How to get the logical right binary shift in python

No, there isn’t. The right shift in python is arithmetical.

Numpy provides the right_shift() function that does this:

>>> import numpy
>>> numpy.right_shift(1000, 3)
125

You can do a bitwise shift padding with zeros with the bitstring module using the >>= operator:

>>> a = BitArray(int=-1000, length=32)
>>> a.int
-1000
>>> a >>= 3
>>> a.int
536870787

Here’s a spinoff of aix’s answer. The normal right-shift operator will work if you feed it a positive value, so you’re really looking for a conversion from signed to unsigned.

def unsigned32(signed):
    return signed % 0x100000000

>>> unsigned32(-1000) >> 3
536870787L

Trying to flip the sign bit of a negative number by masking it with 0x100000000 is fundamentally misconceived as it makes hard assumptions about the word length. In my time as a programmer I have worked with 24-, 48-, 16-, 18-, 32-, 36- and 64-bit numbers. I have also heard of machines that work on odd-lengths, such as 37 and others that use ones-complement, and not twos-complement, arithmetic. Any assumptions you make about the internal representation of numbers, beyond that they are binary, is dangerous.

Even the binary assumption is not absolutely safe, but I think we’ll allow that. 🙂

You need to remember that if the number is negative, the top bit is set and with each shift right you need to make the top bit set as well.

Here is my implementation:

def rshift(val, n):
    s = val & 0x80000000
    for i in range(0,n):
        val >>= 1
        val |= s
    return val

A solution that works without a modulo:

>>> def rshift(val,n): return (val>>n) & (0x7fffffff>>(n-1))

This works since 7fffffff is a positive number and right shifting that will add zeros to the left.

You could also use floor division:

def rshift(val, n):
    if val > 0:
        return val >> n
    return val // -(2^n)

This is not an efficient approach but works just as expected

def _toBinary(x):
    x=int(x)
    binary = []
    while x > 0:
        binary.append(str(x%2))
        x=int(x/2)
    return "".join(binary[::-1])

def _fromBinary(xs):
    ans = 0
    for i,x in enumerate(xs[::-1]):
        if x == '1':
            ans += 2**i
    return ans

def leftLogicalShift(x,n=1):
    if not type(x) == int:
        return x
    xs = _toBinary(x)
    xs = [x for x in xs]
    for _ in range(n):
        xs.pop(0)
        xs.append('0')
    return _fromBinary("".join(xs))

def rightLogicalShift(x,n=1):
    if not type(x) == int:
        return x
    xs = _toBinary(x)
    xs = [x for x in xs]
    for _ in range(n):
        xs.pop()
        xs.insert(0,'0')
    return _fromBinary("".join(xs))

def leftArithmeticShift(x,n=1):
    return leftLogicalShift(x,n)

def rightArithmeticShift(x,n=1):
    if not type(x) == int:
        return x
    xs = _toBinary(x)
    xs = [x for x in xs]
    for _ in range(n):
        tmp = xs[0]
        xs.pop()
        xs.insert(0,tmp)
    return _fromBinary("".join(xs))

lls = leftLogicalShift(10,2) 
print(lls) # 8

rls = rightLogicalShift(10,2) 
print(rls) # 2

las = leftArithmeticShift(10,2)
print(las) # 8

ras = rightArithmeticShift(10,2)
print(ras) # 14

references:

https://open4tech.com/logical-vs-arithmetic-shift/

https://www.interviewcake.com/concept/java/bit-shift

I think a logical right binary shift is not available in Python directly. Instead you can use Javascript in Python as in:

import js2py
rshift = js2py.eval_js('function $(a, b){ return a >>> b}')
print (rshift(244, 324)) #15

The top-voted answer produces WRONG results for val < 0 and n == 0!

def rshift(val, n):
    if (val >= 0): return val >> n
    elif (n == 0): return val
    else: return (val + 0x10000000) >> n

>>> rshift(-1, 0)
-1