apply condition on Matrix?

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Ali Mukhtar
Ali Mukhtar el 9 de Sept. de 2019
Comentada: Ali Mukhtar el 9 de Sept. de 2019
if i have a matrix B=[1111000] and another integer T=4 i want to apply while loop if number of One's in B>=T how should i write it to get desire condition
  12 comentarios
madhan ravi
madhan ravi el 9 de Sept. de 2019
Why not use a for loop? Is loop necessary?
Ali Mukhtar
Ali Mukhtar el 9 de Sept. de 2019
not it is not neccesary this was last what i tried so sent it to you

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Respuesta aceptada

madhan ravi
madhan ravi el 9 de Sept. de 2019
Editada: madhan ravi el 9 de Sept. de 2019
Just use logical indexing "==" to see the values equal 1 and use nnz() and then use >= T if you get 1 it's true else false.
help ==
help nnz
help >=
  5 comentarios
madhan ravi
madhan ravi el 9 de Sept. de 2019
nnz(B==1) >= T
%^^^^^^^^--- counts the number of ones
Ali Mukhtar
Ali Mukhtar el 9 de Sept. de 2019
we can use it with while loop
while or if nnz(B==1)>= T ?

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Más respuestas (1)

David Hill
David Hill el 9 de Sept. de 2019
If B is a 1 by x vector of 1's and 0's (B = [1,1,1,1,1,0,0,0,0,0])
while sum(B)>T
if B is a 1 by x character array (B = '1111100000')
while sum(double(B)-48)>T
Your matrix B is not described well above, it looks like a single number.
  3 comentarios
David Hill
David Hill el 9 de Sept. de 2019
If your array is a character array '111110000', then I converted to a number array which is a ascii representative of each character and subtract 48=='0' to get an array of 1's and 0's. If you already have an array of 1's and 0's you don't need to do this.
Ali Mukhtar
Ali Mukhtar el 9 de Sept. de 2019
but i dnt know if my value is exactly 111110000 so subtracting it from 48 is not fesible . if you show result by compiling it i can use it also

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