Why comparing matrices is not evaluated as boolean in Octave?
up vote
0
down vote
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Im new to Octave and playing around with the console.
why when comparing matrices, the expression is not evaluates as boolean :
example:
>> A=[1,2;3,4]; % creating 2x2 matrix
>> 5 == 5 % sample comparison returns true (1)
ans = 1
>> A(1,1) == A(1,1) % single element comparison returns true (1)
ans = 1
>> A == A % returns 2x2 matrix ???
ans =
1 1
1 1
>> size(A == A) % prove that the above returns 2x2 matrix
ans =
2 2
arrays matlab matrix octave
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
add a comment |
up vote
0
down vote
favorite
Im new to Octave and playing around with the console.
why when comparing matrices, the expression is not evaluates as boolean :
example:
>> A=[1,2;3,4]; % creating 2x2 matrix
>> 5 == 5 % sample comparison returns true (1)
ans = 1
>> A(1,1) == A(1,1) % single element comparison returns true (1)
ans = 1
>> A == A % returns 2x2 matrix ???
ans =
1 1
1 1
>> size(A == A) % prove that the above returns 2x2 matrix
ans =
2 2
arrays matlab matrix octave
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im new to Octave and playing around with the console.
why when comparing matrices, the expression is not evaluates as boolean :
example:
>> A=[1,2;3,4]; % creating 2x2 matrix
>> 5 == 5 % sample comparison returns true (1)
ans = 1
>> A(1,1) == A(1,1) % single element comparison returns true (1)
ans = 1
>> A == A % returns 2x2 matrix ???
ans =
1 1
1 1
>> size(A == A) % prove that the above returns 2x2 matrix
ans =
2 2
arrays matlab matrix octave
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
Im new to Octave and playing around with the console.
why when comparing matrices, the expression is not evaluates as boolean :
example:
>> A=[1,2;3,4]; % creating 2x2 matrix
>> 5 == 5 % sample comparison returns true (1)
ans = 1
>> A(1,1) == A(1,1) % single element comparison returns true (1)
ans = 1
>> A == A % returns 2x2 matrix ???
ans =
1 1
1 1
>> size(A == A) % prove that the above returns 2x2 matrix
ans =
2 2
arrays matlab matrix octave
arrays matlab matrix octave
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
edited Nov 11 at 23:15
Luis Mendo
92.2k1054120
92.2k1054120
asked Nov 11 at 19:50
chenchuk
1,96231322
asked Nov 11 at 19:50
chenchuk
1,96231322
asked Nov 11 at 19:50
chenchuk
1,96231322
1,96231322
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
== is for element-wise comparison of two matrices. To check whether two matrices are same or not, use isequal.
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
add a comment |
up vote
0
down vote
Sardar's answer is correct, but when it comes to computational time, I think that my alternative answer is better: You can as well check that all elements of the boolean matrix A == A are 1, i.e., that the sum of 1s in the matrix A==A equals the number of elements of A, i.e:
sum((A == A)(:)) == numel(A)
ans = 1
Where the operator (:) simply vectorizes the matrix A==A so that it can be added with sum(). Compare the two answers when you matrix is quite big, for instance by defining A = rand(1e4), the computation time is considerably different...
answered Nov 29 at 9:59
nukimov
415
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
== is for element-wise comparison of two matrices. To check whether two matrices are same or not, use isequal.
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
add a comment |
up vote
4
down vote
accepted
== is for element-wise comparison of two matrices. To check whether two matrices are same or not, use isequal.
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
== is for element-wise comparison of two matrices. To check whether two matrices are same or not, use isequal.
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
== is for element-wise comparison of two matrices. To check whether two matrices are same or not, use isequal.
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
answered Nov 11 at 19:55
Sardar Usama
14.4k82244
14.4k82244
add a comment |
add a comment |
up vote
0
down vote
Sardar's answer is correct, but when it comes to computational time, I think that my alternative answer is better: You can as well check that all elements of the boolean matrix A == A are 1, i.e., that the sum of 1s in the matrix A==A equals the number of elements of A, i.e:
sum((A == A)(:)) == numel(A)
ans = 1
Where the operator (:) simply vectorizes the matrix A==A so that it can be added with sum(). Compare the two answers when you matrix is quite big, for instance by defining A = rand(1e4), the computation time is considerably different...
answered Nov 29 at 9:59
nukimov
415
add a comment |
up vote
0
down vote
Sardar's answer is correct, but when it comes to computational time, I think that my alternative answer is better: You can as well check that all elements of the boolean matrix A == A are 1, i.e., that the sum of 1s in the matrix A==A equals the number of elements of A, i.e:
sum((A == A)(:)) == numel(A)
ans = 1
Where the operator (:) simply vectorizes the matrix A==A so that it can be added with sum(). Compare the two answers when you matrix is quite big, for instance by defining A = rand(1e4), the computation time is considerably different...
answered Nov 29 at 9:59
nukimov
415
add a comment |
up vote
0
down vote
up vote
0
down vote
Sardar's answer is correct, but when it comes to computational time, I think that my alternative answer is better: You can as well check that all elements of the boolean matrix A == A are 1, i.e., that the sum of 1s in the matrix A==A equals the number of elements of A, i.e:
sum((A == A)(:)) == numel(A)
ans = 1
Where the operator (:) simply vectorizes the matrix A==A so that it can be added with sum(). Compare the two answers when you matrix is quite big, for instance by defining A = rand(1e4), the computation time is considerably different...
answered Nov 29 at 9:59
nukimov
415
Sardar's answer is correct, but when it comes to computational time, I think that my alternative answer is better: You can as well check that all elements of the boolean matrix A == A are 1, i.e., that the sum of 1s in the matrix A==A equals the number of elements of A, i.e:
sum((A == A)(:)) == numel(A)
ans = 1
Where the operator (:) simply vectorizes the matrix A==A so that it can be added with sum(). Compare the two answers when you matrix is quite big, for instance by defining A = rand(1e4), the computation time is considerably different...
answered Nov 29 at 9:59
nukimov
415
answered Nov 29 at 9:59
nukimov
415
answered Nov 29 at 9:59
nukimov
415
answered Nov 29 at 9:59
nukimov
415
415
add a comment |
add a comment |
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