I have a matrix and I wanted to find all non-zero rows in the matrix and the all(A, 2) function did this but I was wondering if there is a way to list the corresponding row number alongside the value?
Use find(all(A,2). all(A,2) gives you a vector with a 1 where there is a row of ones, and a 0 otherwise. find gives you the indices of non-zeros elements of an array. Putting them together gives the required result:
A=[0 0 1 0
1 1 1 1
0 1 1 0
0 1 0 1]
find(all(A,2))=2
Related
2D array of 1s and 0s. How to label every group of 1s with a unique number?
I’m stuck on this problem for a while now. 1s can be grouped vertically, horizontally and diagonally. How can you go about solving this? For example,
0 0 1 1 0
0 1 1 0 0
0 0 0 0 1
0 0 0 1 0
Should be transformed to
0 0 x x 0
0 x x 0 0
0 0 0 0 y
0 0 0 y 0
x, y can be any unique numbers.
Appreciate it.
Here is what I have so far for iterative: https://i.imgur.com/oCmYC02.png
But the result is a bit off because it only checks for immediate adjacent 1's: https://i.imgur.com/DAtTBmM.png
Anyone have any idea how to fix this?
I'd do it like this:
Scan 2D array sequentially, row by row, column by column
If 1 found, use variation of the flood fill algorithm, which moves in 8 directions instead of 4, from that starting point (see normal 4-direction algorithm at https://en.wikipedia.org/wiki/Flood_fill), since you have diagonal example with "y", each time using new filler number.
Repeat 1 and 2 until no more ones left.
I've got logical array(zeros and ones) 1500x700
I want to find "1" in every column and when there are more than one "1" in column i should choose the middle one.
Is that possible to do it? I know how to find "1", but don't know how to extract the middle "1" if there's couple of "1" in one column.
The find function returns the indices of your ones.
>> example=[1,0,0,1,0,1,1];
>> indices=find(example)
indices =
1 4 6 7
>> indices(floor(numel(indices)/2))
ans =
4
Do this for each column and you have a solution.
You can
Get the row and column indices of ones with find;
Apply accumarray with a custom function to get the middle row index for each column.
x = [1 0 0 0 0; 0 0 1 0 0; 1 0 1 0 0; 1 0 0 1 0]; % example
[ii, jj] = find(x); % step 1
result = accumarray(jj, ii, [size(x,2) 1], #(x) x(ceil(end/2)), NaN); % step 2
Note that:
For an even number of ones this gives the first of the two middle indices. If you prefer the average of the two middle indices replace #(x) x(ceil(end/2)) by #median.
For a column without ones this gives NaN as result. If you prefer a different value, replace the input fifth argument of accumarray by that.
Example:
x =
1 0 0 0 0
0 0 1 0 0
1 0 1 0 0
1 0 0 1 0
result =
3
NaN
2
4
NaN
I'm trying to remove the rows which has duplicates in sequence. I have only 2 possible values which are 0 and 1. I have nXm which n shows possible number of bits and m is not important for my question. My goal is to find an matrix which is nX(m-a). The rows a which has the property which includes duplicates in sequence. For example:
My matrix is :
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0]
I want to remove the rows has t duplicates in sequence for 0. In this question let's assume t is 3. So I want the matrix which:
B=[0 1 0 1 0 1;
0 0 1 0 0 1;
0 1 0 0 1 0]
2nd and 5th rows are removed.
I probably need to use diff.
So you want to remove rows of A that contain at least t zeros in sequence.
How about a single line?
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==-t, 2),:);
How this works:
Transform A to bipolar form (2*A-1)
Convolve each row with a sequence of t ones (conv2(...))
Keep only rows for which the convolution does not contain -t (~any(...)). The presence of -t indicates a sequence of t zeros in the corresponding row of A.
To remove rows that contain at least t ones, just change -t to t:
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==t, 2),:);
Here is a generalized approach which removes any rows which has given number of consecutive duplicates (not just zero. could be any number).
t = 3;
row_mask = ~any(all(~diff(reshape(im2col(A,[1 t],'sliding'),t,size(A,1),[]))),3);
out = A(row_mask,:)
Sample Run:
>> A
A =
0 1 0 1 0 1
0 0 1 5 5 5 %// consecutive 3 5's
0 0 1 0 0 1
0 1 0 0 1 0
1 1 1 0 0 1 %// consecutive 3 1's
>> out
out =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
How about an approach using strings? This is certainly not as fast as Luis Mendo's method where you work directly with the numerical array, but it's thinking a bit outside of the box. The basis of this approach is that I consider each row of A to be a unique string, and I can search each string for occurrences of a string of 0s by regular expressions.
