Fast and efficient way to convert string into integer

Question

I want to convert four character string text into 32bit integer number, e.g. four space string should return 0x20202020. All string elements are ASCII. I know that something like this works

number = ord(text[0]) << 24 | ord(text[1]) << 16 | ord(text[2]) << 8 | ord(text[3]) << 0

but it is rather slow and inefficient. Is there any faster way?

EDIT: This is meant to part of the algorithm sending a string to LCD display via C/Assembler script. String has to be split into list containing numbers that represent four characters each.

Regards.

maybe id(text) might be an option, depending on your application — mcsoini
– mcsoini, Commented Dec 8, 2019 at 11:34
@schwobaseggl To send data to C/Assembler script which then sends it to LCD display — Pygmalion
– Pygmalion, Commented Dec 8, 2019 at 11:46

John Coleman · Accepted Answer · 2019-12-08 12:44:10Z

4

You could use text => bytes => int:

int.from_bytes(text.encode(),'big')

A quick time comparison:

import timeit

print(timeit.timeit("int.from_bytes(text.encode(),'big')","text = 'abcd'"))
print(timeit.timeit("ord(text[0]) << 24 | ord(text[1]) << 16 | ord(text[2]) << 8 | ord(text[3]) << 0","text = 'abcd'"))

Output:

0.21847990699999997
0.4186516309999999

So it seems that int.from_bytes is roughly twice as fast as the explicit bit shifting.

edited Dec 8, 2019 at 12:44

answered Dec 8, 2019 at 12:34

John Coleman

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