I have an array of integer numbers, and I want to 开发者_StackOverflow社区split this array where 0 comes and a function that give me points of split.
Example: Array : 0 0 0 1 2 4 5 6 6 0 0 0 0 0 22 4 5 6 6 0 0 0 4 4 0
The function must return these numbers:
[ 3 10 ;14 20 ;22 25 ]
These numbers are index of start and end of nonzero numbers.
Here's a simple vectorized solution using the functions DIFF and FIND:
>> array = [0 0 0 1 2 4 5 6 6 0 0 0 0 0 22 4 5 6 6 0 0 0 4 4 0]; %# Sample array
>> edgeArray = diff([0; (array(:) ~= 0); 0]);
>> indices = [find(edgeArray > 0)-1 find(edgeArray < 0)]
indices =
3 10
14 20
22 25
The above code works by first creating a column array with ones indicating non-zero elements, padding this array with zeroes (in case any of the non-zero spans extend to the array edges), and taking the element-wise differences. This gives a vector edgeArray
with 1
indicating the start of a non-zero span and -1
indicating the end of a non-zero span. Then the function FIND is used to get the indices of the starts and ends.
One side note/nitpick: these aren't the indices of the starts and ends of the non-zero spans like you say. They are technically the indices just before the starts and just after the ends of the non-zero spans. You may actually want the following instead:
>> indices = [find(edgeArray > 0) find(edgeArray < 0)-1]
indices =
4 9
15 19
23 24
Try this
a = [0 0 0 1 2 4 5 6 6 0 0 0 0 0 22 4 5 6 6 0 0 0 4 4 0];
%#Places where value was zero and then became non-zero
logicalOn = a(1:end-1)==0 & a(2:end)~=0;
%#Places where value was non-zero and then became zero
logicalOff = a(1:end-1)~=0 & a(2:end)==0;
%#Build a matrix to store the results
M = zeros(sum(logicalOn),2);
%#Indices where value was zero and then became non-zero
[~,indOn] = find(logicalOn);
%#Indices where value was non-zero and then became zero
[~,indOff] = find(logicalOff);
%#We're looking for the zero AFTER the transition happened
indOff = indOff + 1;
%#Fill the matrix with results
M(:,1) = indOn(:);
M(:,2) = indOff(:);
%#Display result
disp(M);
On the theme, but with a slight variation:
>>> a= [0 0 0 1 2 4 5 6 6 0 0 0 0 0 22 4 5 6 6 0 0 0 4 4 0];
>>> adjust= [0 1]';
>>> tmp= reshape(find([0 diff(a== 0)])', 2, [])
tmp =
4 15 23
10 20 25
>>> indices= (tmp- repmat(adjust, 1, size(tmp, 2)))'
indices =
4 9
15 19
23 24
As gnovice
already pointed out on the positional semantics related to indices
, I'll just add that, with this solution, various schemes can be handled very straightforward manner, when calculating indices
. Thus, for your request:
>>> adjust= [1 0]';
>>> tmp= reshape(find([0 diff(a== 0)])', 2, []);
>>> indices= (tmp- repmat(adjust, 1, size(tmp, 2)))'
indices =
3 10
14 20
22 25
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