Reputation: 1101
I have been doing some performance testing and got rather curious at my latest findings.
>>> timeit("import timeit")
0.8010718822479248
>>> timeit("from timeit import timeit")
1.3421258926391602
How can importing the whole module be faster, than importing just a specific part ?
Based on the answers, I have been doing some tests and I came across the following:
>>> timeit("x = timeit.timeit", setup="import timeit")
0.09205102920532227
>>> timeit("x = timeit", setup="from timeit import timeit")
0.0244600772857666
Regarding performance, if you plan on using a class/function/submodule a lot, it takes less time if you specify where to import from and can offset or even make up for the time lost in the import.
Upvotes: 3
Views: 390
Reputation: 107297
Because when you want to import one/some part of a module, all the searching through the module's namespace, storing the object in stack and pop from it take time, while for importing the module at once python just does one step, binding the module to its name.
For a better demonstration you can use dis
module for checking the bytecode for two recipes separately:
In [10]: def import_all():
import timeit
....:
In [11]: def import_one():
....: from timeit import timeit
....:
In [12]: import dis
In [13]: dis.dis(import_all)
2 0 LOAD_CONST 1 (0)
3 LOAD_CONST 0 (None)
6 IMPORT_NAME 0 (timeit)
9 STORE_FAST 0 (timeit)
12 LOAD_CONST 0 (None)
15 RETURN_VALUE
In [14]: dis.dis(import_one)
2 0 LOAD_CONST 1 (0)
3 LOAD_CONST 2 (('timeit',))
6 IMPORT_NAME 0 (timeit)
9 IMPORT_FROM 0 (timeit)
12 STORE_FAST 0 (timeit)
15 POP_TOP
16 LOAD_CONST 0 (None)
19 RETURN_VALUE
As you can see in second case we have an IMPORT_FROM
and POP_TOP
more than the first one.
Upvotes: 2
Reputation: 1947
import timeit
will get the module directly while from timeit import timeit
will take time browsing entire timeit module. Hence the results.
Upvotes: 2