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0];
t = 3;
B = sprintfc('%s', char('0' + A));
ind = cellfun('isempty', regexp(B, repmat('0', [1 t])));
B(~ind) = [];
B = double(char(B) - '0');
We get:
B =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
Explanation
Line 1: Convert each line of the matrix A into a string consisting of 0s and 1s. Each line becomes a cell in a cell array. This uses the undocumented function sprintfc to facilitate this cell array conversion.
Line 2: I use regular expressions to find any occurrences of a string of 0s that is t long. I first use repmat to create a search string that is full of 0s and is t long. After, I determine if each line in this cell array contains this sequence of characters (i.e. 000....). The function regexp helps us perform regular expressions and returns the locations of any matches for each cell in the cell array. Alternatively, you can use the function strfind for more recent versions of MATLAB to speed up the computation, but I chose regexp so that the solution is compatible with most MATLAB distributions out there.
Continuing on, the output of regexp/strfind is a cell array of elements where each cell reports the locations of where we found the particular string. If we have a match, there should be at least one location that is reported at the output, so I check to see if any matches are empty, meaning that these are the rows we don't want to remove. I want to turn this into a logical array for the purposes of removing rows from A, and so this is wrapped with a cellfun call to determine the cells that are empty. Therefore, this line returns a logical array where a 0 means that remove this row and a 1 means that we don't.
Line 3: I take the logical array from Line 2 and invert it because that's what we really want. We use this inverted array to index into the cell array and remove those strings.
Line 4: The output is still a cell array, so I convert it back into a character array, and finally back into a numerical array.
I have a table of rows which consist of zeros and numbers like this:
A B C D E F G H I J K L M N
0 0 0 4 3 1 0 1 0 2 0 0 0 0
0 1 0 1 4 0 0 0 0 0 1 0 0 0
9 5 7 9 10 7 2 3 6 4 4 0 1 0
I want to calculate an average of the numbers including zeros, but starting from the first nonzero value and put it into column after tables end. E.g. for the first row first value is 4, so average - 11/11; for the second - 7/13; the last one is 67/14.
How could I using excel formulas do this? Probably OFFSET with nested IF?
This still needs to be entered as an array formula (ctrl-shift-enter) but it isn't volatile:
=AVERAGE(INDEX(($A2:$O2),MATCH(TRUE,$A2:$O2<>0,0)):$O2)
or, depending on location:
=AVERAGE(INDEX(($A2:$O2);MATCH(TRUE;$A2:$O2<>0;0)):$O2)
The sum is the same no matter how many 0's you include, so all you need to worry about is what to divide it by, which you could determine using nested IFs, or take a cue from this: https://superuser.com/questions/671435/excel-formula-to-get-first-non-zero-value-in-row-and-return-column-header
Thank you, Scott Hunter, for good reference.
I solved the problem using a huge formula, and I think it's a bit awkward.
Here it is:
=AVERAGE(INDIRECT(CELL("address";INDEX(A2:O2;MATCH(TRUE;INDEX(A2:O2<>0;;);0)));TRUE):O2)
Imagine the following binary image exemplified by the matrix below. This is a simplified version of the images I'll be working with:
0 1 0 1
0 1 1 1
0 0 0 1
0 1 1 1
I want to construct a graph that will represent the randomness of each column. My thought is to develop a random index = the total transitions between each value in the column / by the total possible transitions. In the matrix above, each column could have a total possible of 3 transitions.
For the example above:
Column 1 would have a random index of 0% (0/3)
Column 2 would have a random index of 66.7% (2/3)
Column 3 = 100% (3/3)
Column 4 = 0% (0/3) even though they are 1's and not 0's. Doesn't matter, I just want the transitions.
Can I draw a boundary around all the 1 values and then have MATLAB sum all of the boundaries?
To calculate what you are suggesting you can just do:
sum( diff(A) ~= 0 )
The diff(A) will take the forward difference down the columns and the sum will count the number of non-zero changes. So if you do this you will get:
ans =
0 2 3 0
Let your image be defined as
im = [ 0 1 0 1
0 1 1 1
0 0 0 1
0 1 1 1 ];
The random index you want can be computed as
result = sum(diff(im)~=0) / (size(im,1)-1);
Explanation: diff computes the difference between consecutive elemtents down each column. The result is compared against zero (~=0), and all nonzero values within each row are added (with sum). Finally, the result is divided by the maximum number os transitions, which is the number of rows minus 1 (size(im,1)-1)
Equivalently, you could use xor between consecutive rows:
result = sum(xor(im(1:end-1,:), im(2:end,:))) / (size(im,1)-1)