在Python中联接字符串的最有效方法

Question

During the interview, I was given a piece of Python code:

"".join([x for x in "\x63\x6c\x75\x62"])

The interviewer asked me: Can you make this piece of code faster with just a tiny tweak to the existing code? The output should stay exactly the same.

Enough to say, I failed this question. I'm wondering if anything could help me understand how to optimize this code?

Thanks a lot.

Answer 1

"".join([x for x in "\x63\x6c\x75\x62"])

is a very redundant way of doing:

"\x63\x6c\x75\x62"

In general, you should not do

x for x in "\x63\x6c\x75\x62"

this is useless because you don't transform or filter x in any way (and it's of course slower).

One last thing, if the interviewer meant to remove the [] he/she was also wrong.

"".join(x for x in "\x63\x6c\x75\x62")

is slower that with the listcomp because join has to build a list anyway (useful to remember this when creating useful comprehension, unlike that one, see Joining strings. Generator or list comprehension? )

Conclusion: let's not join the club

Answer 2

Nothing to do:

>>> res = "\x63\x6c\x75\x62"
>>> res
'club'

It just a different way to enter the sting.

Answer 3

好的，这个版本呢？

"".join(* ["\x63\x6c\x75\x62"])

Answer 4

%%timeit
"".join([x for x in "\x63\x6c\x75\x62"])
289 ns ± 3.54 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Outputs club .

%%timeit 
"".join(["\x63\x6c\x75\x62"])  
66.8 ns ± 0.46 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

Outputs club .

[x for x...] is useless here. Removing it makes the code running 4 times faster.