Python 3 store efficiently values in a list

Question

this is maybe a silly question but I know that must exist a more efficient way how to use a loop in this scenario to create a list that later on I will be able to iterate to compare values.

I have a bunch of r and q, I want to compose a list with these to later compare the values in sort of key values. Please ideas:

here is my code:

r1 = "a"
r2 = "b"
r3 = "c"
r4 = "d"
r5 = "e"

q1 = "a"
q2 = "b"
q3 = "c"
q4 = "d"
q5 = "e"

l = []
record1 = {"q": q1, "r": r1}
record2 = {"q": q2, "r": r2}
record3 = {"q": q3, "r": r3}
record4 = {"q": q4, "r": r4}
record5 = {"q": q5, "r": r5}

l.append(record1)
l.append(record2)
l.append(record3)
l.append(record4)
l.append(record5)

score = 0
for i in l:
    if i['q'] == i['r']:
        score +=1

print(score)

You said yourself there is a better way with a loop, why not try it? — Olivier Melançon, Oct 14 '18 at 22:33
@OlivierMelançon because working with community gives you a wider perspective to improve than try you alone. thanks — Andres Urrego Angel, Oct 14 '18 at 23:33

score 0 · Answer 1 · answered Oct 14 '18 at 23:09

According to Ami Marowka Article on Parallel software, list comprehension is approximately 2 times faster than normal loops

new_list = [expression(i) for i in old_list if filter(i)]

And an example by Wes McKinney in Python for Data Analysis shows that numpy arrays are 10 - 100 times faster than nested loops like you have in your code above. Both numpy and list comprehension are equally useful to learn and hope this helps.

Python 3 store efficiently values in a list

1 Answers1