[Numpy-discussion] Histograms via indirect index arrays
Tim Hochberg
tim.hochberg at cox.net
Thu Mar 30 08:53:05 CST 2006
Norbert Nemec wrote:
>Sorry, Travis, I missed your answer for a week...
>
>I confess, my efficiency concerns about the histogram routine are to
>some extent, of theoretical nature. However, given the widespread use of
>the function and considering the simplicity of the solution, I think it
>is still worth thinking about it:
>
>The current histogram routine is based on a "sort" of the incoming data.
>Quicksort is usually O(N log N), worst case O(N**2). Doing the histogram
>by simply counting is a simple O(N) procedure. The absolute cost of the
>quicksort algorithm is low, but the simple counting should be even
>lower, if done efficiently in C.
>
>
I'm not so sure about this analysis: histogram allows aribtrarily spaced
bins, so I think the actual comparison is O(N log N) versus O(N log M)
where M is the number of bins. The number of bins will typically be
much lower than N it's true, but the log will tend to wash that out
some. Take, for example, 1024 items into 32 bins. Then log N is 10
versus log M being 5. This difference is small enough that probably the
differences in constant factors will dominate.
>What probably weighs heavier though, is the memory usage: sorting can
>either be done in place, destroying the original data, or requires a
>copy.
>
I wouldn't worry about the memory usage itself so much. We all have
oodles of RAM and virtual memory these days anyway. In addition, you'll
probably end up copying discontiguous data anyway, which one ends up
with rather frequently. However, the counting algorithm should be more
cache friendly than the sort based on. I find that operations on
matrices that don't fit in the cache are dominated by memory access
time, not CPU time. On my machine that happens somewhere between 10k and
100k elements. For arrays with 100k or so elements, the counting
algorithm might be a big win.
>Counting could be done even on an iterator without storing all the
>data.
>
>
This is true. However, if you go through the iterator interface, it will
be much slower than directly addressing the array memory. Such a thing
might be cool to have somewhere (itertools or something), but probably
isn't appropriate for numpy.
>Lacking a comparable implementation of the counting algorithm, I have
>not done any actual speed comparisons. Why nobody ever found this
>inefficiency, I cannot say. Probably nobody ever had a real comparison
>how fast the histogram could be. Unless you are dealing with really huge
>data, the difference will not show up.
>
>
This is the crux of the issue I think. The implementaion of histogram
looks fine for what it does. For reasonably large arrays, the overhead
of it being implemented in Python will be negligible. Without someone
who actually needs something different, a new implementation is liable
to wander off into the realm of pointless microoptimization. So, my
advice is to leave it alone till it actually causes you problems.
>Anyway, as I said, a solution should be simple to code once it is
>decided how to do it. Simply recoding the histogram routine itself in C
>would be the minimal solution. Implementing a more flexible counting
>routine that can be used in other contexts as well would certainly be
>more desirable, but I have no clear vision what that should look like.
>
>
Well if you were going to do this. I would code a drop in replacement
for the current histogram. No iterators (they belong in a different
library). Variable spaces bins allowed, etc. It doesn't sound like we
need a different histogram function and the more overlapping functions
that there are, the less each individual one gets used and tested and
the more likely that bugs and bitrot set in. What I'd acutally do is
implent something that behaved like this:
def histogram_core(data, bins):
# I hope I grabbed the right stuff out histogram_core
# bins is an actual list of bins edges as per histogram
# which could really us a docstring
n = sort(a).searchsorted(bins)
n = concatenate([n, [len(a)]])
n = n[1:]-n[:-1]
return n
Except written in C and using a counting algorithm. Then the histogram
can be just a thin wrapper over histogram_core. Then test it with a
bunch of different array sizes and bin counts and see how it does. My
guess is that the sort based algoritm will do better in some cases
(arrays with 10k or fewer elements and lots of bins) while the counting
algorithm does better in others (few bins or 100k or more elements). But
we'll see. Or maybe we won't.
Regards,
-tim
>Greetings,
>Norbert
>
>
>
>Travis Oliphant wrote:
>
>
>
>>Norbert Nemec wrote:
>>
>>
>>
>>>I have a very much related problem: Not only that the idea described by
>>>Mads Ipsen does not work, but I could generally find no efficient way to
>>>do a "counting" of elements in an array, as it is needed for a
>>>histogram.
>>>
>>>
>>>
>>This may be something we are lacking.
>>
>>It depends on what you mean by efficient, I suppose. The sorting
>>algorithms are very fast, and the histogram routines that are
>>scattered all over the place (including the histogram function in
>>numpy) has been in use for a long time, and you are the first person
>>to complain of its efficiency. That isn't to say your complaint may
>>not be valid, it's just that for most people the speed has been
>>sufficient.
>>
>>
>>
>>>What would instead be needed is a function that simply gives the count
>>>of occurances of given values in a given array:
>>>
>>>
>>>
>>I presume you are talking of "integer" arrays, since
>>equality-comparison of floating-point values is usually not very
>>helpful so most histograms on floating-point values are given in terms
>>of bins. Thus, the searchsorted function uses bins for it's
>>"counting" operation.
>>
>>
>>
>>>
>>>
>>>
>>>
>>>>>>[4,5,2,3,2,1,4].count([0,1,2,3,4,5])
>>>>>>
>>>>>>
>>>>>>
>>>[0,1,2,1,1,2]
>>>
>>>
>>>
>>>
>>A count function for integer arrays could certainly be written using a
>>C-loop. But, I would first just use histogram([4,5,2,3,2,1,4],
>>[0,1,2,3,4,5]) and make sure that's really too slow, before worrying
>>about it too much.
>>
>>Also, I according to the above function, the right answer is:
>>
>>[0, 1, 2, 1, 2, 1]
>>
>>
>>Best,
>>
>>
>>-Travis
>>
>>
>>
>>
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>
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