Fill a matrix with same values

2012 10 20
4 个回答
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0 个投票

Hi! I have a matrix like this:
[ 1 8 9 7 7 4]
[ 0 0 0 0 0 0]
[ 0 0 0 0 0 0]
[ 9 8 6 5 5 1]
[ 0 0 0 0 0 0]
And I want to fill it like this
[ 1 8 9 7 7 4]
[ 1 8 9 7 7 4]
[ 1 8 9 7 7 4]
[ 9 8 6 5 5 1]
[ 9 8 6 5 5 1]
It seems to be very easy, but I cannot realize how to do it. Thank you in advice!

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 采纳的回答

Matt J
Matt J 2012-10-20
编辑：Matt J 2012-10-20

1 个投票

Another way,
e=find(any(A,2));
[h,b]=histc(1:size(A,1),[e;inf]);
A=A(e(b),:);

7 个评论
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Matt Fig
Matt Fig 2012-10-20
编辑：Matt Fig 2012-10-20
Nice!
I think this would be somewhat faster:
I = any(A,2);
F = find(I);
A = A(F(cumsum(I)),:);
Image Analyst
Image Analyst 2012-10-20
Of course not faster than
A =[ 1 8 9 7 7 4
1 8 9 7 7 4
1 8 9 7 7 4
9 8 6 5 5 1
9 8 6 5 5 1]
Matt Fig
Matt Fig 2012-10-20
Counting typing time? Ha! What about for:
N = 50000;
A = randi(10,N,10);
A(randperm(N,floor(3*N/4)),:) = 0;
A(1) = 9; % First row is not zero
Matt J
Matt J 2012-10-20
|I think this would be somewhat faster:
I = any(A,2);
F = find(I);
A = A(F(cumsum(I)),:);|
Yes, perhaps!
Image Analyst
Image Analyst 2012-10-20
Well like my answer below says, I'm not making any assumptions about being general, as in it might be different sizes or numbers than what he explicitly put there. Actually, this was a gentle hint to Alex and others who post this stuff all the time: "I have this, and I want that". And they don't say what might vary from one case/situation to another. So I say "just make that directly." like I did when I pasted what he wanted. If that doesn't work for them then they should say what allowances need to be made for the algorithm to work for them, like the number of rows or columns or values can vary, etc. It can waste people's time. For example if someone says I have [-2 -2] and I need [2 2] and someone says just take abs(A). Then they come back and say, "No, that doesn't work because when I put in [10 20] I get [10 20] when it should really be [14 24]" Well it did work for the one situation given but not for the later cases, only after which hearing do we learn that they really wanted A+4, not abs(A). Anyway, Matt - I'm sure you know all this.
Matt Fig
Matt Fig 2012-10-20
I knew what you were saying, IA. I agree completely with your main point - a point we have discussed before (but it does bear repeating!).
Cheers!
Azzi Abdelmalek
Azzi Abdelmalek 2012-10-20
+1 to Fig's comment-answer

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更多回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-10-20
编辑：Azzi Abdelmalek 2012-10-20

1 个投票

A= [1 8 9 7 7 4
0 0 0 0 0 0
0 0 0 0 0 0
9 8 6 5 5 1
0 0 0 0 0 0]
for k=find(~all(A,2))'
A(k,:)=A(k-1,:)
end
Image Analyst
Image Analyst 2012-10-20

1 个投票

You're all making assumptions of criteria Alex did not give. Making no assumptions whatsoever, this is the fastest way I can think of:
A =[ 1 8 9 7 7 4
1 8 9 7 7 4
1 8 9 7 7 4
9 8 6 5 5 1
9 8 6 5 5 1]

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