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##### Activity
Hans Dembinski
@HDembinski
Hey all, iminuit 1.3.7 is out! đźĄł This time really with wheels, so it is installed in a flash and you don't need a compiler. Big thanks go to @henryiii who developed the Azure Pipeline that generates the wheels, which was a big amount of work. Special thanks go to @eduardo-rodrigues who tested the packages before the release.
Luke Kreczko
@kreczko

stupid question. I have two numpy arrays, a, and b and I want to group elements in b by the unique elements in a, i.e.

a = [1, 12, 1, 50]
b = [10, 20, 30, 40]
result == [[10, 30], [20], [50]]

Is there a numpy function that does that? It seems I am missing the right keyword in my searches

Luke Kreczko
@kreczko
The current solution is using a for-loop which I would like to get rid off:
unique_a = np.unique(a)
result = []
for u in unique_a:
result.append(b[a == u].tolist())
Chris Burr
@chrisburr
@kreczko In pandas this is known as groupby:
import numpy as np
import pandas as pd
a = np.arange(10)+0.1
b = np.random.randint(4, size=10)
df = pd.DataFrame({'a': a, 'b': b})
list(df.groupby('b')['a'].apply(list))
I can't think of a numpy equivalent but I'd guess searching numpy groupby will get you there
Luke Kreczko
@kreczko
thanks @chrisburr, yes, that's what I thought at first which lead me to https://stackoverflow.com/questions/38013778/is-there-any-numpy-group-by-function which I then lead me to my for-loop solution.
pandas groupby is nice, I will need to test perfomance, I guess.
Alternatively, there is always numba
benkrikler
@benkrikler

I think awkward array might be able to help instead of pandas (not tested):

reorder = np.argsort(a)
_, counts = np.unique(a[reorder], return_counts=True)
result = awkard.JaggedArray.fromcounts(counts, b[reorder])

The only thing I'm unsure of there is the order of the unique counts, I'm assuming the unique method returns things in the order they're first seen, but I suspect that's not true.

Luke Kreczko
@kreczko
np.unique returns them sorted
benkrikler
@benkrikler
Then I guess you need to use the inverse array somehow (with return_inverse)
Luke Kreczko
@kreczko
that's a good point, did not realize my end-result is basically a awkward.JaggedArray
ok, I will finish the unit tests and then try to change the implementation
Jelle Aalbers
@JelleAalbers

For a pure numpy solution,

a = np.array([1, 12, 1, 1, 1, 12, 12, 50])
b = np.arange(len(a))

si = np.argsort(a)
np.split(b[si], np.where(np.diff(a[si]))[0] + 1)

seems to work. Haven't tested it in detail though.

Eduardo Rodrigues
@eduardo-rodrigues

Dear colleague,

We are pleased to announce the second "Python in HEP" workshop organised by the HEP Software Foundation (HSF). The PyHEP, "Python in HEP", workshops aim to provide an environment to discuss and promote the usage of Python in the HEP community at large.
PyHEP 2019 will be held in Abingdon, near Oxford, United Kingdom, from 16-18 October 2019.

The workshop will be a forum for the participants and the community at large to discuss developments of Python packages and tools, exchange experiences, and steer where the community needs and wants to go. There will be ample time for discussion.

The agenda will be composed of plenary sessions, a highlight of which is the following:
1) A keynote presentation from the Data Science domain.
2) A topical session on histogramming including a talk and a hands-on tutorial.
3) Lightning talks from participants.
4) Presentations following up from topics discussed at PyHEP 2018.

We encourage community members to propose presentations on any topic (email: pyhep2019-organisation@cern.ch). We are particularly interested in new(-ish) packages of broad relevance.

The agenda will be made available on the workshop indico page (https://indico.cern.ch/event/833895/) in due time. It is also linked from the PyHEP WG homepage http://hepsoftwarefoundation.org/activities/pyhep.html.

Registration will open very soon, and we will provide detailed travel and accommodation information at that time.
Travel funds may be available at a modest level. To be confirmed once registration opens.

