apply condition on Matrix?

조회 수: 8 (최근 30일)
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
댓글: Ali Mukhtar 2019년 9월 9일
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
  댓글 수: 12
madhan ravi
madhan ravi 2019년 9월 9일
Why not use a for loop? Is loop necessary?
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
not it is not neccesary this was last what i tried so sent it to you

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채택된 답변

madhan ravi
madhan ravi 2019년 9월 9일
편집: madhan ravi 2019년 9월 9일
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 >=
  댓글 수: 5
madhan ravi
madhan ravi 2019년 9월 9일
nnz(B==1) >= T
%^^^^^^^^--- counts the number of ones
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
we can use it with while loop
while or if nnz(B==1)>= T ?

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추가 답변 (1개)

David Hill
David Hill 2019년 9월 9일
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.
  댓글 수: 3
David Hill
David Hill 2019년 9월 9일
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 2019년 9월 9일
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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