Avoiding unpacking an array when altering its dimension
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
$endgroup$
add a comment |
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
$endgroup$
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}]
.
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago
add a comment |
$begingroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
$endgroup$
I am trying to flatten a matrix from 3-dimensional to 2-dimensional one.
The original one has dimension for example {10,120,120}:
list1 = RandomReal[1, {10, 120, 120}];
I want to convert it to a new matrix with {10,120*120} dimension. I am using:
On["Packing"] (*For checking if exist unpacked arrays *)
Flatten[Map[Flatten, {list1}, {-3}], 1];
But this will generate unpacked array. How should I avoid this issue?
list-manipulation packed-arrays
list-manipulation packed-arrays
edited 1 hour ago
m_goldberg
87.2k872197
87.2k872197
asked 2 hours ago
cj9435042cj9435042
35416
35416
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}]
.
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago
add a comment |
2
$begingroup$
TryFlatten[list1, {{1}, {2, 3}}]
.
$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago
2
2
$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}]
.$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}]
.$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An alternative to using Flatten
as in the comments is to use ArrayReshape
. For large arrays, ArrayReshape
should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192623%2favoiding-unpacking-an-array-when-altering-its-dimension%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An alternative to using Flatten
as in the comments is to use ArrayReshape
. For large arrays, ArrayReshape
should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
An alternative to using Flatten
as in the comments is to use ArrayReshape
. For large arrays, ArrayReshape
should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
$endgroup$
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
An alternative to using Flatten
as in the comments is to use ArrayReshape
. For large arrays, ArrayReshape
should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
$endgroup$
An alternative to using Flatten
as in the comments is to use ArrayReshape
. For large arrays, ArrayReshape
should be significantly faster. For example:
list1 = RandomReal[1, {10, 1200, 1200}];
Comparison:
r1 = Flatten[list1, {{1}, {2, 3}}]; //RepeatedTiming
r2 = ArrayReshape[list1, {10, 1200 1200}]; //RepeatedTiming
r1 === r2
{0.120, Null}
{0.025, Null}
True
answered 1 hour ago
Carl WollCarl Woll
68.9k391177
68.9k391177
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
$begingroup$
I'm too slow..... :)
$endgroup$
– Michael E2
1 hour ago
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192623%2favoiding-unpacking-an-array-when-altering-its-dimension%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Try
Flatten[list1, {{1}, {2, 3}}]
.$endgroup$
– J. M. is computer-less♦
1 hour ago
$begingroup$
Thanks, it works!
$endgroup$
– cj9435042
1 hour ago
$begingroup$
Flatten /@ list1
$endgroup$
– Okkes Dulgerci
31 mins ago