Removing certain Rows from a matrix
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I am producing a number of columns, which are pairs of x and y coords. I.e, the following code shows two sets of x&y coordinates. I need to plot these coordinates, but somehow need to remove all of the rows where two 0's are present.
I was wondering if anyone could help me with a way to extract a set of 2 columns at a time, and only return the coordinates up to the point where the 0 0 starts.
5.0000 5.0000 5.0000 5.0000
5.0000 5.3626 2.4232 5.0000
3.5471 5.3626 0.8831 5.0000
2.9189 5.3626 0.8831 6.1958
2.9189 3.4729 0.8831 6.2884
2.9189 5.3171 0 6.2884
3.0675 5.3171 0 0
3.6450 5.3171 0 0
3.6450 5.9336 0 0
3.6450 6.5008 0 0
3.6450 8.4064 0 0
5.2608 8.4064 0 0
5.2608 6.9090 0 0
5.2608 5.5735 0 0
5.2608 7.0445 0 0
2.6390 7.0445 0 0
3.2644 7.0445 0 0
3.2644 5.1236 0 0
3.8823 5.1236 0 0
3.6361 5.1236 0 0
3.6361 5.5497 0 0
3.6361 7.4126 0 0
3.6361 7.2564 0 0
1.4501 7.2564 0 0
1.4501 7.0662 0 0
0 7.0662 0 0
The data is produced from a for loop, and the loop does not always run for the same amount of time steps.
回答 (2 件)
Paulo Silva
2011 年 9 月 3 日
n=[m(1:size(m,1),[1 2]); m(1:size(m,1),[3 4])]; %two columns
n=n(any(~ismember(n(:,[1 2]),[0 0]),2),:) %one column not 0
%n=n(all(~ismember(n(:,[1 2]),[0 0]),2),:) %both columns not 0
in response to Daniel comments, one way to have 4 columns again
n=[n;nan(1,2)] %you have to add nan values if size(n,1) isn't even
%there was one row missing, now size(n,1) is even
n=[n(1:size(n)/2,:) n((size(n)/2+1):end,:)]
13 件のコメント
Niki
2011 年 9 月 3 日
this is for when we know for example the two last columns are included zero , but if we do not know , then?
Paulo Silva
2011 年 9 月 3 日
I don't understand the question, please rephrase it
Niki
2011 年 9 月 3 日
my mean for example we have a matrix that we do not know exactly which columns are included of zeros (more than half of the rows)
then we want to recognize them and or remove them,
Paulo Silva
2011 年 9 月 3 日
Mohammad please provide me with a simple example of a matrix where that happens and what's the resulting matrix after doing what you need
Niki
2011 年 9 月 3 日
X=[2.47 0 133.717 0 0 57 0 17 20 4.554 5.361
2.377 0 124.799 0 0 34 0 10 19 4.489 5.293
2.377 0 124.799 0 34 0 10 0 19 4.489 5.293
2.398 0 148.023 0 77 0 23 0 22 4.673 5.485
2.445 0 136.51 0 0 40 0 12 20 4.533 5.333
2.676 0 138.08 0 0 46 0 14 20 4.533 5.323
2.483 0 129.093 0 0 20 0 6 19 4.466 5.252
2.514 0 121.675 0 0 0 0 0 18 4.394 5.165
2.445 0 136.51 0 40 0 12 0 20 4.533 5.333
2.465 0 160.208 0 89 0 27 0 23 4.71 5.517
2.633 0 145.498 0 20 32 13 18 21 4.595 5.398
2.445 0 136.51 0 0 32 0 18 20 4.533 5.333
]
and then i want to know which columns have a lot of zero( for example half of the number of row)
for example in this case columns 2, 4, 5 7,
Walter Roberson
2011 年 9 月 3 日
zerolike_columns = find(mean(X == 0) >= 1/2);
Niki
2011 年 9 月 3 日
walter Thanks, it works, I have another question ,
when I have a binary matrix (zero and one) I want to recognize in each rows which cell has value 1 , could you please also tell me that
like this Matrix
X=[0 0 0
0 0 0
0 0 0
1 0 0
0 0 1
0 1 1
1 0 0
0 1 0
0 0 1
1 0 0
]
Daniel
2011 年 9 月 3 日
Paulo Silva
2011 年 9 月 3 日
Another solution for the previous question
%m is the array, I prefer to use m instead of X for the array name
c=arrayfun(@(n)sum(m(:,n)==0),1:size(m,2)); %count the zeros in each column
z=5; %number of zeros
find(c>z) %find what columns have more zeros than the number you selected (z)
Niki
2011 年 9 月 3 日
thanks Paulo
for the second question i wrote this, but does not work , because i want to know in each row which cells are included of one
[m,n]=size(X)
for i=1:m
onevlue=find(X==1)
end
Paulo Silva
2011 年 9 月 3 日
solution for Mohammad second question
[row,col]=find(X==1);
result=sortrows([row col])
%first column is the rows where the ones are and second column indicates the %column of the ones, also works for rows with multiple ones
Paulo Silva
2011 年 9 月 3 日
I hope to have answered all questions the correct way, Mohammad please create new questions for your own problems, don't hijack other people questions, that makes it hard for us to help, must go now, have a nice day :)
Niki
2011 年 9 月 3 日
thanks, have a nice day
Andrei Bobrov
2011 年 9 月 3 日
Hi friends! My variant
n = reshape(permute(reshape(m,size(m,1),[],2),[1 3 2]),[],2);
out = n(any(n,2),:);
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