The multiplication of list of matrices
3
$begingroup$
I need to multiply 3 lists of matrices (b.a.b) as the following code
a = {{{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`,
5.298073701591974`}}, {{-17.812203521929003`,
-1.5013126607114478`}, {-1.5013126574896714`,
4.384050851253119`}}, {{-17.801677045750512`,
-1.4055541329078751`}, {-1.405554138172727`,
3.869511752542245`}}};
b = {{{0.8409518416651456`, 0}, {0,
0.1274293000222242`}}, {{0.8409815693580924`, 0}, {0,
0.14187218616724442`}}, {{0.841011296000238`, 0}, {0,
0.15290209433231844`}}};
I used the following:
mat = b.a.b
But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication.
a1 = {{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`, 5.298073701591974`}};
b1 = {{0.8409518416651456`, 0}, {0, 0.1274293000222242`}};
mat1 = b1.a1.b1
{{-12.6042, -0.177516}, {-0.177516, 0.0860313}}
Thanks.
list-manipulation matrix
asked Mar 18 at 20:56
GhadyGhady
696
$endgroup$
add a comment |
3
$begingroup$
I need to multiply 3 lists of matrices (b.a.b) as the following code
a = {{{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`,
5.298073701591974`}}, {{-17.812203521929003`,
-1.5013126607114478`}, {-1.5013126574896714`,
4.384050851253119`}}, {{-17.801677045750512`,
-1.4055541329078751`}, {-1.405554138172727`,
3.869511752542245`}}};
b = {{{0.8409518416651456`, 0}, {0,
0.1274293000222242`}}, {{0.8409815693580924`, 0}, {0,
0.14187218616724442`}}, {{0.841011296000238`, 0}, {0,
0.15290209433231844`}}};
I used the following:
mat = b.a.b
But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication.
a1 = {{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`, 5.298073701591974`}};
b1 = {{0.8409518416651456`, 0}, {0, 0.1274293000222242`}};
mat1 = b1.a1.b1
{{-12.6042, -0.177516}, {-0.177516, 0.0860313}}
Thanks.
list-manipulation matrix
asked Mar 18 at 20:56
GhadyGhady
696
$endgroup$
1
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}]or#[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].
$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago
add a comment |
3
3
3
$begingroup$
I need to multiply 3 lists of matrices (b.a.b) as the following code
a = {{{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`,
5.298073701591974`}}, {{-17.812203521929003`,
-1.5013126607114478`}, {-1.5013126574896714`,
4.384050851253119`}}, {{-17.801677045750512`,
-1.4055541329078751`}, {-1.405554138172727`,
3.869511752542245`}}};
b = {{{0.8409518416651456`, 0}, {0,
0.1274293000222242`}}, {{0.8409815693580924`, 0}, {0,
0.14187218616724442`}}, {{0.841011296000238`, 0}, {0,
0.15290209433231844`}}};
I used the following:
mat = b.a.b
But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication.
a1 = {{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`, 5.298073701591974`}};
b1 = {{0.8409518416651456`, 0}, {0, 0.1274293000222242`}};
mat1 = b1.a1.b1
{{-12.6042, -0.177516}, {-0.177516, 0.0860313}}
Thanks.
list-manipulation matrix
asked Mar 18 at 20:56
GhadyGhady
696
$endgroup$
I need to multiply 3 lists of matrices (b.a.b) as the following code
a = {{{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`,
5.298073701591974`}}, {{-17.812203521929003`,
-1.5013126607114478`}, {-1.5013126574896714`,
4.384050851253119`}}, {{-17.801677045750512`,
-1.4055541329078751`}, {-1.405554138172727`,
3.869511752542245`}}};
b = {{{0.8409518416651456`, 0}, {0,
0.1274293000222242`}}, {{0.8409815693580924`, 0}, {0,
0.14187218616724442`}}, {{0.841011296000238`, 0}, {0,
0.15290209433231844`}}};
I used the following:
mat = b.a.b
But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication.
a1 = {{-17.8227277373099`, -1.6565234964560602`},
{-1.6565234954649242`, 5.298073701591974`}};
b1 = {{0.8409518416651456`, 0}, {0, 0.1274293000222242`}};
mat1 = b1.a1.b1
{{-12.6042, -0.177516}, {-0.177516, 0.0860313}}
Thanks.
list-manipulation matrix
list-manipulation matrix
asked Mar 18 at 20:56
GhadyGhady
696
asked Mar 18 at 20:56
GhadyGhady
696
asked Mar 18 at 20:56
GhadyGhady
696
asked Mar 18 at 20:56
GhadyGhady
696
asked Mar 18 at 20:56
GhadyGhady
696
696
1
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}]or#[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].
