# Finding Sequences of 1's values

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I have a matrix of values, for example: [0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 0 ...]

In practice this is a vector (1 row by somewhere around 80000 columns). I need to output a vector with the duration of these 'bursts' of 1's. For example, the output for the above would be: [3 2 5 ...]

Background: I have a time series where all the 1's are values above a threshold and all the 0's are values below a threshold. So by doing this I'm trying to investigate the duration of these extreme events.

Can anyone tell me how I can do this?

##### 0 Comments

### Accepted Answer

Image Analyst
on 21 Sep 2011

If you have the Image Processing Toolbox, just call bwconncomp() or bwlabel(), then call regionprops() asking for 'Area' and then extract the areas from the structure. Three lines.

labeledMatrix = bwlabel(data);

measurements = regionprops(labeledMatrix, 'Area');

allAreas = [measurements.Area];

##### 1 Comment

Jan
on 22 Sep 2011

data = round(rand(1, 80000))

regionprops (Image Analyst): 0.55 sec

If you follow the MLint advice: BWLABEL->LOGICAL: 0.18 sec

Vectorized FINDSTR(DIFF): 0.0081 sec

Cleaned loop (Derek + Jan): 0.0048 sec

(Measured under Matlab 2009a, WinXP, 2.3GHz Core2Duo)

### More Answers (8)

Derek O'Connor
on 21 Sep 2011

If you don't have any toolboxes then this plain Matlab function may help. It is based loosely on the WordCount procedure in Kernighan & Plauger [1981], Software Tools in Pascal, ©Bell Labs, Addison-Wesley, page 17.

For n = 10^8 it takes about 4 secs (single core) 2.3GHz, R2008a 64bit on Windows 7.

function [start, len, k1] = ZeroOnesCount(v)

%

% Determine the position and length of each 1-string in v,

% where v is a vector of 1s and 0s

% Derek O'Connor 21 Sep 2011

%

n = length(v);

start = zeros(1,n); % where each 1-string starts

len = zeros(1,n); % length of each 1-string

k1= 0; % count of 1-strings

inOnes = false;

for k = 1:n

if v(k) == 0 % not in 1-string

inOnes = false;

elseif ~inOnes % start of new 1-string

inOnes = true;

k1 = k1+1;

start(k1) = k;

len(k1) = 1;

else % still in 1-string

len(k1) = len(k1)+1;

end

end

%------------- ZeroOnesCount ----------------------------

>> v = [0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 0];

>> [start,len,k1] = ZeroOnesCount(v);disp([start(1:k1); len(1:k1)]);

7 15 21

3 2 5

##### 1 Comment

Jan
on 22 Sep 2011

To get the answer needed by the OP, collecting "start" is not needed, but occupies n*8 bytes in the memory.

But the idea to create a loop is efficient. +1

Derek O'Connor
on 22 Sep 2011

This is a more general and simpler solution than my previous answer.

%-------------------------------------------------------

function [start, val, len, rn] = RunsCount(v)

%

% Determine the length of each val-run in v,

% where v is a vector of 'values'

% Example 1: RunsCount([2 2 2 1 1 9 8 8 3 3 3 3]) gives

% start 1 4 6 7 9

% val 2 1 9 8 3

% len 3 2 1 2 4

% Example 2: RunsCount(['aaaabbaacccbbb']);

% start 1 5 7 9 12

% val 97 98 97 99 98

% len 4 2 2 3 3

%

% Derek O'Connor 22 Sep 2011

%

n = length(v);

val = zeros(1,n); % run value

len = zeros(1,n); % run length

start = zeros(1,n); % pos. in v where run starts

start(1) = 1;

rk = 1; % number in run

rn = 1; % number of runs

for k = 2:n

if v(k) == v(k-1) % in run

rk = rk+1;

else % end of run

val(rn) = v(k-1);

len(rn) = rk;

rk = 1; % v(k) is start of

rn = rn+1; % next run

start(rn) = k; % position of start in v

end

end

val(rn) = v(n); % last run

len(rn) = n - sum(len);

%

%------------- End RunsCount ----------------------------

EDIT : Pre-allocated all output arrays. Changed loop counter from i to k.

##### 2 Comments

Callula Killingly
on 30 Jun 2022

Jan
on 22 Sep 2011

A modified version of Derek's solution:

function len = CountOnes(v)

n = length(v);

len = zeros(1,n); % length of each 1-string

k1 = 0; % count of 1-strings

inOnes = false;

s = 0;

for k = 1:n

if v(k) % not in 1-string

s = s + 1; % increase the counter

inOnes = true;

elseif inOnes % leave the ones block

k1 = k1 + 1;

len(k1) = s;

s = 0; % reset the counter

inOnes = false;

end

end

len = len(1:k1);

And this is a vectorized approach:

function len = CountOnes_vectorized(v)

v = diff([0, x, 0]);

len = findstr(v, -1) - findstr(v, 1);

It looks nice, but is 45% slower than the loop method for the 80'000 elements needed by the OP.

[EDITED] New version with a nested loop:

function len = CountOnes2(v)

n = length(v);

len = zeros(1, ceil(n/2), 'uint32');

j = 0;

k = 1;

while k <= n

if v(k)

a = k;

k = k + 1;

while k <= n && v(k)

k = k + 1;

end

j = j + 1;

len(j) = k - a;

end

k = k + 1;

end

len = len(1:j);

##### 5 Comments

Image Analyst
on 14 Jan 2022

My solution above (the top, accepted one) also works for matrices. If it doesn't read this

and post your image and question in a new question.

Derek O'Connor
on 22 Sep 2011

Jan,

I'm using the Add-an-Answer window because the comment window is hard to use for all but short, text-only replies.

