# How can I count the largest number of repeated numbers in a double?

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Jon Mick on 13 Sep 2019
Commented: Adam Danz on 14 Sep 2019
I have a double that is 5644x1 and it consists of nothing but 0's and 1's.
ex: 00000000101100000011110000110000
I want to count the longest length of repeated 0's in this double, which in the case of this example would be 8. Can you please help?

John D'Errico on 13 Sep 2019
Edited: John D'Errico on 13 Sep 2019
First, this is NOT a double vector.
00000000101100000011110000110000
It might be a string. But if you tried to write that vector as a double, it would be just one number, in decimal form. So you might have it stored as a vector binary elements. I'll assume you have a character string, but the answer is trivially changed if not.
V = '00000000101100000011110000110000';
We are looking for the longest length of repeated 0's.
st10 = strfind(['1',V],'10') + 1;
st01 = strfind([V,'1'],'01');
st01 - st10 + 2
ans =
8 1 6 4 4
So the length of the LONGEST such string of zeros is just...
max(st01 - st10 + 2)
ans =
8
If your vector is actually a numeric vector, perhaps like this, strfind still works
V=[0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0];
st10 = strfind([1,V],[1 0]) + 1;
st01 = strfind([V,1],[0 1]);
max(st01 - st10 + 2)
ans =
8

Adam Danz on 14 Sep 2019
I didn't know strfind() does that, too. Nice!
John D'Errico on 14 Sep 2019
I guess it is one of those features that date back to the dark ages of MATLAB, when there was sort of a duality beween strings and numeric integer vectors containing ASCII code values. The reason why you can still do things like this:
C = 'ABCDE';
+C % ascii code values
ans =
65 66 67 68 69
C - 'A'
ans =
0 1 2 3 4
N = '012345';
N - '0'
ans =
0 1 2 3 4 5
Adam Danz on 14 Sep 2019
char([73 39 108 108 32 104 97 118 101 32 102 117 110 32 119 105 116 104 32 116 104 105 115 33])

the cyclist on 13 Sep 2019

Image Analyst on 13 Sep 2019
You can use regionprops(), if you have the Image Processing Toolbox
signal = [0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0];
props = regionprops(logical(signal == 0), 'Area'); % Measure lengths of each run of zeros
longestStretch = max([props.Area]) % Find the length of the longest one.
You get:
longestStretch =
8

#### 1 Comment

Jon Mick on 13 Sep 2019
This might work but I don't have the Image Processing Toolbox :(