So basically I want to count the number of occurrences a floating point appears in a given list. For example: a list of grades (all scores out of 100) are inputted by the u
This method uses bisect which can be more efficient, but it requires that you sort the scores first.
from bisect import bisect
import random
scores = [random.randint(0,100) for _ in xrange(100)]
bins = [20, 40, 60, 80, 100]
scores.sort()
counts = []
last = 0
for range_max in bins:
i = bisect(scores, range_max, last)
counts.append(i - last)
last = i
I wouldn't expect you to install numpy just for this, but if you already have numpy you can use numpy.histogram
.
UPDATE
First, using bisect is more flexible. Using [i//n for i in scores]
requires that all the bins are the same size. Using bisect allows the bins to have arbitrary limits. Also i//n
means the ranges are [lo, hi). Using bisect the ranges are (lo, hi] but you can use bisect_left if you want [lo, hi).
Second bisect is faster, see timings bellow. I've replaced scores.sort() with the slower sorted(scores) because the sorting is the slowest step and I didn't want to bias the times with a pre-sorted array, but the OP says his/her array is already sorted so bisect could make even more sense in that case.
setup="""
from bisect import bisect_left
import random
from collections import Counter
def histogram(iterable, low, high, bins):
step = (high - low) / bins
dist = Counter(((x - low + 0.) // step for x in iterable))
return [dist[b] for b in xrange(bins)]
def histogram_bisect(scores, groups):
scores = sorted(scores)
counts = []
last = 0
for range_max in groups:
i = bisect_left(scores, range_max, last)
counts.append(i - last)
last = i
return counts
def histogram_simple(scores, bin_size):
scores = [i//bin_size for i in scores]
return [scores.count(i) for i in range(max(scores)+1)]
scores = [random.randint(0,100) for _ in xrange(100)]
bins = range(10, 101, 10)
"""
from timeit import repeat
t = repeat('C = histogram(scores, 0, 100, 10)', setup=setup, number=10000)
print min(t)
#.95
t = repeat('C = histogram_bisect(scores, bins)', setup=setup, number=10000)
print min(t)
#.22
t = repeat('histogram_simple(scores, 10)', setup=setup, number=10000)
print min(t)
#.36
If you are fine with using the external library NumPy, then you just need to call numpy.histogram()
:
>>> data = [82, 85, 90, 91, 70, 87, 45]
>>> counts, bins = numpy.histogram(data, bins=10, range=(0, 100))
>>> counts
array([0, 0, 0, 0, 1, 0, 0, 1, 3, 2])
>>> bins
array([ 0., 10., 20., 30., 40., 50., 60., 70., 80.,
90., 100.])
To group the data, divide it by the interval width. To count the number in each group, consider using collections.Counter. Here's a worked out example with documentation and a test:
from collections import Counter
def histogram(iterable, low, high, bins):
'''Count elements from the iterable into evenly spaced bins
>>> scores = [82, 85, 90, 91, 70, 87, 45]
>>> histogram(scores, 0, 100, 10)
[0, 0, 0, 0, 1, 0, 0, 1, 3, 2]
'''
step = (high - low + 0.0) / bins
dist = Counter((float(x) - low) // step for x in iterable)
return [dist[b] for b in range(bins)]
if __name__ == '__main__':
import doctest
print doctest.testmod()
decs = [int(x/10) for x in scores]
maps scores from 0-9 -> 0, 10-19 -> 1, et cetera. Then just count the occurrences of 0, 1, 2, 3, and so on (via something like collections.Counter
), and map back to ranges from there.