how to find the indices of the numbers in an vector WITHOUT using loops and find function?

for example in an vector a ={100,182,100,487,4,100,44,66,9}

the indices of 100 should be 0 2 5.

[Jan]

Without loops and FIND?! Tell your teacher that such homework questions help to learn how to use Matlab in the worst way. And the forum will have to cleanup this junk afterwards. Sorry, I cannot resist to solve the "WITHOUT builtin function" homework question. Please be sure, that you mention the forum as assistence, when you provide the solutions to your teacher.

Answer 1

[Jan]

a = [100,182,100,487,4,100,44,66,9];
b = 0:length(a) - 1;
result = b(a == 100);

[Albert Yam]

b = 1:length(a);  % unless you mean to use zero indexing on purpose

[Oleg Komarov]

@Albert: the OP requires it.

[Albert Yam]

Oh. My mistake

Answer 2

[Jan]

a = [100,182,100,487,4,100,44,66,9];
result = strfind(a, 100) - 1

[Oleg Komarov]

Uhm...I read a "find" in your solution, didn't you read the request carefully? He doesn't want to "find" it!

[Jan]

No, Oleg, the name of this command is comming from STRuggle For INDex. And, yes, I know that this name is what we call in German "behumst", because actually you get the INDex Of SUBstringS such that it should be called INDOSUBS. But the strange naming convention has been caused by the former insecure FINDSTR command, which, in fact, meant FIND STRing.

Do you find a REGEXP version?

Answer 3

[Tomas Jurena]

Hi, your problem could be solved by following function, which takes into account different indexing of elements in an array (in MATLAB, the very first element of an array is indexed by 1, whereas in C programming language it has index 0):

function ind = find_index(A,value,firstElemIndex)
i = firstElemIndex:length(A)+firstElemIndex-1;
v = (A==value);
ind = i(v);
end

Function call for your example:

a=[100,182,100,487,4,100,44,66,9];
ind = find_index(a,100,0);

Tomas

[marc]

Input argument "firstElemIndex" is undefined....

Answer 4

[Daniel Shub]

Despite thinking this is a homework assignment, I sometimes like to solve problems with silly constraints. Taking up Jan's challenge for a REGEXP solution gives

a = [100,182,100,487,4,100,44,66,9];
regexp(num2str([a==100]')', '1')-1;

but I think using regexp and strfind are both cheating (even more than asking us to to the homework).

Whenever I hear no loops, I think recursion:

function ii = isx(x, y)
  ii = [];
  if length(x) > 1
    ii = isx(x(2:end), y)+1;
  end
  if x(1) == y
    ii = [1, ii];
  end
end

a = [100,182,100,487,4,100,44,66,9];
isx(a, 100)-1