blocks|key|1509155|text|下面是使用out执行此操作(以及其他二进制数据操作)的快速方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1509156|out+=+logical([1+1+1+0+1+0+1+0+0+1+1+1]);
d+=+diff([out(1)+out]);+%25+find+all+switches+between+1+to+0+or+0+to+1
len+=+1:length(out);+%25+make+a+list+of+all+indices+in+'out'
idx+=+[len(d~=0)-1+length(out)];+%25+the+index+of+the+end+each+group
counts+=+[idx(1)+diff(idx)];+%25+the+number+of+elements+in+the+group
elements+=+out(idx);+%25+the+type+of+element+(0+or+1)
singles+=+idx(counts==1+&+elements==1)|code-block|syntax|javascript|1509157|你将会得到：|1509158|singles+=

+++5+++7|1509159|从这里开始，您可以继续并根据需要创建输出：|1509160|out+=+false(size(out));+%25+create+an+output+vector
out(singles)+=+true+%25+fill+with+'1'+by+singles|1509161|然后你会得到：|1509162|out+=

+++0+++0+++0+++0+++1+++0+++1+++0+++0+++0+++0+++0|1509163|entityMap^0|5|3|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|16|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|X|$]]

Here is a quick way to do it (and other manipulations binary data) using your <code>out</code>:

<pre><code>out = logical([1 1 1 0 1 0 1 0 0 1 1 1]);
d = diff([out(1) out]); % find all switches between 1 to 0 or 0 to 1
len = 1:length(out); % make a list of all indices in 'out'
idx = [len(d~=0)-1 length(out)]; % the index of the end each group
counts = [idx(1) diff(idx)]; % the number of elements in the group
elements = out(idx); % the type of element (0 or 1)
singles = idx(counts==1 &amp; elements==1)
</code></pre>

and you will get:

<pre><code>singles =

 5 7
</code></pre>

from here you can continue and create the output as you need it:

<pre><code>out = false(size(out)); % create an output vector
out(singles) = true % fill with '1' by singles
</code></pre>

and you get:

<pre><code>out =

 0 0 0 0 1 0 1 0 0 0 0 0
</code></pre>

blocks|key|1509165|text|另一种解决方案，这完全避免了你的初始逻辑矩阵，我认为你不需要它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1509166|A+=+{123+13232+132+0+56+0+12+0+0+[]+[]+[]};
N+=+length(A);

B+=+A;+++++++++++++++++++++++++++++++++%25+helper+array
for+I+=+1+:+N
+++if+isempty+(B{I}),+B{I}+=+nan;+end;+%25+convert+empty+cells+to+nans
end++++++++++++++++++++++++++++++++++++++++++++
B+=+[nan,+B{:},+nan];++++++++++++++++++%25+pad,+and+collect+into+array

C+=+zeros+(1,+N);++++++++++++++++++++++%25+preallocate+your+answer+array
for+I+=+1+:+N;+
++if+~any+(isnan+(B(I:I%2B2)))+&&+isequal+(logical+(B(I:I%2B2)),+logical+([0,1,0]))
++++C(I)+=+1;+
++end
end
C+=+logical(C)|code-block|syntax|javascript|1509167|C+=
+++++0++0++0++0++1++0++1++0++0++0++0++0|1509168|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another solution, this completely avoids your initial logical matrix though, I don't think you need it.

<pre><code>A = {123 13232 132 0 56 0 12 0 0 [] [] []};
N = length(A);

B = A; % helper array
for I = 1 : N
 if isempty (B{I}), B{I} = nan; end; % convert empty cells to nans
end 
B = [nan, B{:}, nan]; % pad, and collect into array

C = zeros (1, N); % preallocate your answer array
for I = 1 : N; 
 if ~any (isnan (B(I:I+2))) &amp;&amp; isequal (logical (B(I:I+2)), logical ([0,1,0]))
 C(I) = 1; 
 end
end
C = logical(C)
</code></pre>



<pre><code>C =
 0 0 0 0 1 0 1 0 0 0 0 0
</code></pre>

blocks|key|1548330|text|您可以使用conv查找邻居为0的元素(请注意，该~已从isequal中删除)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1548331|out+=+cellfun(@(c)(isequal(c,0)),+A);++++++%25+find+0+elements
out+=+double(out);+++++++++++++++++++++++++%25+cast+to+double+for+conv
%25+elements+that+have+more+than+one+0+neighbor
between0+=+conv(out,+[1+-1+1],+'same')+>+1;

between0+=

+++0+++0+++0+++0+++1+++0+++1+++0+++0+++0+++0+++0|code-block|syntax|javascript|1548332|(卷积内核已更正，以修复@TasosPapastylianou发现的错误，其中3个连续的零将导致True。)|1548333|如果你想要一个逻辑向量的话。如果需要索引，只需添加find|1548334|between0+=+find(conv(out,+[1+-1+1],+'same')+>+1);+++

between0+=

+++5+++7|1548335|entityMap^0|5|4|O|1|R|7|0|0|0|P|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

You can use <code>conv</code> to find the elements with 0 neighbors (notice that the <code>~</code> has been removed from <code>isequal</code>):

<pre><code>out = cellfun(@(c)(isequal(c,0)), A); % find 0 elements
out = double(out); % cast to double for conv
% elements that have more than one 0 neighbor
between0 = conv(out, [1 -1 1], 'same') &gt; 1;

between0 =

 0 0 0 0 1 0 1 0 0 0 0 0
</code></pre>

(Convolution kernel corrected to fix bug found by @TasosPapastylianou where 3 consecutive zeros would result in True.)

That's if you want a logical vector. If you want the indices, just add <code>find</code>:

<pre><code>between0 = find(conv(out, [1 -1 1], 'same') &gt; 1); 

between0 =

 5 7
</code></pre>

I have a cell array <code>(11000x500)</code> with three different type of elements. 
1) Non-zero doubles
2) zero
3) Empty cell

I would like to find all occurances of a non-zero number between two zeros.

E.g. <code>A = {123 13232 132 0 56 0 12 0 0 [] [] []};</code>

I need the following output 
<code>out = logical([0 0 0 0 1 0 1 0 0 0 0 0]);</code>

I used <code>cellfun</code> and <code>isequal</code> like this

<pre><code>out = cellfun(@(c)(~isequal(c,0)), A);
</code></pre>

and got the follwoing output

<pre><code>out = logical([1 1 1 0 1 0 1 0 0 1 1 1]);
</code></pre>

I need help to perform the next step where i can ignore the consecutive <code>1's</code> and only take the '1's' between two <code>0's</code>

Could someone please help me with this?

Thanks!

How to find a non-zero number between two zeros in a cell array in matlab

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个单元数组(11000x500)，它有三种不同类型的元素。1)非零双精度2)零3)空单元格我想找出所有出现在两个零之间的非零数。例如，A = {123 13232 132 0 56 0 12 0 0 [] [] []};我需要以下输出out = logical([0 0 0 0 1 0 1 0 0 0 0 0])...

问如何在matlab中找到单元数组中两个零之间的非零数
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问如何在matlab中找到单元数组中两个零之间的非零数EN