I was trying to code a loop generation of prime in python using filter to learn a bit about the function when I tumbled on what I believe to be a weird behavior in Python3 (not tested in other version).
Two code which, I believe, should return the same thing as they are the same code (written or in a loop). I have spend the last hour trying to understand why the second implementation does not yield the same results but can't seem to figure out why.
l = infinite()
n = next(l)
odd = filter(lambda x: (x%n != 0), l)
t = next(odd)
no3 = filter(lambda x: (x%t != 0), odd)
f = next(no3)
no5 = filter(lambda x: (x%f != 0), no3)
s = next(no5)
print(n, t, f, s)
Yields
2 3 5 7
But this code yield different results:
li = []
l = infinite()
for i in range(4):
n = next(l)
l = filter(lambda x: (x%n != 0), l)
li += [n]
print(*li)
2 3 4 5
infinite() simply return all integers starting from 2 with the following code
def infinite(count = 2):
while True:
yield count
count += 1
Here are some variation of that code and their results:
def _not_divisible(n):
return lambda x: x % n != 0
li = []
l = infinite()
for i in range(4):
n = next(l)
l = filter(_not_divisible(n), l)
li += [n]
print(*li)
2 3 5 7
li = []
l = infinite()
for i in range(4):
n = next(l)
def j(x):
return x%n != 0
l = filter(j, l)
li += [n]
print(*li)
2 3 4 5
Are filters on list stored by reference and thus changing the filter at the reference affect the filtering ?
I would also really appreciate if someone can point me in a direction to understand why this execute like this as i could not find any satisfactory answer on my own.
