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

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madhan ravi
madhan ravi 2019년 9월 9일
편집: madhan ravi 2019년 9월 9일
What have you tried for your homework?
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
i apply a for loop and put while condition until B matrix is have 1's but that is not correct
madhan ravi
madhan ravi 2019년 9월 9일
편집: madhan ravi 2019년 9월 9일
Post the code that you tried.
youcha
youcha 2019년 9월 9일
Can you please explain more of what do you want to do with your while loop? It should be in a form of : I want to do action B while the condition A is true. Your condition in this ase is B>=T while your action is not clear
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
if i can count the number of 1's in B matrix i can apply the condition
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
for i=1
for j=1:n %n= number of one in matrix B
while B(i,j)>=T
end
madhan ravi
madhan ravi 2019년 9월 9일
hint: If one add one or else continue
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
dnt understand in which one i have to add one? can i count number of 1's in B matrix?
madhan ravi
madhan ravi 2019년 9월 9일
Your goal is to count the number of ones in B and then finally check if it's greater than or equal to T , is that right?
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
yes exactly
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

1 개 추천

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 >=

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Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
nnz()?
help nnz % will tell you what it actually does [hint: this is to count the no of ones present in B]
Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
can you please write few code lines it become more and more complex
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

0 개 추천

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.

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Ali Mukhtar
Ali Mukhtar 2019년 9월 9일
while sum(double(B)-48)>T
B is array which contains 1's and 0's . in above code line can you explain why you use 48 and subtract it from B
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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