Cast a 10 bit unsigned integer to a signed integer in python

Question

I have a list of numbers in the range 0-1023. I would like to convert them to integers such that 1023 maps to -1, 1022 maps to -2 etc.. while 0, 1, 2, ....511 remain unchanged.

I came up with a simple:

def convert(x):
    return (x - 2**9) % 2**10 - 2**9

is there a better way?

In some programming languages, the % operator yields a negative (or zero) value for a negative left operand, so the code could perhaps be made clearer by changing (x - 2**9) to (x + 2**9). Aside from that, this seems like a reasonable implementation. — ashastral
– ashastral, Commented Jan 28, 2014 at 23:52

Brave Sir Robin · Accepted Answer · 2014-01-28 23:52:16Z

5

Naivest possible solution:

def convert(x):
    if x >= 512:
        x -= 1024
    return x

answered Jan 28, 2014 at 23:52

Brave Sir Robin

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3 Comments

John1024 Over a year ago

Alternatively, reducing three lines to one: return x if x < 512 else x - 1024

Brave Sir Robin Over a year ago

or return x - 1024 * (x >= 512)

ashastral Over a year ago

@rmartinjak, multiplying by a boolean requires Python to do an extra cast, and as a result that solution is a bit slower than the other ones. Your original solution and John1024's seem to be roughly equivalent. I haven't done any rigorous testing yet, though.

falsetru · Accepted Answer · 2014-01-29 15:57:14Z

2

def convert(x):
    return x - (x >> 9) * 1024

Using ctypes:

from ctypes import c_short

def convert(x):
    return c_short(x << 6).value >> 6

edited Jan 29, 2014 at 15:57

answered Jan 28, 2014 at 23:56

falsetru

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1 Comment

yota Over a year ago

perfect answer, I may add some explanations

Ilan Olkies · Accepted Answer · 2018-06-08 17:37:12Z

0

 max_uint_value = 512

 def unassigned_to_int(uint):
      return uint - max_uint_value if uint >= max_uint_value else uint

answered Jun 8, 2018 at 17:37

Ilan Olkies

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