Effectively counts ranges of non-zero values

Question

I am dealing with timing of precipitation volumes, for which I want to calculate the lengths and volumes of individual precipitation events, where the “event” is a sequence of non-zero timestamps. I deal with multiple timings of ~ 60k timesteps, and my current approach is rather slow.

I currently have the following:

import numpy as np

def count_events(timeseries):
    start = 0  
    end = 0
    lengths = []
    volumes = []
    # pad a 0 at the edges so as to include edges as "events"
    for i, val in enumerate(np.pad(timeseries, pad_width = 1, mode = 'constant')):

        if val > 0 and start==0:
            start = i
        if val == 0 and start>0:
            end = i

            if end - start != 1:
                volumes.append(np.sum(timeseries[start:end]))
            elif end - start == 1:
                volumes.append(timeseries[start-1])

            lengths.append(end-start)
            start = 0

    return np.asarray(lengths), np.asarray(volumes)

Expected Result:

testrain = np.array([1,0,1,0,2,2,8,2,0,0,0.1,0,0,1])
lengths, volumes = count_events(testrain)
print lengths
[1 1 4 1 1]
print volumes
[  1.    1.   12.    0.1   1. ] # 12 should actually be 14, my code returns wrong results.

I imagine a much better way to do this using numpy efficiency, but nothing comes to mind ...

EDIT:

Comparison of various solutions:

testrain = np.random.normal(10,5, 60000)
testrain[testrain<0] = 0 

My solution (produces incorrect results, not entirely accurate):

%timeit count_events(testrain)
#10 loops, best of 3: 129 ms per loop

@ Dawg's:

%timeit dawg(testrain) # using itertools
#10 loops, best of 3: 113 ms per loop
%timeit dawg2(testrain) # using pure numpy
#10 loops, best of 3: 156 ms per loop

@ Dsm's:

%timeit DSM(testrain)
#10 loops, best of 3: 28.4 ms per loop

@ DanielLenz's:

%timeit DanielLenz(testrain)
#10 loops, best of 3: 316 ms per loop

Answer 1

numpy, numpy pandas. volume groupby, numpy, pandas.

:

>>> tr = pd.Series(testrain)
>>> nonzero = (tr != 0)
>>> group_ids = (nonzero & (nonzero != nonzero.shift())).cumsum()
>>> events = tr[nonzero].groupby(group_ids).agg([sum, len])
>>> events
    sum  len
1   1.0    1
2   1.0    1
3  14.0    4
4   0.1    1
5   1.0    1

Answer 2

:

import numpy as np
from itertools import groupby

testrain = np.array([1,0,1,0,2,2,8,2,0,0,0.1,0,0,1])

lengths=[]
volumes=[]
for k, l in groupby(testrain, key=lambda v: v>0):
    if k:
        li=list(l)
        lengths.append(len(li))
        volumes.append(sum(li))

print lengths     
print volumes

[1, 1, 4, 1, 1]
[1.0, 1.0, 14.0, 0.10000000000000001, 1.0]

- numpy:

def find_runs(arr):
    subs=np.split(testrain, np.where(testrain== 0.)[0])
    arrs=[np.delete(sub, np.where(sub==0.)) for sub in subs]
    return [(len(e), sum(e)) for e in arrs if len(e)]

>>> find_runs(testrain)    
[(1, 1.0), (1, 1.0), (4, 14.0), (1, 0.10000000000000001), (1, 1.0)]
>>> length, volume=zip(*find_runs(testrain))

Answer 3

Here is my approach using labelsfrom scipy.ndimage.measurements:

import numpy as np
from scipy.ndimage.measurements import label

testrain = np.array([1,0,1,0,2,2,8,2,0,0,0.1,0,0,1])
labels, nlabels = label(testrain)
labels
>> array([1, 0, 2, 0, 3, 3, 3, 3, 0, 0, 4, 0, 0, 5], dtype=int32)

def sum_and_length(n):
    obj = np.array(testrain[labels==n])
    return [np.sum(obj), obj.size]

sums, lengths = np.array(map(sum_and_length, range(1, nlabels+1))).T
sums
>> array([  1. ,   1. ,  14. ,   0.1,   1. ])
lenghts
>> array([ 1.,  1.,  4.,  1.,  1.])

This is not the most beautiful approach, given that this problem is ideal for pandas, but it can make you look into measurementswhich is a very powerful set of tools.