On 2019-04-03 22:42, Sayth Renshaw wrote:
In an email, I received this question as part of a newsletter.
def fetch_squares ( max_root ):
squares = []
for x in range ( max_root ):
squares . append (x **2)
return squares
MAX = 5
for square in fetch_squares (MAX ):
do_something_with ( square )
1) Do you see a memory bottleneck here? If so, what is it?
2) Can you think of a way to fix the memory bottleneck?
Want to know if I am trying to solve the correct bottleneck.
I am thinking that the bottleneck is to create a list only to iterate the same
list you created sort of doubling the time through.
Is that the correct problem to solve?
If it is then I thought the best way is just to supply the numbers on the fly,
a generator.
def supply_squares(max_root):
for x in max_root:
yield x
MAX = 5
So then I set up a loop and do whatever is needed. At this time I am creating
generator objects. But is this the correct way to go? More of a am I thinking
correctly questino.
item = 0
while item < MAX:
print(supply_squares(item))
item += 1
<generator object supply_squares at 0x0000000004DEAC00>
<generator object supply_squares at 0x0000000004DEAC00>
<generator object supply_squares at 0x0000000004DEAC00>
<generator object supply_squares at 0x0000000004DEAC00>
<generator object supply_squares at 0x0000000004DEAC00>
You should create a single generator that will yield the squares. The
'for' loop should remain the same.
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