Good point; I agree that defaulting to online SGD (single example per
iteration) would be a poor UX due to performance.
On Fri, Aug 7, 2015 at 12:44 PM, Meihua Wu
wrote:
> Feynman, thanks for clarifying.
>
> If we default miniBatchFraction = (1 / numInstances), then we will
> only hit one row fo
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Meihua Wu schreef
>Feynman, thanks for clarifying.
>
>If we default miniBatchFraction = (1 / numInstances), then we will
>only hit one row for every iteration of SGD regardless the number of
>partitions and executors. In other words the par
Feynman, thanks for clarifying.
If we default miniBatchFraction = (1 / numInstances), then we will
only hit one row for every iteration of SGD regardless the number of
partitions and executors. In other words the parallelism provided by
the RDD is lost in this approach. I think this is something w
Yep, I think that's what Gerald is saying and they are proposing to default
miniBatchFraction = (1 / numInstances). Is that correct?
On Fri, Aug 7, 2015 at 11:16 AM, Meihua Wu
wrote:
> I think in the SGD algorithm, the mini batch sample is done without
> replacement. So with fraction=1, then all
I think in the SGD algorithm, the mini batch sample is done without
replacement. So with fraction=1, then all the rows will be sampled
exactly once to form the miniBatch, resulting to the
deterministic/classical case.
On Fri, Aug 7, 2015 at 9:05 AM, Feynman Liang wrote:
> Sounds reasonable to me,
Sounds reasonable to me, feel free to create a JIRA (and PR if you're up
for it) so we can see what others think!
On Fri, Aug 7, 2015 at 1:45 AM, Gerald Loeffler <
gerald.loeff...@googlemail.com> wrote:
> hi,
>
> if new LinearRegressionWithSGD() uses a miniBatchFraction of 1.0,
> doesn’t that mak
hi,
if new LinearRegressionWithSGD() uses a miniBatchFraction of 1.0,
doesn’t that make it a deterministic/classical gradient descent rather
than a SGD?
Specifically, miniBatchFraction=1.0 means the entire data set, i.e.
all rows. In the spirit of SGD, shouldn’t the default be the fraction
that r
