CODEWITHSUNDEEP
matlabsparse-matrix
hhh
hhh

Reputation: 52850

Difference between similar sparse structures

My code works with this kind of structures

K>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; mlf

mlf =

                  (1,1)                       1

but it fails with this kind of inputs below where I choose the terms in mlf that are larger than zero (I cannot understand how this selection makes the input different)

K>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; mlf(mlf>0)

ans =

   (1,1)        1

where the only visual difference is some tabs/spaces.

Please, explain how they are different.

Upvotes: 0

Views: 83

Answers (2)

hhh
hhh

Reputation: 52850

Rody Oldenhuis recommended the whos

>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; mlf=mlf(mlf>0)

mlf =

   (1,1)        1


>> whos mlf
  Name      Size            Bytes  Class     Attributes

  mlf       1x1                32  double    sparse    

>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; mlf

mlf =

                  (1,1)                       1

>> whos mlf
  Name               Size            Bytes  Class     Attributes

  mlf       2147483649x1                32  double    sparse

which shows the key problem: the size of the structures have changed. chappjc provided a way to solve this problem by introducing new variables.

Upvotes: 0

chappjc
chappjc

Reputation: 30589

I think the answer is the size of the resulting array, as Rody suggested:

>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; size(mlf(mlf>0))
ans =
     1     1
>> mlf=sparse([],[],[],2^31+1,1); mlf(1)=1; size(mlf)
ans =
                2147483649                         1

*EDIT 1: Indexing works properly:

>> mlf(mlf>0) = 2
mlf =
                  (1,1)                       2

This is functionally equivalent to using find:

>> mlf(find(mlf)) = 2
mlf =
                  (1,1)                       2

It seems like a good conclusion that display is formatting the output with enough space for an element at (2147483649,1), but only when you are indexing for an assignment to that element (think lvalue vs rvalue).

*EDIT 2: If you are going after those elements in a full (not sparse) variable, use full:

>> full(mlf(mlf>0))
ans =
     1

*EDIT 3: To assign to the last element according to the dimensions of mlf rather than to the last non-zero element,

>> mlf(numel(mlf))=77
mlf =
                  (1,1)                       1
         (2147483649,1)                      77

*EDIT 4: To remove negative values:

mlf(mlf<0)=0; % or mlf(find(mlf<0)) = 0;

If you want to make a copy and remove the negatives:

mlf2 = mlf;
mlf2(mlf2<0) = 0;
mlf3 = mlf;
mlf3(mlf3>0) = 0;

Then you have mlf with all values, mlf2 with only positives, and mlf3 with only negatives. The key thing with this is that the size stays the same as with the original mlf so you are able to use the things such as end in the original way based on the size of the sparse, hurray!

Upvotes: 2

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