Avoid if else statement in tuples and make loop faster-CodePudding

import string

a , b , c = string.ascii_lowercase , string.ascii_uppercase , string.digits

l1 = [ (i , a ,b , c ,) for i in range(100,1000000) ] #all

l2 = [ (i,) for i in range(150,340192) ] # used

x = [

    item for item in l1 if (item[0],) not in l2
]

print(len(x)) # it costs at least 50 seconds to run

I’m looking to improve my code or make it faster.

CodePudding user response：

l2 is a list (a very large one at that) of singleton tuples and you check for membership. Since you're checking for membership, making it a set will be much more efficient.

# construct a set from the range
s_l2 = set(range(150,340192))
# check for set membership
x = [item for item in l1 if item[0] not in s_l2]

Benchmark:

%timeit x = [item for item in l1 if (item[0],) not in l2]
# 6.55 s ± 326 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit y = [item for item in l1 if item[0] not in range(150,340192)]
# 4.22 s ± 56.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit s_l2 = set(range(150,340192)); z = [item for item in l1 if item[0] not in s_l2]
#141 ms ± 3.51 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

CodePudding user response：

The simple idea is to directly judge whether each element is in the range without generating l2:

>>> r = range(150, 340192)
>>> timeit(lambda: [item for item in [(i, a, b, c) for i in range(100,1000000)] if item[0] not in r], number=7)
1.5238849000015762

It can also be simplified directly by comparison:

>>> timeit(lambda: [item for item in [(i, a, b, c) for i in range(100,1000000)] if not 150 <= item[0] < 340192], number=7)
1.341257099993527

The simplest way is to build the target range in advance, and then generate the target list in once loop:

>>> from itertools import chain
>>> timeit(lambda: [(i, a, b, c) for i in chain(range(100, 150), range(340192, 1000000))], number=7)
0.5550074999919161