Originally, it was this technique was used to estimate a sampling distribution.
Think of the drawing with replacement as work-around for generating *new* data
from a population that is simulated by this repeated sampling from the given
dataset with replacement.
For more details, I’d recommend
So what is the point of having duplicate entries in your training set? This
seems just a pure overhead. Sorry but you will again have to help me here.
On Tue, Oct 4, 2016 at 1:29 AM, Sebastian Raschka
wrote:
> > Hi,
> >
> > That helped a lot. Thank you very much. I have
> Hi,
>
> That helped a lot. Thank you very much. I have one more (silly?) doubt though.
>
> Won't an n-sized bootstrapped sample have repeated entries? Say we have an
> original dataset of size 100. A bootstrap sample (say, B) of size 100 is
> drawn from this set. Since 32 of the original
Hi,
That helped a lot. Thank you very much. I have one more (silly?) doubt
though.
Won't an n-sized bootstrapped sample have repeated entries? Say we have an
original dataset of size 100. A bootstrap sample (say, B) of size 100 is
drawn from this set. Since 32 of the original samples are left
Congrats, hope to see lot's more ;)
On 10/03/2016 12:09 PM, Raghav R V wrote:
Thanks everyone! Looking forward to contributing more :D
On Mon, Oct 3, 2016 at 5:40 PM, Ronnie Ghose > wrote:
congrats! :)
On Mon, Oct 3, 2016 at
Or maybe more intuitively, you can visualize this asymptotic behavior e.g., via
import matplotlib.pyplot as plt
vs = []
for n in range(5, 201, 5):
v = 1 - (1. - 1./n)**n
vs.append(v)
plt.plot([n for n in range(5, 201, 5)], vs, marker='o',
markersize=6,
alpha=0.5,)
Great.
Thanks for your time Manoj
Cheers,
Klo
On Mon, Oct 3, 2016 at 8:20 PM, Manoj Kumar
wrote:
> Let's say you would like to generate just the first feature of 1000
> samples with label 0.
>
> The distribution of the first feature conditioned on label 1
Say the probability that a given sample from a dataset of size n is *not* drawn
as a bootstrap sample is
P(not_chosen) = (1 - 1\n)^n
Since you have a 1/n chance to draw a particular sample (since bootstrapping
involves drawing with replacement), which you repeat n times to get a n-sized
Hi,
>From docs
http://scikit-learn.org/stable/auto_examples/ensemble/plot_ensemble_oob.html
:
The RandomForestClassifier is trained using bootstrap aggregation, where
each new tree is fit from a bootstrap sample of the training observations
z_i = (x_i, y_i). The out-of-bag (OOB) error is the
> From whatever little knowledge I gained last night about Random Forests, each
> tree is trained with a sub-sample of original dataset (usually with
> replacement)?.
Yes, that should be correct!
> Now, what I am not able to understand is - if entire dataset is used to train
> each of the
Let's say you would like to generate just the first feature of 1000 samples
with label 0.
The distribution of the first feature conditioned on label 1 follows a
Bernoulli distribution (as suggested by the name) with parameter
"exp(feature_log_prob_[0, 0])". You could then generate the first
Hi Manoj,
thanks for your reply.
Sorry to say, but I don't understand how to generate new feature.
In this example I have `X` with shape (1000, 64) with 5 unique classes.
`feature_log_prob_` has shape (5, 64)
I can generate for example uniform data with `r = np.random.rand(64)`
Now how can I
Congrats Raghav!
On Mon, Oct 3, 2016 at 10:06 AM, Sebastian Raschka
wrote:
> Congrats Raghav! And thanks a lot for all the great work on the
> model_selection module!
>
> > On Oct 3, 2016, at 12:53 PM, Siddharth Gupta <
> siddharthgupta...@gmail.com> wrote:
> >
> >
Congrats Raghav! And thanks a lot for all the great work on the model_selection
module!