You are encouraged to register to the PyHEP WG Gitter channel (https://gitter.im/HSF/PyHEP) and/or to the HSF forum (https://groups.google.com/forum/#!forum/hsf-forum) to receive further information concerning the organisation of the workshop.

Eduardo Rodrigues & Ben Krikler, for the organising committee

Luke Kreczko
@kreczko
R.I.P. rootpy documentation:
http://rootpy.org/
NOTICE: This domain name expired on 7/11/2019 and is pending renewal or deletion.
Eduardo Rodrigues
@eduardo-rodrigues
Hi @kreczko, try getting in touch with Noel Dawe, see https://github.com/ndawe. He's probably still responsible for the site - just my best guess.
Nicholas Smith
@nsmith-
@kreczko here's an option that also delays the reindex of b and preserves the order of first-seen values:
import numpy as np, awkward
a = np.array([1, 12, 1, 10, 50, 10])
b = np.array([10, 20, 30, 40, 50, 60])
arg = a.argsort(kind='stable')
offsets, = np.where(np.r_[True, np.diff(a[arg]) > 0])
output = awkward.JaggedArray.fromoffsets(offsets.flatten(), awkward.IndexedArray(arg, b))
in other news, np.where([0, 1, 0, 0, 1])[0].base is surprisingly 2d (hence the flatten)
Nicholas Smith
@nsmith-
thanks to that nerd-snipe I fixed a bug in awkward
Luke Kreczko
@kreczko
nice!
Luke Kreczko
@kreczko

since the knowledge in this channel proved invaluable before, another question :)
I have

group_1 = np.array([(1, 2), (3, 3), (5, 7), (4, 4)])
test_elements = np.array([(1, 2), (3, 3), (3, 5)])

and would like to test if the elements in test_elements are in group_1. I expect the result

[True, True, False]

as I take the tuples as unique objects.

Numpy has the function isin where

np.isin(group_1, test_elements)

will return

[[True, True], [True, True], [True, False], [False, False]]

OK, so this is inverse to what I want, fine.

np.isin(group_1, test_elements)
# returns
[[True, True], [True, True], [True, True]]

Clearly it compares element by element and since both 3 and 5 are contained, therefore (3,5) should be as well, right?
Well, not in my case. Is there a way to do this comparison for each 2-vector instead of element-wise? For loop (even with numba) is quite slow

Jonas Eschle
@jonas-eschle

Yes, you can do that. The idea is: make a comparison of all possible combinations of each element with each other element. This gives you a rank three boolean object with: number of elements in the group, number of elements to test, dimension of an element. Then make two reduce operations: 1. reduce all on the axis of the tuple, requiring that true is if in a tuple everything is true and 2. a reduce any on the axis of all the possible combinations, since at least one tuple has to be fully contained.

For example (may change the axis for convenience):

test_elements_expanded = np.expand_dims(test_elements, axis=1)
entries_equal = group_1 == test_elements_expanded
tuple_equal = np.all(entries_equal, axis=2)
tuple_contained = np.any(tuple_equal, axis=0)
Luke Kreczko
@kreczko
interesting. Thank you, I will try this out!
The dimension of entries_equal is off :(
Luke Kreczko
@kreczko

For the first example the result is

E         - [True, True, False, False]
E         ?             -------
E         + [True, True, False]

for the 2nd (group_2 = np.array([(0, 0), (1, 2), (2, 2), (3, 3)])):

E         - [False, True, False, True]
E         + [True, True, False]
Jonas Eschle
@jonas-eschle
ah sorry, wrong axis! the last reduce is along the possible combinations which are in axis 1, not 0. So change the last line to:
tuple_contained = np.any(tuple_equal, axis=1)
Axis 0 lists all test samples. In axis 1 are the possible combinations. In axis 2 is the tuple itself. Reducing axis 2 with all means that an entry is true where all elements in a tuple are true, otherwise false, reducing axis 1 with any means that an entry with at least one matching tuple is true. So your left with the axis 0.
Luke Kreczko
@kreczko
that works!
thanks!
Jelle Aalbers
@JelleAalbers

You can also map the tuples to scalars (easy if you have some idea what the values are going to be), e.g:

def squash(x):
return 10000 * x[:,0] + x[:,1]

np.in1d(squash(test_elements), squash(group_1))

This should be more memory-efficient and faster if both arrays are large.

