Which is the most efficient way to iterate through a list in python?
Solution 1
for item in list:
its obviously the one with fewer function calls.
If you want to get the index of items as you go use enumerate like this
for pos, item in enumerate(collection):
Solution 2
def loop_1(data):
for i in range(len(data)):
print(data[i])
def looper_2(data):
for val in data:
print(val)
Checking with dis gives us the following bytecode for loop_1:
12 0 SETUP_LOOP 40 (to 43)
3 LOAD_GLOBAL 0 (range)
6 LOAD_GLOBAL 1 (len)
9 LOAD_FAST 0 (data)
12 CALL_FUNCTION 1
15 CALL_FUNCTION 1
18 GET_ITER
>> 19 FOR_ITER 20 (to 42)
22 STORE_FAST 1 (i)
13 25 LOAD_GLOBAL 2 (print)
28 LOAD_FAST 0 (data)
31 LOAD_FAST 1 (i)
34 BINARY_SUBSCR
35 CALL_FUNCTION 1
38 POP_TOP
39 JUMP_ABSOLUTE 19
>> 42 POP_BLOCK
>> 43 LOAD_CONST 0 (None)
46 RETURN_VALUE
The bytecode for loop_2 looks like this:
17 0 SETUP_LOOP 24 (to 27)
3 LOAD_FAST 0 (data)
6 GET_ITER
>> 7 FOR_ITER 16 (to 26)
10 STORE_FAST 1 (val)
18 13 LOAD_GLOBAL 0 (print)
16 LOAD_FAST 1 (val)
19 CALL_FUNCTION 1
22 POP_TOP
23 JUMP_ABSOLUTE 7
>> 26 POP_BLOCK
>> 27 LOAD_CONST 0 (None)
30 RETURN_VALUE
The second version is obviously better.
Solution 3
Another possible solution would be to use numpy
which would be very efficient, for large lists perhaps even more efficient than a list comprehension or a for loop.
import numpy as np
a = np.arange(5.0) # a --> array([0., 1., 2., 3., 4.])
# numpy operates on arrays element by element
#
b =3.*a # b --> array([0., 3., 6., 9., 12.])
This is a pretty simple operation but you can get more complex using an array as simply an argument in a formula. For large arrays this can be much faster than a list comprehension and it makes the code cleaner and easier to read (no need to create a function to map in a list comprehension). You can also use indexing and slicing to tailor what you want to do:
If you want to have access to the actual index positions use ndenumerate
# b is as above
for i, x in np.ndenumerate(b):
print i, x
The output of this for loop is:
(0,) 0.0
(1,) 3.0
(2,) 6.0
(3,) 9.0
(4,) 12.0
NOTE: the index returned as a tuple by numpy to handle additional dimensions. Here we only have a single dimension so you'd have to unpack the tuple to get the index of the element.
Solution 4
Obviously for i in range(len(list)):
will be slower - in python 2, it's equivalent to this:
list2 = range(len(list))
for i in list2:
...
If that were faster, then this would be even faster, right?
list2 = range(len(list))
list3 = range(len(list2))
list4 = range(len(list3))
for i in list4:
...
mankand007
Updated on August 24, 2022Comments
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mankand007 over 1 year
Say I have a list of items:
x = [1, 2, 3, 4, 5]
I need to perform some functions for each of these items. In a certain case, I need to return the index of an item.
Which is the best and most efficient way?
for item in list: ....
or
for i in range(len(list)): ....