🌐 Mailman 3 February 2017 - scikit-learn - python.org

Announcing: Docathon, week of 6 March 2017
by Nelle Varoquaux Feb. 16, 2017

Feb. 16, 2017

Hi everyone, I don't really think scikit-learn's documentation is lacking, but here is an announcement for an event we are organizing called the "Docathon". Several of us will be meeting up to sprint on documentation or documentation-related projects at Berkeley, New York and Seattle. If you are interested in joining us, either remotely or on campus, don't hesitate to join! Cheers, Nelle *What's a Docathon?* It's a week-long sprint where we focus our efforts on improving the state of documentation in the open-source and open-science world. This means writing better documentation, building tools, and sharing skills. *Who’s this for?* Anyone who is interested in improving the understandability, accessibility, and clarity of software! This might mean developers with a particular project, or individuals who would like to contribute to a project. You don’t need to use a specific language (though there will be many Python and R developers) and you don’t need to be a core developer in order to help out. *Where can I sign up?* Check out the *Docathon website* <https://bids.github.io/docathon/>. You can sign up as a *participant* <https://goo.gl/forms/AaW2b24mMxOutxt02>, *suggest a project* <https://goo.gl/forms/0cPpw01zehrEyDDE3> to work on, or sign up *to host your own* <https://bids.github.io/docathon/pages/hosting.html> remote Docathon wherever you like. You don’t have to use a specific language - we’ll be as accommodating as possible! *When is the Docathon?* The Docathon will be held *March 6 through March 10*. For those coming to BIDS at UC Berkeley, on the first day we'll have tutorials about documentation and demos of documentation tools, followed by a few hours of hacking. During the middle of the week, we'll set aside a few hours each afternoon for hacking as a group at BIDS. On the last day, we'll have a wrap-up event to show off what everybody worked on. *Where will the Docathon take place?* There are a *few docathons being held simultaneously* <https://bids.github.io/docathon/pages/hosts.html>, each with their own schedule. At Berkeley we'll have a physical presence at BIDS over the week, and we encourage you to show up for the hours we set aside for doc hacking. However, it is totally fine to work remotely; we will coordinate people via email/GitHub, too. *Where can I get more information?* Check out an updated schedule, list of tutorials, and more information at our website here: *bids.github.io/docathon* <http://bids.github.io/docathon> . *Contact* If you have any questions, open an issue on our *GitHub repo* <https://github.com/BIDS/docathon>. We look forward to hearing from you! Please feel free to forward this email to anyone who may be interested. We'd love for other institutions/groups to get involved.

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Need help to start contributing
by Soumyo Dey Feb. 16, 2017

Feb. 16, 2017

Hello, I want to start contributing to the project, help me get started with an easyfix. I was able to setup git repository. Now I would like to start contributing with some code. Thank you, Soumyo Dey Twitter : @SoumyoDey<https://link.nylas.com/link/z82y9ejfckg74qf8wv68racn/local-f41ee7f5-b444/3?…> Website: http://ace139.com/<https://link.nylas.com/link/z82y9ejfckg74qf8wv68racn/local-f41ee7f5-b444/4?…>

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A quick question regarding permutation_test_score
by Afarin Famili Feb. 16, 2017

Feb. 16, 2017

Hi folks, I have a question regarding how to use permutation_test_Score. Given data X (predictor) and Y (target), I hold aside 20% of my data for testing (Xtest and Ytest) and would then Perform hyperparameter-tuning on the rest (using Xtrain and Ytrain). This way I can get the best parameters via RandomizedSearchCV. I now want to call permutation_test_score to compute the score, as well as the p-value of the model prediction. But the question is what X and Y should I send as input arguments to this function? I could send in X and Y but then my hyperparameter parameters were already tuned to Xtrain and Ytrain, which are a part of X and Y and that would bias the output values. Any help would be greatly appreciated. Thanks, Afarin ________________________________ UT Southwestern Medical Center The future of medicine, today.

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PLSRegression cross validates poorly when scaling
by Paul Anton Letnes Feb. 14, 2017

Feb. 14, 2017

Hi! I've noticed that PLSRegression seems to cross validate incredibly poorly when scale=True. Could there be a bug here, or is there something I'm not getting this time, too? I noticed the very small (i.e. large negative) cross validation scores on a dataset that was far from unit variance; there, too, cross validation was extremely poor: around 0.4 in score when scaling was disabled, but (for example) -54422617.41005663 when scaling was enabled! In [1]: import numpy as np In [2]: from sklearn import cross_decomposition In [3]: x = np.random.random((10,17)) In [4]: y = np.random.random((10, 3)) In [5]: pls = cross_decomposition.PLSRegression(scale=True) In [6]: pls.fit(x,y) Out[6]: PLSRegression(copy=True, max_iter=500, n_components=2, scale=True, tol=1e-06) In [7]: from sklearn import model_selection In [8]: model_selection.cross_val_score(pls, x, y) Out[8]: array([-10.1680294 , -12.94229352, -13.39506559]) In [9]: pls = cross_decomposition.PLSRegression(scale=False) In [10]: model_selection.cross_val_score(pls, x, y) Out[10]: array([-0.5904095 , -1.16551493, -1.71555855]) Cheers Paul