I posted an older version of RunsCount (no pre-allocation) by mistake. At that time I wasn't sure how long the output arrays should be. The size varies with the input but the worst is n, so pre-allocate the output arrays of length n.

I take your point that the array start, etc., are not necessary, but I suspect this information might be needed in other applications of these counters. Anyway, to compare your modification with ZeroOnesCount and RunsCount, I stripped out all but the output array len(1,n).

RC3 is RunsCount with start and val array and their ops stripped out

ZOC2 is ZeroOnesCount with start array and its ops stripped out.

COJ is CountOnes, Jan's modification of ZOCorig

The table below are the normalized average times, averaged over 100 runs The final row gives the actual average times for n = 10^8.

Normalized Average Times

n RC3 ZOCorig ZOC2 COJ

-----------------------------------------------

10^3 1.53 3.84 1 1.16

10^4 1.26 1.49 1 1.11

10^5 1.19 1.24 1 1.15

10^6 1.15 1.25 1 1.31

10^7 1.15 1.24 1 1.32

10^8 1.14 1.26 1 1.32

-----------------------------------------------

Actual(secs)10^8 3.84 4.23 3.36 4.44

I was surprised by this result: ZOC2 is uniformly better than COJ.

I profiled both functions with n = 10^8 and compared statement counts (timing info is ignored because it is useless for comparisons).

COJ ZOC2

1 10 for k = 1:n 1 11 for k = 1:n

100000000 11 if v(k) 100000000 12 if v(k) == 0

49999019 12 s = s + 1; 50000981 13 inOnes = false;

49999019 13 inOnes = true; 49999019 14 elseif ~inOnes

50000981 14 elseif inOnes 25002599 15 inOnes = true;

25002599 15 k1 = k1 + 1; 25002599 16 k1 = k1+1;

25002599 16 len(k1) = s; 25002599 17 len(k1) = 1;

25002599 17 s = 0; 24996420 18 else

25002599 18 inOnes = false; 24996420 19 len(k1) = len(k1)+1;

25002599 19 end 24996420 20 end

100000000 20 end 100000000 21 end

At first glance there seems to be very little difference in the counts: if-elseif about the same, inOnes = true-false, about the same.

The only difference I can see is that COJ does about 75x10^6 ops on the counter s (lines 12,17) which are not done by ZOC2. However, ZOC2 does about 25x10^6 ops extra on len (line 19) which are not done by COJ.

Thus COJ is doing about 50x10^6 extra assignment operations. This would seem to account for the difference between the two functions.

##### 4 Comments

Jos (10584)
on 18 Mar 2014

Another idea, not tested for speed:

A = [0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 0]

B = cumsum(a)

tf = B(2:end) == B(1:end-1) & A(1:end-1)==1

result = diff([0 B(tf)])

##### 0 Comments

Jan
on 8 Jan 2018

Edited: Jan
on 8 Jan 2018

And for completeness a C-Mex function also: https://de.mathworks.com/matlabcentral/fileexchange/41813-runlength

[B, N] = RunLength(x)

Result = N(B == 1);

##### 0 Comments

Jan
on 30 Jun 2022

Some timings of the methods provided in this thread:

x = randi([0, 1], 1, 80000);

format long g

disp(timeit(@() Jos(x)))

disp(timeit(@() CountOnes2(x)))

disp(timeit(@() CountOnes_vectorized(x)))

disp(timeit(@() CountOnes(x)))

disp(timeit(@() ZeroOnesCount(x)))

disp(timeit(@() RunsCount(x)))

disp(timeit(@() ImgProcMethod(x)))

function r = Jos(x)

B = cumsum(x);

tf = B(2:end) == B(1:end-1) & x(1:end-1);

r = diff([0 B(tf)]);

end

function len = CountOnes2(v)

n = length(v);

len = zeros(1, ceil(n/2), 'uint32');

j = 0;

k = 1;

while k <= n

if v(k)

a = k;

k = k + 1;

while k <= n && v(k)

k = k + 1;

end

j = j + 1;

len(j) = k - a;

end

k = k + 1;

end

len = len(1:j);

end

function len = CountOnes_vectorized(x)

v = diff([0, x, 0]);

len = findstr(v, -1) - findstr(v, 1);

end

function len = CountOnes(v)

n = length(v);

len = zeros(1,n); % length of each 1-string

k1 = 0; % count of 1-strings

inOnes = false;

s = 0;

for k = 1:n

if v(k) % not in 1-string

s = s + 1; % increase the counter

inOnes = true;

elseif inOnes % leave the ones block

k1 = k1 + 1;

len(k1) = s;

s = 0; % reset the counter

inOnes = false;

end

end

len = len(1:k1);

end

function len = RunsCount(v)

n = length(v);

len = zeros(1,n); % run length

rk = 1; % number in run

rn = 1; % number of runs

for k = 2:n

if v(k) == v(k-1) % in run

rk = rk+1;

else % end of run

len(rn) = rk;

rk = 1; % v(k) is start of

rn = rn+1; % next run

end

end

len(rn) = n - sum(len);

end

function len = ZeroOnesCount(v)

n = length(v);

len = zeros(1,n); % length of each 1-string

k1= 0; % count of 1-strings

inOnes = false;

for k = 1:n

if v(k) == 0 % not in 1-string

inOnes = false;

elseif ~inOnes % start of new 1-string

inOnes = true;

k1 = k1+1;

len(k1) = 1;

else % still in 1-string

len(k1) = len(k1)+1;

end

end

end

function allAreas = ImgProcMethod(data)

labeledMatrix = bwlabel(data);

measurements = regionprops(labeledMatrix, 'Area');

allAreas = [measurements.Area];

end

##### 0 Comments

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