Replace a zeros row with one under a condition very important for me
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if Matrix A (M x N) like this matrix as input of function
A =[0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1
0 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1
1 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1
0 1 0 1 1 1 0 0 0 0 0 1 1 1 1 1
0 1 1 1 1 0 1 1 0 1 1 1 1 1 0 1
0 0 1 1 0 1 1 0 1 1 1 0 1 0 0 1
1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1];
then the function do many things
- First one
count number of ones in each row and in each column , For example
the first row is [0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1] then the first row in the row_matrix will be [1 1 1 1 1 1 1]
the second row is [ 0 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1] then the second row in row_matrix will be [2 8 1 1]
the last row is [0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1] then in the last row in row_matrix will be [5 1]
- this Number in row_matrix compute by if [1 1 0 1 0 1 1] then solution will be [2 1 2]
- And do the same thing for the Column in matrix A
- Second one
a) compute number of ones in each row in A and put save it in vector sum_row , for example
sum_row = [ 7
12
8
9
12
9
11
13
16
6 ]
b) compute number of ones in each column in A and put the result in sum_col like this
sum_column = [ 3 7 6 6 7 6 8 6 7 7 6 7 8 7 3 9]
- third one
a) count number of element in rows of row_matrix and put it in count_row like this example
the first row in row_matrix = [1 1 1 1 1 1 1] then the first row in count_row will be [7]
the second row in row_matrix = [2 8 1 1] then the second row in count_row will be [4]
and so on ...
b) do the same thing for the column
- the last thing
Generate a (M x N) zeros_matrix then do this equation If N == (sum_row(k) + (count_row(k) - 1) where k is number of row in each matrix , for example
example one :
the 7th row in A matrix is [1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1]
then the 7th row in row_matrix will be [1 1 1 4 3 1]
and the 7th row in sum_row will be [11]
and the 7th row in count_row will be [6]
then we apply the equation N== (sum_row(k) + (count_row(k) - 1))
N== (sum_row(7) + (count_row(7)-1) >>> 16 == (11 + (6-1) >> then 16 ==16 (True)
example 2 : the 9th row in A matrix is [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
then the 9th row in row_matrix will be [16]
and the 9th row in sum_row will be [16]
and the 9th row in count_row will be [1]
then when we apply the equation
N== (sum_row(9) + (count_row(9)-1) >>> 16 == (16 + (1-1) >> then 16 ==16 (True)
then in 7th and 9th row in A empty_matrix will be like this
empty_matrix = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]
and do the same thing for a column in this equation M==(sum_column(k) + (count_column(k) - 1)
*Note i need this function to cont my graduation project *
4 Comments
Answers (2)
dpb
on 2 Apr 2016
Edited: dpb
on 2 Apr 2016
On the first, I know there is a submittal on File Exchange to count runs; I think it may be from Bruno Luong...but, consider what happens if you diff a row or column and what that tells you about start/stop locations...it's not really very difficult problem.
The rest are pretty simple and looking through the doc and "Getting Started" tutorial should lead you there pretty quickly...
sum_r=sum(M,2); % sum rows
sum_c=sum(M); % sum columns
The count vector is simply length(eachRow) in the answer to Part 1.
The last is convoluted enough I don't have time to dig thru the explanation to the result but Walter's advice is good...
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Roger Stafford
on 2 Apr 2016
I'll just answer for your "First one" in getting "the same thing for the columns in A". Let x be a column in A.
s = cumsum(diff([0;x])==1);
t = (x==1);
a = accumarray(s(t),1);
The 'a' will be a column of the various contiguous sums. In your A example for the second column you would get a = [4;2].
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