Astan Chee wrote:
Hi,
I have a list that has a bunch of numbers in it and I want to get the
most common number that appears in the list. This is trivial because i
can do a max on each unique number. What I want to do is to have a
tolerance to say that each number is not quite unique and if the
difference from other numbers is small, than it can be counted together.
This is what I have to get the common number (taken from the internet
somewhere):
l = [10,30,20,20,11,12]
d = {}
tolerance = 5
for elm in l:
d[elm] = d.get(elm, 0) + 1
counts = [(j,i) for i,j in d.items()]
This of course returns a list where each of them is unique
[(1, 12), (1, 10), (1, 11), (2, 20), (1, 30)]
but I am expecting a list that looks like this:
[(3, 10), (2, 20), (1, 30)]
Maybe check for this number: tolerance * ( X / tolerance)
or a variation if tolerance is non-integer?
Eg. (rough):
from itertools import groupby
def tmax(seq, alpha):
longest = []
for k, g in groupby(sorted(seq), lambda x: alpha * (x / alpha)):
g = list(g)
if len(g) > len(longest):
longest = g
return longest[0]
a = [10, 30, 20, 20, 11, 12]
assert tmax(a, 5) == 10
assert tmax(a, 1) == 20
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