How to find position of array

For example i have data like this [11111100001111]

Now i need to find out the position of first 1 and last 1 of first group of 1s and then position of first 1 and last 1 of second group of 1s. How do i do this??

i have used' find' to find the position of all 1s but i dont want all...instead 1 want only the first and last position of groupof 1s

[madhan ravi]

[1 0 1 1 0 0 0 0 1 1 1 1] - the result would be?

[Wind flower]

for my data [11111100001111], the answer would be 1 and 6 for first set , 11 and 14 for the second set. Afer that i want to save that in array like this:

1 6

11 14

Answer 1

[Debasish Samal]

You can use regular expressions to do this.

x = '11111100001111';
expr = '1*';
[sInd,eInd] = regexp(x, expr)

This gives the following result:

sInd = 1×2    
     1    11

eInd = 1×2    
     6    14

You just need to convert the input data into a string.

[madhan ravi]

Don't ever name a variable as exp it would hinder the in-built function exp().

[Debasish Samal]

Thank You. Point noted :-)

[Wind flower]

@Debasish Samal your answer is good. But i have that data in column form in excel and reading it using xlsread so how will i convert that into a string (My actual data is very very big)

[madhan ravi]

Wind flower: Did you see the answer below? I don’t know why my answer was marked Accepted mistakenly.

[Debasish Samal]

@Wind flower as far as I know xlsread returns a matrix of integers(if you are reading integers). So very basic approach would be using num2str to convert the num type to string type. Then iterate over each cell to get the result. I am not sure how your data looks like after reading it using xlsread. But I am sure you can figure out a way to optimise the conversion and get the desired result. Thanks.

[Wind flower]

@madhan ravi: Your answer was correct. Sorry first i got some error then i got to know. Your answer is simpler and correct. Sorry for the mistake

Thank You

Answer 2

[madhan ravi]

Simpler without any conversions:

s=[1 0 1 1 0 0 0 0 1 1 1 1]; % example data
x=s~=0;
Start = strfind([0,x],[0 1])
End = strfind([x,0],[1 0])

[madhan ravi]

s = 11111100001111; % if your data is indeed like below then
m = floor(log10(s));  
D = mod(floor(s ./ 10 .^ (m:-1:0)), 10);
x=D~=0;
Start = strfind([0,x],[0 1])
End = strfind([x,0],[1 0])

[Rik]

Note that using numeric arrays as inputs to strfind is currently undocumented (and it even returns an error in GNU Octave). It would be nice if Mathworks changed the documentation. It would be even better if they extended this function to work on arrays of any object type.

[Wind flower]

@madhan ravi : Thank You