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cross validation scores seem off for PLSRegression
by Paul Anton Letnes Feb. 14, 2017

Feb. 14, 2017

Hi! Versions: sklearn 0.18.1 numpy 1.11.3 Anaconda python 3.5 on ubuntu 16.04 What range is the cross_val_score supposed to be in? I was under the impression from the documentation, although I cannot find it stated explicitly anywhere, that it should be a number in the range [0, 1]. However, it appears that one can get large negative values; see the ipython session below. Cheers Paul In [2]: import numpy as np In [3]: y = np.random.random((10, 3)) In [4]: x = np.random.random((10, 17)) In [5]: from sklearn.cross_decomposition import PLSRegression In [6]: pls = PLSRegression(n_components=3) In [7]: from sklearn.cross_validation import cross_val_score In [8]: from sklearn.model_selection import cross_val_score In [9]: cross_val_score(pls, x, y) Out[9]: array([-32.52217837, -4.17228083, -5.88632365]) PS: This happens even if I cheat by setting y to the predicted value, and cross validate on that. In [29]: y = x @ pls.coef_ In [30]: cross_val_score(pls, x, y) /home/paul/anaconda3/envs/wp3-paper/lib/python3.5/site-packages/sklearn/cross_decomposition/pls_.py:293: UserWarning: Y residual constant at iteration 5 warnings.warn('Y residual constant at iteration %s' % k) /home/paul/anaconda3/envs/wp3-paper/lib/python3.5/site-packages/sklearn/cross_decomposition/pls_.py:293: UserWarning: Y residual constant at iteration 6 warnings.warn('Y residual constant at iteration %s' % k) /home/paul/anaconda3/envs/wp3-paper/lib/python3.5/site-packages/sklearn/cross_decomposition/pls_.py:293: UserWarning: Y residual constant at iteration 6 warnings.warn('Y residual constant at iteration %s' % k) Out[30]: array([-35.01267353, -4.94806383, -5.9619526 ]) In [34]: np.max(np.abs(y - x @ pls.coef_)) Out[34]: 0.0

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Logistic regression doesnt converge?
by Mahesh Chandra Feb. 11, 2017

Feb. 11, 2017

>reg = 0.1 lr = LogisticRegression(C=1/reg,max_iter=100, fit_intercept=True,solver='lbfgs').fit(X_train, y_train) ytrain_hat = lr.predict_proba(X_train) loss = log_loss(y_train,ytrain_hat) print loss print loss + 0.5*reg*LA.norm(lr.coef_) Maybe i am doing it wrong

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Re: [scikit-learn] Preparing a scikit-learn 0.18.2 bugfix release
by Andrew Howe Feb. 9, 2017

Feb. 9, 2017

How many current deprecations are expected in the next release? Andrew On Jan 12, 2017 00:53, "Gael Varoquaux" <gael.varoquaux(a)normalesup.org> wrote: On Thu, Jan 12, 2017 at 08:41:51AM +1100, Joel Nothman wrote: > When the two versions deprecation policy was instituted, releases were much > more frequent... Is that enough of an excuse? I'd rather say that we can here decide that we are giving a longer grace period. I think that slow deprecations are a good things (see titus's blog post here: http://ivory.idyll.org/blog/2017-pof-software-archivability.html ) G > On 12 January 2017 at 03:43, Andreas Mueller <t3kcit(a)gmail.com> wrote: > On 01/09/2017 10:15 AM, Gael Varoquaux wrote: > instead of setting up a roadmap I would rather just identify bugs > that > are blockers and fix only those and don't wait for any feature > before > cutting 0.19.X. > I agree with the sentiment, but this would mess with our deprecation cycle. > If we release now, and then release again soonish, that means people have > less calendar time > to react to deprecations. > We could either accept this or change all deprecations and bump the removal > by a version? > _______________________________________________ > scikit-learn mailing list > scikit-learn(a)python.org > https://mail.python.org/mailman/listinfo/scikit-learn > _______________________________________________ > scikit-learn mailing list > scikit-learn(a)python.org > https://mail.python.org/mailman/listinfo/scikit-learn -- Gael Varoquaux Researcher, INRIA Parietal NeuroSpin/CEA Saclay , Bat 145, 91191 Gif-sur-Yvette France Phone: ++ 33-1-69-08-79-68 http://gael-varoquaux.info http://twitter.com/GaelVaroquaux _______________________________________________ scikit-learn mailing list scikit-learn(a)python.org https://mail.python.org/mailman/listinfo/scikit-learn

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Need Corresponding indices array of values in each split of a DesicisionTreeClassifier
by Nixon Raj Feb. 8, 2017

