Note: I\'m a Ruby developer trying to find my way in Python.
When I wanted to figure out why some scripts use mylist[:]
instead of list(mylist)
Use the timeit module in python for testing timings.
from copy import *
a=range(1000)
def cop():
b=copy(a)
def func1():
b=list(a)
def slice():
b=a[:]
def slice_len():
b=a[0:len(a)]
if __name__=="__main__":
import timeit
print "copy(a)",timeit.timeit("cop()", setup="from __main__ import cop")
print "list(a)",timeit.timeit("func1()", setup="from __main__ import func1")
print "a[:]",timeit.timeit("slice()", setup="from __main__ import slice")
print "a[0:len(a)]",timeit.timeit("slice_len()", setup="from __main__ import slice_len")
Results:
copy(a) 3.98940896988
list(a) 2.54542589188
a[:] 1.96630120277 #winner
a[0:len(a)] 10.5431251526
It's surely the extra steps involved in a[0:len(a)]
are the reason for it's slowness.
Here's the byte code comparison of the two:
In [19]: dis.dis(func1)
2 0 LOAD_GLOBAL 0 (range)
3 LOAD_CONST 1 (100000)
6 CALL_FUNCTION 1
9 STORE_FAST 0 (a)
3 12 LOAD_FAST 0 (a)
15 SLICE+0
16 STORE_FAST 1 (b)
19 LOAD_CONST 0 (None)
22 RETURN_VALUE
In [20]: dis.dis(func2)
2 0 LOAD_GLOBAL 0 (range)
3 LOAD_CONST 1 (100000)
6 CALL_FUNCTION 1
9 STORE_FAST 0 (a)
3 12 LOAD_FAST 0 (a) #same up to here
15 LOAD_CONST 2 (0) #loads 0
18 LOAD_GLOBAL 1 (len) # loads the builtin len(),
# so it might take some lookup time
21 LOAD_FAST 0 (a)
24 CALL_FUNCTION 1
27 SLICE+3
28 STORE_FAST 1 (b)
31 LOAD_CONST 0 (None)
34 RETURN_VALUE