Sorry... I mean penalized likelihood, not large weight penalization.
Here's the reference I was thinking of
http://m.statisticalhorizons.com/?task=get&pageid=1424858329
On Thu, Dec 15, 2016 at 9:12 PM wrote:
> just some generic comments, I don't have any experience with penalized
> estimation n
just some generic comments, I don't have any experience with penalized
estimation nor did I go through the math.
In unregularized Logistis Regression or Logit and in several other models
the estimator satisfies some aggregation properties so that in sample or
training set proportions match between
Here's a discussion
http://stats.stackexchange.com/questions/6067/does-an-unbalanced-sample-matter-when-doing-logistic-regression
See the Zheng and King reference.
It would be nice to have these methods in scikit.
On Thu, Dec 15, 2016 at 7:05 PM Rachel Melamed wrote:
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Stuart,
Yes the data is quite imbalanced (this is what I meant by p(success) < .05 )
To be clear, I calculate
\sum_i \hat{y_i} = logregN.predict_proba(design)[:,1]*(success_fail.sum(axis=1))
and compare that number to the observed number of success. I find the predicted
number to always be higher
I don't mean that scikit-learn's modeling is non-deterministic -- I mean
the pickle library. Same input different serialized bytes output. It was my
recollection that dictionaries were inconsistently ordered when serialized,
or some the object ID was included in the serialization -- anyhow I don't
Sorry just saw you are not using the liblinear solver, agree with
Sebastian, you should subtract mean not median
On 15 Dec 2016 11:02 pm, "Sean Violante" wrote:
> The problem is the (stupid!) liblinear solver that also penalises the
> intercept (in regularisation) . Use a different solver or
The problem is the (stupid!) liblinear solver that also penalises the
intercept (in regularisation) . Use a different solver or change the
intercept_scaling parameter
On 15 Dec 2016 10:44 pm, "Sebastian Raschka" wrote:
> Subtracting the median wouldn’t result in normalizing the usual sense,
>
Subtracting the median wouldn’t result in normalizing the usual sense, since
subtracting a constant just shifts the values by a constant. Instead, for
logistic regression & most optimizers, I would recommend subtracting the mean
to center the features at mean zero and divide by the standard devi
LR is biased with imbalanced datasets. Is your dataset unbalanced? (e.g. is
there one class that has a much smaller prevalence in the data that the
other)?
On Thu, Dec 15, 2016 at 1:02 PM, Rachel Melamed
wrote:
> I just tried it and it did not appear to change the results at all?
> I ran it as f
I just tried it and it did not appear to change the results at all?
I ran it as follows:
1) Normalize dummy variables (by subtracting median) to make a matrix of about
1 x 5
2) For each of the 1000 output variables:
a. Each output variable uses the same dummy variables, but not all settings o
Could you try to normalize dataset after feature dummy encoding and see if
it is reproducible behavior?
2016-12-15 22:03 GMT+03:00 Rachel Melamed :
> Thanks for the reply. The covariates (“X") are all dummy/categorical
> variables. So I guess no, nothing is normalized.
>
> On Dec 15, 2016, at 1
Thanks for the reply. The covariates (“X") are all dummy/categorical
variables. So I guess no, nothing is normalized.
On Dec 15, 2016, at 1:54 PM, Alexey Dral
mailto:aad...@gmail.com>> wrote:
Hi Rachel,
Do you have your data normalized?
2016-12-15 20:21 GMT+03:00 Rachel Melamed
mailto:mela
Hi Rachel,
Do you have your data normalized?
2016-12-15 20:21 GMT+03:00 Rachel Melamed :
> Hi all,
> Does anyone have any suggestions for this problem:
> http://stackoverflow.com/questions/41125342/sklearn-
> logistic-regression-gives-biased-results
>
> I am running around 1000 similar logistic
Hi all,
Does anyone have any suggestions for this problem:
http://stackoverflow.com/questions/41125342/sklearn-logistic-regression-gives-biased-results
I am running around 1000 similar logistic regressions, with the same covariates
but slightly different data and response variables. All of my re
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