I have two vectors A and B. A contains zeros and ones, whereas B contains the vectors at which A is zero or 1. I want to know the range of B where A is 1

for eg.

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

B=[0.0359624956468314 0.143849982587326 0.323662460821483 0.575399930349303 0.899062391170786 1.29464984328593 1.76216228669474 2.30159972139721 2.91296214739335 3.59624956468314 4.35146197326660 5.17859937314373 6.07766176431451 7.04864914677896 8.09156152053707 9.20639888558885 10.3931612419343 11.6518485895734 12.9824609285061 14.3849982587326 15.8594605802527 17.4058478930664 19.0241601971738 20.7143974925749 22.4765597792696 24.3106470572581 26.2166593265401 28.1945965871158 30.2444588389852 32.3662460821483 34.5599583166050 36.8255955423554 39.1631577593994 41.5726449677371 44.0540571673685 46.6073943582935 49.2326565405122 51.9298437140246 54.6989558788306 57.5399930349303 60.4529551823236 63.4378423210106 66.4946544509913 69.6233915722656 72.8240536848336 76.0966407886953 79.4411528838506 82.8575899702996 86.3459520480423 89.9062391170786 93.5384511774085 97.2425882290322 101.018650271949 104.866637306160 108.786549331665 112.778386348463 116.842148356555 120.977835355941 125.185447346620 129.464984328593 133.816446301860 138.239833266420 142.735145222274 147.302382169422 151.941544107863 156.652631037598 161.435642958626 166.290579870949 171.217441774564 176.216228669474 ]

How to find the range of B where A is 1

The answer should come as

range1=[19.02 36.83 ]

range2=[101 147.3]

range3=[171.2 176.2]

Answer 1

[Stephan]

result = range(B(A==1))

You edited the question - here's the edited answer:

% your arrays
A=[0	0	0	0	0	0	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	0	0	0	0	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	0	0	0	0	1	1];
B=[0.0359624956468314	0.143849982587326	0.323662460821483	0.575399930349303	0.899062391170786	1.29464984328593	1.76216228669474	2.30159972139721	2.91296214739335	3.59624956468314	4.35146197326660	5.17859937314373	6.07766176431451	7.04864914677896	8.09156152053707	9.20639888558885	10.3931612419343	11.6518485895734	12.9824609285061	14.3849982587326	15.8594605802527	17.4058478930664	19.0241601971738	20.7143974925749	22.4765597792696	24.3106470572581	26.2166593265401	28.1945965871158	30.2444588389852	32.3662460821483	34.5599583166050	36.8255955423554	39.1631577593994	41.5726449677371	44.0540571673685	46.6073943582935	49.2326565405122	51.9298437140246	54.6989558788306	57.5399930349303	60.4529551823236	63.4378423210106	66.4946544509913	69.6233915722656	72.8240536848336	76.0966407886953	79.4411528838506	82.8575899702996	86.3459520480423	89.9062391170786	93.5384511774085	97.2425882290322	101.018650271949	104.866637306160	108.786549331665	112.778386348463	116.842148356555	120.977835355941	125.185447346620	129.464984328593	133.816446301860	138.239833266420	142.735145222274	147.302382169422	151.941544107863	156.652631037598	161.435642958626	166.290579870949	171.217441774564	176.216228669474 ];

% call function
result = findRange(A,B)

function range = findRange(A,B)
    % indices of A which are equal to 1
    res = find(A==1);
    % Starting index for a new block in res
    idx_start = [1 find(diff(find(A~=0))~=1)+1];
    % Ending index for blocks
    idx_end = [find(diff(find(A~=0))~=1) numel(res)];
    % preallocate
    blocks = cell(1,numel(idx_start));
    range = nan(numel(idx_start),2);
    % Building blocks and evaluating range;
    for k = 1:numel(idx_start)
        blocks(k) = {B(res(idx_start(k):idx_end(k)))};
        range(k,:) = [min(blocks{k}) max(blocks{k})];
    end
end

result is:

result =

   19.0242   36.8256
  101.0187  147.3024
  171.2174  176.2162

Best regards

Stephan

[RAJAN PRASAD]

Thanks for the answer. Can it it be done if there are multiple ranges and the range starting and ending point is required.

[Stephan]

See my edited answer.

[madhan ravi]

@RAJAN it‘s not appropriate to edit the question after someone has answered.

[RAJAN PRASAD]

Thank you very much.The code works great. I would start the question in next thread from now on. My sincere apolgies for editing the question.