how to get rid of error using find too many input arguments?

[find(stock.permno)]==notrading(1);

for i=1:18522
end
  for j=10137

    newt(find(toremove(i)==c),:)=[];

    end

[Stephen23]

Note that find is not required: just use logical indexing.

Answer 1

[Walter Roberson]

In the command

[find(stock.permno)]==notrading(1);

if stock is a non-scalar structure, then it will undergo structure expansion; https://www.mathworks.com/help/matlab/matlab_prog/comma-separated-lists.html#br2js35-6 becoming the equivalent of

[find(stock(1).permno, stock(2).permno, stock(3).permno  ..., stock(end).permno)] == notrading(1);

The find command cannot have more than three arguments; https://www.mathworks.com/help/matlab/ref/find.html so if stock() is a structure array with more than three elements, the structure expansion would lead to more than three arguments in the find() call.

Possibly you want

find([stock.permno])==notrading(1);

This would be the equivalent of

find(horzcat(stock.permno))==notrading(1);

which would still be structure expansion but horzcat() allows any number of arguments and tries to put them together into rows. The [stock.permno] would create rows or 2D arrays, depending on the shape of the permno elements.

find() applied to that would give the indices of all of the non-zero values of that list.

The == operator applied to that list would compare all of those indices to the scalar value notrading(1) . A logical result, true or false, would result for each index.

Having computed the list of true and false values, you then throw all of them away without storing them or displaying them. So unless you are doing timing tests, you might as well not have computed this value.

[LM]

thank you, used first option

find([stock.permno])==notrading(1);

but now i have another error on

newt(find(toremove(i)==c),:)=[];

[Walter Roberson]

You have not posted anything about the size of newt, so we do not know.

As Stephen indicates it is better to use logical indexing:

newt(toremove(i)==c, :) = [];

Did you notice that you have

for i=1:18522
end

which is the empty "for" loop? After that loop, the value of "i" will be left at its last value, 18522. Then your "for j"

for j=10137

    newt(find(toremove(i)==c),:)=[];

    end

is going to be equivalent to

for j = 10137
   newt(toremove(18522)==c,:) = [];
end

Does toremove have that many elements? And notice that if you were to extend that for, like

for j = 1 : 10137
   newt(toremove(18522)==c,:) = [];
end

then you would be removing (or not removing) the same location in newt many times, each time with the elements afterwards "falling down" to fill the hole, like columns in tetris, until eventually there was nothing left afterwards to delete...

Answer 2

[LM]

Thank you for all your help! I have fixed it!