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help with my home work plzzzz

조회 수: 1 (최근 30일)
Bader Herzallah
Bader Herzallah 2020년 5월 13일
댓글: Steven Lord 2020년 5월 13일
4. Generate the matrix X= [-10:10; 1:2:42;-10:3:52], then do the following
  • Multiply the elements that have a value less than or equal zero by 4.
  • Generate a new Matrix Y that have value 1000 if the value of X is less than or equal zero and zero otherwise.
  • Calculate number of 1000 in new matrix
%this is for a
clc
clear all
x=[-10:10;1:2:42;-10:3:52];
[r c]=size(x);
for i=1;r;
for j=1:c;
if x(i,j)<0
x(i,j)=4.*x(i,j);
else
x(i,j)=x(i,j);
end
end
end
x
-40 -36 -32 -28 -24 -20 -16 -12 -8 -4 0 1 2 3 4 5 6 7 8 9 10
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41
-10 -7 -4 -1 2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50
what i am doing wron why the third row do not get Multiply by 4

답변 (1개)

Geoff Hayes
Geoff Hayes 2020년 5월 13일
편집: Geoff Hayes 2020년 5월 13일
Bader - look closely at the line
for i=1;r;
You have a semi-colon between the 1 and the r. Replace with a colon and it will work as expected:
for i=1:r
By the way, you can remove the else body as it doesn't add any value (since it is changing the value at the given indices).
  댓글 수: 8
Geoff Hayes
Geoff Hayes 2020년 5월 13일
Bader - the code looks reasonable though you may want to consider pre-sizing the y array. When you run the code, do you get the expected output?
Steven Lord
Steven Lord 2020년 5월 13일
This works for the x you've been given, but it won't work for all x arrays. Consider this 3-dimensional x array:
x = randn(2, 3, 4);
The y you create will be 2-by-12. See the description of the sz1,sz2, ... szn output arguments on the documentation page for the size function for an explanation.
You could simplify this code in at least two different ways. If your assignment requires you to use a for loop, preallocate y to be an array the same size as x. There's an example on the documentation page for the ones function that shows how to do this. Then loop over the elements in the array (there's a function on this page that will help you count the number of elements) and use linear indexing to address each in turn.
Alternately you can skip for loops entirely by preallocating y exactly as in the previous approach then using logical indexing to change the appropriate elements.

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