2 
a = ['123b4', '234v5', 'lobf56']
b = [obj1, obj2, obj3] # where each obj is list of object which has attribute called 'serial' which matches serial numbers in list #a

Where obj1.serial is 234v5, obj2.serial is lobf56 and obj3.serial is 123b4

tmplist=list()
for each in a:
    for obj in b:
        if each == obj.serial:
            tmplist.append(obj)

print(tmplist)

output: [obj3, obj1, obj2]

I am currently able to achieve the sorting in above manner. But is there a better way to do it?

Improve this question
1
  • 1
    I added an answer to your question. Commented Apr 6, 2019 at 1:35

1 Answer 1

Reset to default
1

Does a list comprehension helps?

[obj for each in a for obj in b if each == obj.serial]

If you compare the time between both, your approach takes:

1.6 µs ± 25.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

The list comprehension takes:

1.37 µs ± 18.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Therefore, if by "a better way to do it" you mean efficiency. This definitely counts.

Improve this answer
Sign up to request clarification or add additional context in comments.

2 Comments

What code exactly did you time? I'm just curious because the gap is usually not that big.
@Tomothy32, you are right. I made the mistake of including the print statement. I just updated my answer. Great catch!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.