[MRG] ENH: Make normalization an explicit transformation. #649
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@piskvorky @tmylk I've created |
devashishd12
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I've added initial tests for just checking the logic. Input validation testing and testing for the Similarity class constructor still remains. |
| if scipy.sparse.issparse(vec): # convert scipy.sparse to standard numpy array | ||
| vec = vec.tocsr() | ||
| veclen = numpy.sqrt(numpy.sum(vec.data ** 2)) | ||
| if norm == 'l0': | ||
| veclen = len(vec.data) |
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L_0 'norm' is not homogeneous so it is impossible to scale.
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Yes true.... Could you please elaborate more on the scaling part? Will it be affecting efficiency since as the corpus size increases we cannot use the fact that f(ax) = (a^k)f(x)? Should I remove the support for zero "norm"?
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Sorry to confuse you, meant scaling in mathematical not IT sense.
What number do we need to divide vector a=(5,5) by in order to have l_0_norm(a)=1?
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Thanks for the explanation! I think it would be best to remove L0 norm then and keep the normalization just for l1 and l2. Should I proceed with it?
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Yep. Thanks.
On Mon, Apr 4, 2016 at 8:04 PM, Devashish Deshpande <
notifications@github.com> wrote:
In gensim/matutils.py
#649 (comment):if scipy.sparse.issparse(vec): # convert scipy.sparse to standard numpy array vec = vec.tocsr()
veclen = numpy.sqrt(numpy.sum(vec.data *\* 2))if norm == 'l0':veclen = len(vec.data)Thanks for the explanation! I think it would be best to remove L0 norm
then and keep the normalization just for l1 and l2. Should I proceed with
it?—
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https://github.com/piskvorky/gensim/pull/649/files/8b27eb4f46e0854fefdb4b55b6cea6c42c11a14d#r58429753
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@tmylk I've removed L0 norm and added some more tests to |
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@tmylk @piskvorky I've added further tests to cover all input types. I think this pr is ready for review. |
devashishd12
changed the title
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Removed merge conflicts. |
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@dsquareindia Hey, why did you close this pull request? It is a useful feature. |
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Sorry for that; I'll open it again. |
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Removed redundant use of corpus, rebased and squashed. Could someone please review this? Thanks! |
| return list(vec) | ||
| if norm == 'l1': | ||
| length = float(sum(abs(val) for _, val in vec)) | ||
| assert length > 0.0, "Document contains all zero entries" |
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What is the reason for this check?
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Sorry I had just added that to make it symmetric with the L2 case. I should be returning the vector unchanged in case of 0 length right? Should I proceed with the change?
| return ret_normalized_vec(vec, length) | ||
| if norm == 'l2': | ||
| length = 1.0 * math.sqrt(sum(val ** 2 for _, val in vec)) | ||
| assert length > 0.0, "sparse documents must not contain any explicit zero entries" |
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Should I change this to returning the vector unchanged in case of zero length as we are doing with other input types?
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@tmylk I have split the test suite into multiple tests to ease error checking. I have also removed the |
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@piskvorky could you please review this? I think it's ready for merge. |
| """ | ||
| Calculates the norm by calling matutils.unitvec with the norm parameter. | ||
| """ | ||
| logger.info("Normalizing...") |
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Only did a quick scan, looks nicely done, but I'll leave the thorough review to @tmylk. My only nitpick would be |
| self.num_nnz = numnnz | ||
| self.norms = norms | ||
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| def normalize(self, bow): |
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@piskvorky should this rather be a __getitem__() since a transformation here is defined as something which acts like transformation[doc].
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Sure. We can also let users use either -- a named function or [ ].
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@tmylk @piskvorky I've addressed all the comments. Hopefully it's much better now! Could you please review :) |
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@tmylk @piskvorky thanks a ton for the reviews! 🍻 |
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@piskvorky I was thinking of writing a notebook tutorial for this feature however I can't think of a practical use case for it. What would be the best way to go about this? |
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Not sure... @tmylk ? Maybe indexing some corpus with/without normalization, showing how the returned "most similar" docs (or words, or embeddings... basically any vectors) differ? |
Addresses issue #69. What this PR includes:
TODO:
@piskvorky @tmylk Could you please tell me if I'm going on the right track?