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Home/ Questions/Q 6694073
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T06:03:35+00:00 2026-05-26T06:03:35+00:00

I have a large numpy array of signed bytes ( dtype int8 ). It

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I have a large numpy array of signed bytes (dtype int8). It contains values in the full range -128 to +127. I’d like to convert the efficiently to an array of unsigned bytes (dtype uint8) by adding 128 to each element, such that -128 → 0, 0 → 128, +127 → 255, etc. so of course the results still fit into an unsigned byte.

Simple elementwise addition given the correct numerical result, but creates a result array using twice the memory (dtype int16) in addition to the source array, even though only the low bytes of the result elements are needed.

>>> import numpy
>>> a = numpy.array( [-128, -1, 0, 1, 127 ], dtype=numpy.int8)
>>> b = a + 128
>>> b
array([  0, 127, 128, 129, 255], dtype=int16)

Is there a way to control the dtype of the result array to be uint8?

The alternative approach of modifying the values in-place and “casting” the data to a new type, like this:

>>> for i in xrange(0, 5):
...     if a[i] < 0:
...         a[i] -= 128
...     elif a[i] >= 0:
...         a[i] += 128
...
>>> a
array([   0,  127, -128, -127,   -1], dtype=int8)
>>> a.view(dtype=numpy.uint8)
array([  0, 127, 128, 129, 255], dtype=uint8)

is much more space efficient but very costly in time for large arrays with the transformation in Python.

How can I do this transformation in-place and quickly?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T06:03:36+00:00Added an answer on May 26, 2026 at 6:03 am
    import numpy as np
    a = np.array([-128, -1, 0, 1, 127], dtype=np.int8)
    a = a.view(np.uint8)
    a += 128
    print a
    # -> array([  0, 127, 128, 129, 255], dtype=uint8)
    

    This creates no copies, and all operations are in-place.

    EDIT: safer to cast first to uint — unsigned wrap-around is defined.
    EDIT2: s/numpy/np/g;

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