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Finding Sequences of 1's values

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James
James on 21 Sep 2011
Edited: Jan on 8 Jan 2018
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?

Accepted Answer

Image Analyst
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
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)

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More Answers (7)

Derek O'Connor
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
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

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James
James on 21 Sep 2011
Thanks, that worked perfect!

Derek O'Connor
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.
  1 Comment
Jan
Jan on 22 Sep 2011
Without pre-allocating "val", "len" and "start", this code will be very slow for a large input due to the huge memory footprint. Therefore I'd prefer your other solution.

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Jan
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);
  3 Comments
Leendert
Leendert on 18 Mar 2014
@Jan: Thanx a million now it was a piece of cake to rewrite it!

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Derek O'Connor
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
Jan
Jan on 23 Sep 2011
@Walter: I think Derek bantered me with the UINT1 because he thought, that I had changed the topic durign the process of measurement.

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Jos (10584)
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)])

Jan
Jan on 8 Jan 2018
Edited: Jan on 8 Jan 2018
[B, N] = RunLength(x)
Result = N(B == 1);

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