How to index on a loop for first occurrence?

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Chameleon17 2015-6-1

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编辑： Walter Roberson 2015-9-9

Hi,

I have a set of data which I would like to loop through for each row.

I would like for every row, the first occurrence of a one to cause all other subsequent values in that row to be replaced by a zero.

Does anybody have any advice on how I could go about doing this?

Thanks

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Answer 1

Andrei Bobrov 2015-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/221843-how-to-index-on-a-loop-for-first-occurrence#answer_181180

编辑：Andrei Bobrov 2015-6-1

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a = [1     0     1     0     0     0     1
   0     0     0     0     0     0
   0     0     0     1     0     0
   0     0     0     0     0     0
   0     0     1     1     0     0
   0     0     0     0     0     0
   0     0     1     0     0     0
   1     0     0     0     0     0
   0     0     0     1     0     0
   0     0     0     0     0     0
   0     0     0     0     0     0
   0     0     0     1     0     0
   0     0     1     0     0     1
   0     0     1     0     0     0
   0     0     0     0     0     0
   0     0     0     0     0     1
   0     0     0     0     0     0
   0     1     0     0     0     0
   0     0     0     0     0     0
   0     0     0     0     0     0
   0     0     0     0     1     0
   0     0     0     0     0     0
   0     0     0     0     0     1
   0     0     0     0     0     0
   0     1     0     0     1     0
   0     1     0     0     0     0
   0     0     0     0     1     0
   0     0     0     1     0     0]
out = cumsum(a==1,2)==1 & a==1;

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Walter Roberson 2015-9-9

编辑：Walter Roberson 2015-9-9

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The ==1 examines each element independently. It does not process row by row.

The & will result in 1 if both values are true, and 0 otherwise. It is a binary operator. The cumsum(a==1,2)==1 part is going to create a logical matrix the same size as "a" and the a==1 part is going to create a logical matrix the same size as "a", and then the & is going to compare corresponding elements one at a time, returning true only if both are set. Another way of writing the code would be

L1 = cumsum(a==1,2)==1;
L2 = a==1;
out = false(size(L1));
for K = 1 : size(L1,2)
  for J = 1 : size(L1,1)
    out(J,K) = L1(J,K) & L2(J,K);
  end
end

In the case where "a" contains only 0 and 1, then the code could be shorter:

out = cumsum(a,2)==1 & a;

or

out = (cumsum(a,2)==1) .* a;

If "a" can contain values other than 0 and 1, then none of the posted versions in this sub-thread are correct. I would need to think more about a good solution for that case.

Chameleon17 2015-9-9

Thank you! That helps a lot!

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Answer 2

Adam 2015-6-1

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idx = find( x == 1, 1 );
x( (idx + 1):end ) = 0;

will do this for an array x. I will leave it up to you to make the trivial change to do it in a loop.

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Chameleon17 2015-6-1

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Hi! Thank you very much for the help! I get the idea of what it's doing and I am trying to not have it work in a loop. It only seems to work for column one, row one, after that everything is a zero.

Original:

   0     1     0     0     0     1
   0     0     0     0     0     0
   0     0     0     1     0     0
   0     0     0     0     0     0
   0     0     1     1     0     0
   0     0     0     0     0     0
   0     0     1     0     0     0
   1     0     0     0     0     0
   0     0     0     1     0     0
   0     0     0     0     0     0
   0     0     0     0     0     0
   0     0     0     1     0     0
   0     0     1     0     0     1
   0     0     1     0     0     0
   0     0     0     0     0     0
   0     0     0     0     0     1
   0     0     0     0     0     0
   0     1     0     0     0     0
   0     0     0     0     0     0
   0     0     0     0     0     0
   0     0     0     0     1     0
   0     0     0     0     0     0
   0     0     0     0     0     1
   0     0     0     0     0     0
   0     1     0     0     1     0
   0     1     0     0     0     0
   0     0     0     0     1     0
   0     0     0     1     0     0

After:

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Is there a way to make it read through each row individually?

Thank you again!

Adam 2015-6-1

编辑：Adam 2015-6-1

在 MATLAB Online 中打开

Yes, it definitely won't work without the loop, but your original question suggested you wanted to use a loop to do this so something like this (untested on command line so could be a syntax error. nRows and x are just for pseudo code for you to replace with real data)

for n = 1:nRows
idx = find( x(n,:) == 1, 1 );
x( n,(idx + 1):end ) = 0;
end

Andrei's solution appears to work fine for me for not using a for loop and does actually look similar to a solution I gave to a similar question a while ago so it is a solution I like.

In general though, if speed is not an issue then the question of whether to use a loop or some other alternative that doesn't use a loop is a matter of preference where readability is often the most important factor.

If, when you come back to your own code a week later, you can no longer understand what it is doing because it uses unintuitive methods then it probably isn't worth it.

Obviously if the for loop is slow then finding an alternative is important.

Adam 2015-6-1

编辑：Adam 2015-6-1

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The final 1 after the comma is the 2nd argument to the find function which tells it to just find the first occurrence rather than all occurrences (or some other number than 1 if explicitly defined).

If you don't add that argument you will get all indices on the row corresponding to ones.

For example:

a = [0 0 1 1 0 0 1]
a =
     0     0     1     1     0     0     1
>> find( a == 1 )
ans =
     3     4     7
>> find( a == 1, 1 )
ans =
     3

When I have something like this I don't fully understand I find it invaluable to just create a quick example on the command line so I can see for myself what is going on. It is one of the big advantages of Matalb over, for example, C++ where you need to do a lot more work to just test to see what simple syntax alternatives give you.

Chameleon17 2015-6-2

Yes, I think that is really good advice, to trial it on smaller data sets, I've already reduced my data set loads to learn how to do this, but yes, nothing is going to replace understanding all the basics properly.

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