Feb. 8, 2017

For Example, In the below decision tree dot file, I have 223 samples which splits into [174, 49] in the first split and [110, 1] in the 2nd split I would like to get the array of indices for the values of each split like *[174, 49] and their corresponding indices (idx) like [[0, 1 ,5, 7,....,200,221], [3, 4, 6, ....., 199,222,223]]* *[110, 1] and their corresponding indices (idx) like [[0,5,....200,221], [7]]* Please help me node [shape=box] ; 0 [label="X[0] <= 13.9191\nentropy = 0.7597\nsamples = 223\nvalue = [174, 49]"] ; 1 [label="X[1] <= 3.1973\nentropy = 0.0741\nsamples = 111\nvalue = [110, 1]"] ; 0 -> 1 [labeldistance=2.5, labelangle=45, headlabel="True"] ; 2 [label="entropy = 0.0\nsamples = 109\nvalue = [109, 0]"] ; 1 -> 2 ; 3 [label="entropy = 1.0\nsamples = 2\nvalue = [1, 1]"] ; 1 -> 3 ; 4 [label="X[1] <= 3.1266\nentropy = 0.9852\nsamples = 112\nvalue = [64, 48]"] ; 0 -> 4 [labeldistance=2.5, labelangle=-45, headlabel="False"] ; 5 [label="X[2] <= -0.4882\nentropy = 0.7919\nsamples = 63\nvalue = [48, 15]"] ; 4 -> 5 ; 6 [label="entropy = 0.684\nsamples = 11\nvalue = [2, 9]"] ; 5 -> 6 ; 7 [label="X[2] <= 0.5422\nentropy = 0.5159\nsamples = 52\nvalue = [46, 6]"] ; 5 -> 7 ; 8 [label="entropy = 0.0\nsamples = 18\nvalue = [18, 0]"] ; 7 -> 8 ; 9 [label="X[2] <= 0.6497\nentropy = 0.6723\nsamples = 34\nvalue = [28, 6]"] ; 7 -> 9 ; 10 [label="entropy = 0.0\nsamples = 1\nvalue = [0, 1]"] ; 9 -> 10 ; 11 [label="X[2] <= 1.887\nentropy = 0.6136\nsamples = 33\nvalue = [28, 5]"] ; 9 -> 11 ; 12 [label="entropy = 0.0\nsamples = 12\nvalue = [12, 0]"] ; 11 -> 12 ; 13 [label="X[2] <= 2.6691\nentropy = 0.7919\nsamples = 21\nvalue = [16, 5]"] ; 11 -> 13 ; 14 [label="entropy = 0.8113\nsamples = 4\nvalue = [1, 3]"] ; 13 -> 14 ; 15 [label="entropy = 0.5226\nsamples = 17\nvalue = [15, 2]"] ; 13 -> 15 ; 16 [label="X[0] <= 17.3284\nentropy = 0.9113\nsamples = 49\nvalue = [16, 33]"] ; 4 -> 16 ; 17 [label="entropy = 0.9183\nsamples = 6\nvalue = [4, 2]"] ; 16 -> 17 ; 18 [label="X[2] <= 19.7048\nentropy = 0.8542\nsamples = 43\nvalue = [12, 31]"] ; 16 -> 18 ; 19 [label="X[2] <= 5.8511\nentropy = 0.8296\nsamples = 42\nvalue = [11, 31]"] ; 18 -> 19 ; 20 [label="X[0] <= 31.8916\nentropy = 0.878\nsamples = 37\nvalue = [11, 26]"] ; 19 -> 20 ; 21 [label="X[1] <= 3.3612\nentropy = 0.6666\nsamples = 23\nvalue = [4, 19]"] ; 20 -> 21 ; 22 [label="entropy = 0.8905\nsamples = 13\nvalue = [4, 9]"] ; 21 -> 22 ; 23 [label="entropy = 0.0\nsamples = 10\nvalue = [0, 10]"] ; 21 -> 23 ; 24 [label="entropy = 1.0\nsamples = 14\nvalue = [7, 7]"] ; 20 -> 24 ; 25 [label="entropy = 0.0\nsamples = 5\nvalue = [0, 5]"] ; 19 -> 25 ; 26 [label="entropy = 0.0\nsamples = 1\nvalue = [1, 0]"] ; 18 -> 26 ; }

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Permutation-test-score
by Afarin Famili Feb. 5, 2017

Feb. 5, 2017

Hi, Can anyone please tell me what does "permutation_test_score"(and the p_value it returns) do in scikit-learn? I am assuming it outputs the statistical significance of the performance of regression models. I am planning on comparing the performance of various regression models if the performance measure they are reporting is statistically significant. To this end, I wanna output the p-value of the prediction first, and if it was smaller than a certain cut-off, I would then report the performance metrics, such as r2 and MSE. Do p-value and score outputs from "permutation-test-score" not provide me with what I want? Afarin ________________________________ UT Southwestern Medical Center The future of medicine, today.

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Calculate p-value, the measure of statistical significance, in scikit-learn
by Afarin Famili Feb. 5, 2017

Feb. 5, 2017

Hi all, I am aiming at calculating the p-value of regression models using scikit-learn, in order to report their statistical significance. Aside from permutation_test_score in scikit-learn, do you have any suggestions for calculating the p-value of the model? Ultimately, I am interested in computing the coefficient of determination, r2 as well as MSE to indicate the performance of the model for those models that were statistically significant. Thank you, Afarin? ? ________________________________ UT Southwestern Medical Center The future of medicine, today.

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