apply condition on Matrix?

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Ali Mukhtar
Ali Mukhtar 2019-9-9
评论： Ali Mukhtar 2019-9-9
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 2019-9-9
Why not use a for loop? Is loop necessary?
Ali Mukhtar
Ali Mukhtar 2019-9-9
not it is not neccesary this was last what i tried so sent it to you

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采纳的回答

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

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更多回答（1 个）

David Hill
David Hill 2019-9-9
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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David Hill
David Hill 2019-9-9
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 2019-9-9
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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