Serhiy Storchaka added the comment:
Yes, the total time of repeated string concatenation is O(N**2). s3 is twice
larger s2, therefore s3 time about twice large s2 time.
In the first case Python use a special optimization which allows O(N) in some
cases. You can deactivate it:
s5 = """
text = ""
for i in range(1,50000):
text2 = text
text += "12345678901234567890"
"""
print("str cat 20: {0}s".format(timeit.timeit(s5,number=1)))
It's not a bug, it's feature.
----------
nosy: +storchaka
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue15503>
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