Timeline for Efficient linear regression given columns of A rather than rows of A
Current License: CC BY-SA 4.0
19 events
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May 28, 2022 at 7:11 | answer | added | Sextus Empiricus | timeline score: 1 | |
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S Jun 15, 2019 at 11:42 | history | suggested | burning_question | CC BY-SA 4.0 |
Fixed a typo: it should be Nx1 columns, not 1xM columns
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Jun 15, 2019 at 11:40 | comment | added | gung - Reinstate Monica | Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. | |
Jun 15, 2019 at 10:41 | review | Suggested edits | |||
S Jun 15, 2019 at 11:42 | |||||
Jun 15, 2019 at 10:38 | answer | added | burning_question | timeline score: 0 | |
S Sep 7, 2018 at 13:13 | history | suggested | burning_question | CC BY-SA 4.0 |
added more information to address questions raised in the comments
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Sep 7, 2018 at 12:01 | review | Suggested edits | |||
S Sep 7, 2018 at 13:13 | |||||
Sep 7, 2018 at 2:37 | comment | added | jbowman | It's not clear to me what you mean by "I'm only able to load $1 \times M$ columns of $A$"? Evidently you don't mean "I can load all $M$ columns of $A$"... And why would loading row-by-row not be possible? | |
Sep 6, 2018 at 19:48 | comment | added | whuber♦ | Of course: with a vector $x$ in RAM, accumulate the values of $y_i$ and $y_ix_i$ as you read in the vector $y.$ Repeat for all entries of $A^\prime A.$ Indeed, you only need $O(M^2)$ RAM if you're willing to read the columns an average of $(M+3)/2$ times each. | |
Sep 6, 2018 at 18:30 | review | First posts | |||
Sep 6, 2018 at 18:39 | |||||
Sep 6, 2018 at 18:30 | history | asked | burning_question | CC BY-SA 4.0 |