Use bytearrays for building up bytes for I/O. #245
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Bytes are immutable in python, which means that repeatedly appending
to them is _quadratic_ in time. That is, we would like and expect
that deleting 10k objects 10 times would take about as long as
deleting 100k, but because we repeatedly concatenate to a bytes object
in the loop, they are not:
timeit.timeit(lambda: server.delete_multi(["foo"]*10000),number=10)
# 0.5087270829999966
timeit.timeit(lambda: server.delete_multi(["foo"]*100000),number=1)
# 10.650619775999985
Switch to using `bytearray` in all places which handle variable
numbers of bytes. After this change:
timeit.timeit(lambda: server.delete_multi(["foo"]*10000),number=10)
# 0.1197161969999998
timeit.timeit(lambda: server.delete_multi(["foo"]*100000),number=1)
# 0.1269589350000011
jaysonsantos
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Dec 28, 2021
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Bytes are immutable in python, which means that repeatedly appending
to them is _quadratic_ in time. That is, we would like and expect
that deleting 10k objects 10 times would take about as long as
deleting 100k, but because we repeatedly concatenate to a bytes object
in the loop, they are not:
timeit.timeit(lambda: server.delete_multi(["foo"]*10000),number=10)
# 0.5087270829999966
timeit.timeit(lambda: server.delete_multi(["foo"]*100000),number=1)
# 10.650619775999985
Switch to using `bytearray` in all places which handle variable
numbers of bytes. After this change:
timeit.timeit(lambda: server.delete_multi(["foo"]*10000),number=10)
# 0.1197161969999998
timeit.timeit(lambda: server.delete_multi(["foo"]*100000),number=1)
# 0.1269589350000011
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Bytes are immutable in python, which means that repeatedly appending
to them is quadratic in time. That is, we would like and expect
that deleting 10k objects 10 times would take about as long as
deleting 100k, but because we repeatedly concatenate to a bytes object
in the loop, they are not:
Switch to using
bytearrayin all places which handle variablenumbers of bytes. After this change:
This change is