$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago
add a comment |
1
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}]or#[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].
$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago
1
1
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}] or #[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}] or #[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
8
$begingroup$
Using MapThread and Dot:
MapThread[Dot, {b, a, b}]
answered Mar 18 at 21:21
swishswish
4,1761535
$endgroup$
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
add a comment |
6
$begingroup$
If you need it really fast, then use Compile:
n = 1000000;
a = RandomReal[{-1, 1}, {n, 2, 2}];
b = RandomReal[{-1, 1}, {n, 2, 2}];
cf = Compile[{{a, _Real, 2}, {b, _Real, 2}},
b.a.b,
RuntimeAttributes -> Listable,
Parallelization -> True
];
MapThread[Dot, {b, a, b}]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First
1.46343
0.075722
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
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
8
$begingroup$
Using MapThread and Dot:
MapThread[Dot, {b, a, b}]
answered Mar 18 at 21:21
swishswish
4,1761535
$endgroup$
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
add a comment |
8
$begingroup$
Using MapThread and Dot:
MapThread[Dot, {b, a, b}]
answered Mar 18 at 21:21
swishswish
4,1761535
$endgroup$
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
add a comment |
8
8
8
$begingroup$
Using MapThread and Dot:
MapThread[Dot, {b, a, b}]
answered Mar 18 at 21:21
swishswish
4,1761535
$endgroup$
Using MapThread and Dot:
MapThread[Dot, {b, a, b}]
answered Mar 18 at 21:21
swishswish
4,1761535
answered Mar 18 at 21:21
swishswish
4,1761535
answered Mar 18 at 21:21
swishswish
4,1761535
answered Mar 18 at 21:21
swishswish
4,1761535
4,1761535
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
add a comment |
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
2 days ago
add a comment |
6
$begingroup$
If you need it really fast, then use Compile:
n = 1000000;
a = RandomReal[{-1, 1}, {n, 2, 2}];
b = RandomReal[{-1, 1}, {n, 2, 2}];
cf = Compile[{{a, _Real, 2}, {b, _Real, 2}},
b.a.b,
RuntimeAttributes -> Listable,
Parallelization -> True
];
MapThread[Dot, {b, a, b}]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First
1.46343
0.075722
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
add a comment |
6
$begingroup$
If you need it really fast, then use Compile:
n = 1000000;
a = RandomReal[{-1, 1}, {n, 2, 2}];
b = RandomReal[{-1, 1}, {n, 2, 2}];
cf = Compile[{{a, _Real, 2}, {b, _Real, 2}},
b.a.b,
RuntimeAttributes -> Listable,
Parallelization -> True
];
MapThread[Dot, {b, a, b}]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First
1.46343
0.075722
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
add a comment |
6
6
6
$begingroup$
If you need it really fast, then use Compile:
n = 1000000;
a = RandomReal[{-1, 1}, {n, 2, 2}];
b = RandomReal[{-1, 1}, {n, 2, 2}];
cf = Compile[{{a, _Real, 2}, {b, _Real, 2}},
b.a.b,
RuntimeAttributes -> Listable,
Parallelization -> True
];
MapThread[Dot, {b, a, b}]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First
1.46343
0.075722
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
$endgroup$
If you need it really fast, then use Compile:
n = 1000000;
a = RandomReal[{-1, 1}, {n, 2, 2}];
b = RandomReal[{-1, 1}, {n, 2, 2}];
cf = Compile[{{a, _Real, 2}, {b, _Real, 2}},
b.a.b,
RuntimeAttributes -> Listable,
Parallelization -> True
];
MapThread[Dot, {b, a, b}]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First
1.46343
0.075722
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
edited 2 days ago
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
answered Mar 18 at 23:36
Henrik SchumacherHenrik Schumacher
57.4k578158
57.4k578158
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
Thank you very much!
$endgroup$
– Ghady
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
2 days ago
add a comment |
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1
$begingroup$
Table[b[[i]].a[[i]].b[[i]], {i, 1, 3}]or#[[2]].#[[1]].#[[2]] & /@ Transpose[{a, b}].$endgroup$
– corey979
Mar 18 at 21:06
$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
2 days ago