 2 2
4 4 4 3 3 3 3 3 3
5 5 5 5 4 4 4 4 4
6 6 6 6 6 5 5 5 5
7 7 7 7 7 7 6 6 6
8 8 8 8 8 8 8 7 7
9 9 9 9 9 9 9 9 8
How do I make values present in T3(:,:,1) turn to zero in T3(:,:,2)?
e.g. in the first column of T3(:,:,1) the values are 2,4,5. I'd like the first column of T3(:,:,2) to have the the values 2,4,5 as zero.
T3(:,:,2) =
0 0 1 0 0 1 1 1 1
3 0 0 0 0 0 2 2 2
0 0 4 3 0 0 3 3 3
0 0 0 0 0 4 0 0 4
6 0 0 6 0 0 0 0 0
7 7 7 0 0 7 6 0 0
8 8 8 0 0 0 0 0 7
9 9 9 9 0 0 9 0 0
I wonder if there is a way to do this using setdiff or unique.
for y=1:H-1
for z=1:H-1
for h=1:H
for d=1:D-1
if T3(y,h,d+1) == T3(z,h,d)
T3(y,h,d+1)=0;
end
end
end
end
end
I can do it as a loop where H=number of columns (9) and D= number of dimensions (2). There must be a better way :)?
Many thanks guys.