the tranpose of transpose doed not equals the original matrix

Question

Hantao Nie 2023-10-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2040331-the-tranpose-of-transpose-doed-not-equals-the-original-matrix

编辑： James Tursa 2024-1-23

a.mat

I find that on my data(see the attachment) ,the tranpose of transpose doed not equals the original matrix. This is my code. I donot understand why (a')' does not equals a. My matlab version is R2022a.

load("a.mat")
norm((a')' - a, 'fro')
ans = 140.4280

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2023-10-30

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2040331-the-tranpose-of-transpose-doed-not-equals-the-original-matrix#answer_1343236

移动：Walter Roberson 2023-10-30

在 MATLAB Online 中打开

a.mat

You have a broken sparse matrix, probably build by a third-party product, but possibly built by a broken mex file.

See a similar discussion at https://www.mathworks.com/matlabcentral/answers/1954909-sparse-matrix-entries-are-not-displayed-correctly#answer_1225544

load("a.mat")
whos a
  Name          Size             Bytes  Class     Attributes

  a         30084x1             157808  double    sparse    
at = a';
att = at';
whos a at att
  Name          Size                Bytes  Class     Attributes

  a         30084x1                157808  double    sparse    
  at            1x30084            398472  double    sparse    
  att       30084x1                157808  double    sparse    
mask = att ~= a;
locs = find(mask, 1)
locs = 499
a(locs)
ans = 
   All zero sparse: 1×1
at(locs)
ans = 
   (1,1)      -0.1000
att(locs)
ans = 
   (1,1)      -0.1000
d = att - a;
max(d(:))
ans = 
   (1,1)        1
min(d(:))
ans = 
   (1,1)       -1
nnz(imag(d))
ans = 0

2 个评论
显示无隐藏无

James Tursa 2023-10-30

编辑：James Tursa 2023-10-30

I see that Tim Davis's SPOK has been removed from the FEX. I will write a replacement and upload it when I get some free time, maybe this weekend. Should I call it SPOK for consistency in searches or give it a new name?

Walter Roberson 2023-10-30

In view of tomorrow's date, perhaps it should be SPOOK ;-)

请先登录，再进行评论。

Answer 2

James Tursa 2024-1-23

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2040331-the-tranpose-of-transpose-doed-not-equals-the-original-matrix#answer_1395136

编辑：James Tursa 2024-1-23

在 MATLAB Online 中打开

I finally finished my sparse matrix integrity checker. You can find it in the FEX under the name SAREK (I think you will be able to figure out the inside joke):

https://www.mathworks.com/matlabcentral/fileexchange/158161-sarek-sparse-analyzer-real-et-komplex?s_tid=srchtitle

Here is what it reports for the posted matrix:

>> sarek(a)
SAREK -- Sparse Analyzer Real Et Komplex , by James Tursa
Compiled in version R2023a (with -R2018a option)
Running  in version R2023a
Matrix is double ...
Matrix is real ...
M = 30084 >= 0 OK ...
N = 1 >= 0 OK ...
Nzmax = 9862 >= 1 OK ...
Jc[0] == 0 OK ...
Jc[N] = 9862 <= Nzmax = 9862 OK ...
Jc array OK ...
ERROR: There are 9861 Ir entries out of order or out of range
TO FIX: [M,N] = size(A); [I,J,V] = find(A); B = sparse(I,J,V,max(max(I),M),N);
All stored elements nonzero OK ...
    
There were ERRORS found in matrix!
ans =
9861

And if you plot the row indexes, you can see that the matrix was built in reverse order since they are all decreasing!

>> [I,J,V] = find(a);
>> plot(I)

After applying the recommended fix and re-plotting you can see that the row indexing order has been corrected:

>> [M,N] = size(a); [I,J,V] = find(a); B = sparse(I,J,V,max(max(I),M),N);
>> sarek(B)
SAREK -- Sparse Analyzer Real Et Komplex , by James Tursa
Compiled in version R2020a (with -R2018a option)
Running  in version R2020a
Matrix is double ...
Matrix is real ...
M = 30084 >= 0 OK ...
N = 1 >= 0 OK ...
Nzmax = 9862 >= 1 OK ...
Jc[0] == 0 OK ...
Jc[N] = 9862 <= Nzmax = 9862 OK ...
Jc array OK ...
Ir array OK ...
All stored elements nonzero OK ...
    
All sparse integrity checks OK
ans =
0
>> [I,J,K] = find(B);
>> plot(I)