If you are going to do membership testing repeatedly on the same array, it might be even better to convert it to a set, dictionary or some other object backed by a hash table, so membership tests are a constant time operation.

Luke Kreczko
@kreczko
Comparing the two solutions:
yours: 28366.0 microseconds
stackoverflow: 10999.0 microseconds
@JelleAalbers Where does the 10000 come from?
it's interesting how many solutions exist for the same operation
Jelle Aalbers
@JelleAalbers
It's just a placeholder, you can put in whatever the maximum value you expect to be is. If your numbers are e.g. arbitrary-sized floats I guess this solution doesn't work. Though you can probably replace 'squash' with some other function (if you want to go really overboard you can use some cryptographic hash function, though then forget about speed :-)
Jonas Eschle
@jonas-eschle

Comparing the two solutions:
yours: 28366.0 microseconds
stackoverflow: 10999.0 microseconds

So essentially same speed for this exact problem. At this point, it matters, if: you call it once or a million times? How big is your array really? That's when things like presorting can make the difference. My advice: use which ever method you understand/like better (not from the speed, from the concept) and try only to improve on it if it proves to be a bottleneck.

Luke Kreczko
@kreczko
speed is important :). Your solution sped up the function by almost a factor 300 :)
benkrikler
@benkrikler
Exciting news everyone:

Registration is now open for PyHEP 2019, in Abingdon, UK, from the 16th to 18th of October! The registration fee for the 2.5 days has been set at ÂŁ80; it includes the venue, lunches, dinners, and refreshments. We also have about 46 rooms at Cosenerâ€™s House, available on a first-come-first-served basis. The actual payment system will not be online for a few more days, however, so youâ€™ll only be able to complete registration then including the room booking.

The agenda is also shaping up with talks confirmed on topics ranging from histogramming, statistical methods, distributed workflows, visualisation, and even GPU-programming. Several speakers from industry are confirmed, including our keynote speaker on the PyViz library.

Since the PyHEP series is all about growing a â€śPython in High Energy Physicsâ€ť community, this year weâ€™re also including a session of lighting talks where 30 people can present any topic of their choosing for 3 minutes with a single slide as a way for everyone, especially newcomers and early careers researchers, to introduce themselves.

Community members can also propose presentations on any topic (email: pyhep2019-organisation@cern.ch). We are particularly interested in new(-ish) packages of broad relevance.

More details can be found on the indico page (https://indico.cern.ch/e/PyHEP2019) or from the PyHEP WG homepage http://hepsoftwarefoundation.org/activities/pyhep.html. You can also join the HSF forum (https://groups.google.com/forum/#!forum/hsf-forum) to get more information about the workshop and community

Help us spread the word! :slight_smile:
Hans Dembinski
@HDembinski
pyhepmc-ng 0.4.2 was released today
Hans Dembinski
@HDembinski
@benkrikler I tried to complete my registration today, but during checkout I was not offered any payment options. There is a combobox which is supposed to show the options, but it is empty in my case. Is this a problem on my end or ...?
Chris Burr
@chrisburr
we are still finalising the payment system and will let know when this is available at the email address you use to register
Hans Dembinski
@HDembinski
Yes, but that was 12 days ago :) And Ben said: "however, so youâ€™ll only be able to complete registration then including the room booking."
Not true in my case...
I can't book any rooms.
benkrikler
@benkrikler
Thanks Hans and Chris. Chris is correct. The payment system is still not set up (the company we've had to use have been extremely difficult to work with, I've been trying to reach them by the phone every working day for the last week). We'll let you know straight away, I expect to have it in the next day or two.
Hans Dembinski
@HDembinski
I am sorry to hear that Ben :(
It is no problem for me, I was just surprised, thanks for clearing this up!
Eduardo Rodrigues
@eduardo-rodrigues
Kind reminder on the PyHEP workshop: the available slots are being filled up at a nice pace, so don't delay registration too much, if you intend to come and participate - we hope you do! See https://indico.cern.ch/e/PyHEP2019 â€¦