> On Oct 3, 2016, at 12:53 PM, Siddharth Gupta
> wrote:
>
> Congrats Raghav! :D
>
>
> On Oct 3, 2016 10:22 PM, "Aakash Agarwal" wrote:
> Congrats
Congratulations!
On Mon, Oct 3, 2016 at 12:21 PM, Nelle Varoquaux
wrote:
> Congratulation Raghav!
>
> On 3 October 2016 at 08:40, Ronnie Ghose wrote:
> > congrats! :)
> >
> > On Mon, Oct 3, 2016 at 11:28 AM, lin yenchen
Congratulation Raghav!
On 3 October 2016 at 08:40, Ronnie Ghose wrote:
> congrats! :)
>
> On Mon, Oct 3, 2016 at 11:28 AM, lin yenchen
> wrote:
>>
>> Congrats, Raghav!
>>
>> Nelson Liu 於 2016年10月3日 週一 下午11:27寫道:
>>>
>>> Yay!
congrats! :)
On Mon, Oct 3, 2016 at 11:28 AM, lin yenchen
wrote:
> Congrats, Raghav!
>
> Nelson Liu 於 2016年10月3日 週一 下午11:27寫道:
>
>> Yay! Congrats, Raghav!
>>
>> On Mon, Oct 3, 2016 at 8:14 AM, Gael Varoquaux <
>> gael.varoqu...@normalesup.org> wrote:
>>
Congrats Raghav. :)
On Mon, Oct 3, 2016 at 5:28 PM, lin yenchen
wrote:
> Congrats, Raghav!
>
> Nelson Liu 於 2016年10月3日 週一 下午11:27寫道:
>
>> Yay! Congrats, Raghav!
>>
>> On Mon, Oct 3, 2016 at 8:14 AM, Gael Varoquaux <
>> gael.varoqu...@normalesup.org>
Congrats, Raghav!
Nelson Liu 於 2016年10月3日 週一 下午11:27寫道:
> Yay! Congrats, Raghav!
>
> On Mon, Oct 3, 2016 at 8:14 AM, Gael Varoquaux <
> gael.varoqu...@normalesup.org> wrote:
>
> Hi,
>
> We have the pleasure to welcome Raghav RV to the core-dev team. Raghav
> (@raghavrv) has been
Yay! Congrats, Raghav!
On Mon, Oct 3, 2016 at 8:14 AM, Gael Varoquaux <
gael.varoqu...@normalesup.org> wrote:
> Hi,
>
> We have the pleasure to welcome Raghav RV to the core-dev team. Raghav
> (@raghavrv) has been working on scikit-learn for more than a year. In
> particular, he implemented the
Hi,
feature_log_prob_ is an array of size (n_classes, n_features).
exp(feature_log_prob_[class_ind, feature_ind]) gives P(X_{feature_ind} = 1
| class_ind)"
Using the conditional independence assumptions of NaiveBayes, you can use
this to sample each feature independently given the class.
Hope
Hi,
We have the pleasure to welcome Raghav RV to the core-dev team. Raghav
(@raghavrv) has been working on scikit-learn for more than a year. In
particular, he implemented the rewrite of the cross-validation utilities,
which is quite dear to my heart.
Welcome Raghav!
Gaël
On Mon, Oct 3, 2016 at 5:08 PM, klo uo wrote:
> I can see how can I sample from `feature_log_prob_`...
>
I meant I cannot see
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Hm it sounds like "weights" should have been called "weighting" maybe?
Not sure if it's worth changing now, as we released it already.
And I think passing the weighting to the confusion matrix is correct.
There should be tests for weighted metrics to confirm that.
PR welcome.
On 10/03/2016
Hi Klo.
Yes, you could, but as the model is very simple, that's usually not very
interesting.
It stores for each label an independent Bernoulli distribution for each
feature.
these are stored in feature_log_prob_.
I would suggest you look at this attribute, rather than sample from the
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