Question

I’m seeking the fastest way to extract all tuple members from a list under condition(s).

Example:
From a list of tuple (e.g. [(0,0,4),(1,0,3),(1,2,1),(4,0,0)]) I need to extract all members that have more than 3 in first tuple position, then more than 2 in second tuple position, and then more than 1 in last tuple position.
Which should extract in this example (4,0,0) (->first condition), nothing (->second condition) and (0,0,4),(1,0,3) (->last condition). This example is very small, I need to perform that on list of thousands of tuples.

From the code I produced from your answers, here are the results in sec:

my_naive1, like proposed by Emil Vikström? 13.0360000134

my_naive2 110.727999926

Tim Pietzcker 9.8329999446

Don 12.5640001297

import itertools, operator, time, copy
from operator import itemgetter


def combinations_with_replacement_counts(n, r):  #(A, N) in our example.N individuals/balls in A genotypes/boxes
   size = n + r - 1
   for indices in itertools.combinations(range(size), n-1):
       #print indices
       starts = [0] + [index+1 for index in indices]
       stops = indices + (size,)
       yield tuple(map(operator.sub, stops, starts))


xp = list(combinations_with_replacement_counts(3,20))  # a very small case

a1=time.time()
temp=[]
for n in xp:
    for n1 in xp:

        for i in xp:
            if i[0] <= min(n1[0],n[0]) or i[1] <= min(n1[1],n[1]) or i[2] <= min(n1[2],n[2]):
                temp.append(i)


a2=time.time()
for n in xp:
    for n1 in xp:
        xp_copy = copy.deepcopy(xp)
        for i in xp:
            if i[0] > min(n[0],n[0]) or i[1] > min(n[1],n[1]) or i[2] > min(n[2],n[2]):
                xp_copy.remove(i)

a3=time.time()
for n in xp:
    for n1 in xp:
        output = [t for t in xp if t[0]<=min(n[0],n[0]) or t[1]<=min(n[1],n[1]) or t[2]<=min(n[2],n[2])]
a4=time.time()

for n in xp:
    for n1 in xp:
        l1 = sorted(xp, key=itemgetter(0), reverse=True)
        l1_fitered = []
        for item in l1:
            if item[0] <= min(n[0],n[0]):
                break
            l1_fitered.append(item)

        l2 = sorted(l1_fitered, key=itemgetter(1), reverse=True)
        l2_fitered = []
        for item in l2:
            if item[1] <= min(n[1],n[1]):
                break
            l2_fitered.append(item)

        l3 = sorted(l2_fitered, key=itemgetter(2), reverse=True)
        l3_fitered = []
        for item in l3:
            if item[2] <= min(n[2],n[2]):
                break
            l3_fitered.append(item)
a5=time.time()            



print "soluce my_naive1, like proposed by Emil Vikström?",a2-a1
print "soluce my_naive2",a3-a2
print "soluce Tim Pietzcker",a4-a3
print "soluce Don",a5-a4

Answer 1

>>> l = [(0,0,4), (1,0,3), (1,2,1), (4,0,0)]
>>> output = [t for t in l if t[0]>3 or t[1]>2 or t[2]>1]
>>> output
[(0, 0, 4), (1, 0, 3), (4, 0, 0)]

This is fast because t[1]>2 is only evaluated if t[0]>3 is False (same for the third condition). So in your example list, only 8 comparisons are necessary.

You might save time and memory (depending on what you’re doing with the filtered data) if you use a generator expression instead:

>>> l = [(0,0,4), (1,0,3), (1,2,1), (4,0,0)]
>>> for item in (t for t in l if t[0]>3 or t[1]>2 or t[2]>1):
>>>     # do something with that item