Why is a=a*100 almost two times faster than a*=100? [duplicate]

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Numpy in-place operation performance (2 answers)

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Following the question about Chaining *= += operators and the good comment of Tom Wojcik ("Why would you assume aaa *= 200 is faster than aaa = aaa * 200 ?"), I tested it in Jupyter notebook:

%%timeit aaa = np.arange(1,101,1)
    aaa*=100

%%timeit aaa = np.arange(1,101,1)
    aaa=aaa*100

And I was surprised because the first test is longer than the second one: 1530ns and 952ns, respectively. Why these values are so different?

python numpy

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edited Apr 20 at 14:21

Konrad Rudolph

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asked Apr 20 at 14:09

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  If you reverse the order here, what are the results? – jarmod Apr 20 at 14:11
• 3
  This is related to numpy. Doesn't happen with regular ints or floats – Pranav Hosangadi Apr 20 at 14:16
• 1
  possibly related? stackoverflow.com/questions/16034672/… – python_user Apr 20 at 14:19
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  Changing the range to be np.arange(1,10001,1) actually reverses the results: aaa*=100 is faster! So the in-place is still faster as the input grows in size. For small arrays, for some reason, creating a new array is more efficient... – Tomerikoo Apr 20 at 14:22
• 2
  @MaPy I think you missed the point. The one assigning a new array is faster... – Tomerikoo Apr 20 at 14:23

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