2013-01-20 30 views

回答

1

我認爲你可以使用sum

octave:23> m = [1 2 3 4; 2 1 1 1; 2 2 3 1; 0 0 0 0] 
m = 

    1 2 3 4 
    2 1 1 1 
    2 2 3 1 
    0 0 0 0 

octave:24> m(length(m), :) = sum(m) 
m = 

    1 2 3 4 
    2 1 1 1 
    2 2 3 1 
    5 5 7 6 
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