在列表理解上使用.join 是否比 python 中的简单 for 循环更快？

Question

Say I have two codes:

a = [1,2,3,4,5,6,7,8,9,10]

print("\n".join(x for x in a))

a = [1,2,3,4,5,6,7,8,9,10]
for x in a:
    print(a)

Which one is faster and more importantly why?

Answer 1

The first one will be throwing a type error, because you need a list of STRINGS to use join.

But for this purpose, i'll consider the list as string:

eg

a = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']

For time evaluating i'll be using 'time', but i'll give you an example of how to use timeit at the end

So let's create two functions for evaluating:

def first_case(a):
     start = time.time()
     print("\n".join(x for x in a))
     print(time.time() - start)

def second_case(a):
     start=time.time()
     for x in a:
         print(x)
     print(time.time() - start)

>>> first_case(a) 

1
2
3
4
5
6
7
8
9
10
3.147125244140625e-05

>>> second_case(a)
second_case(a)
1
2
3
4
5
6
7
8
9
10
6.270408630371094e-05

So here we're taking more execution time in the loop one.

Also, we can improve the first expression, since we don't actually need for in the join operation.

#We can rewrite the expression as the following 

>>> print("\n".join(a))
1
2
3
4
5
6
7
8
9
10

Not, the join already expect a list, so you don't need to iterate on the list.

Now let's run a third_case test

def third_case(a):
     start = time.time()
     print("\n".join(a))
     print(time.time() - start)

>>> third_case(a)
1
2
3
4
5
6
7
8
9
10
3.0040740966796875e-05

And now we have the fastest version.

You can and should also evaluate the same functions a few more time, since we can't actually take the real results only running one time.

For that, you can use timeit.timeit. From the documentation :

timeit.timeit('"-".join(str(n) for n in range(100))', number=10000)