Compute C's `%` using Python's `%`?

Question

How do I compute C's % using Python's %? The difference between the two is in the way they handle the case of negative arguments.

In both languages, the % is defined in such a way that this relationship (// being integer division) holds:

a // b * b + a % b == a

but the rounding of a // b is different in C and in Python, leading to a different definition of a % b.

For example, in C (where integer division is just / with int operands) we have:

int a = 31;
int b = -3;
a / b;  // -10
a % b;  // 1

while in Python:

a = 31
b = -3
a // b  # -11
a % b  # -2

I am aware of this question, which addresses the opposite (i.e. how to compute Python's % from C's %) and contains additional discussions.

I am also aware of Python 3.7 math module introducing remainder() but its result is a float, not an int and hence it will not enjoy arbitrary precision.

Answer 1

Some ways would be:

def mod_c0(a, b):
    if b < 0:
        b = -b
    return -1 * (-a % b) if a < 0 else a % b

def mod_c1(a, b):
    return (-1 if a < 0 else 1) * ((a if a > 0 else -a) % (b if b > 0 else -b))

def mod_c2(a, b):
    return (-1 if a < 0 else 1) * (abs(a) % abs(b))

def mod_c3(a, b):
    r = a % b
    return (r - b) if (a < 0) != (b < 0) and r != 0 else r

def mod_c4(a, b):
    r = a % b
    return (r - b) if (a * b < 0) and r != 0 else r

def mod_c5(a, b):
    return a % (-b if a ^ b < 0 else b)

def mod_c6(a, b):
    a_xor_b = a ^ b
    n = a_xor_b.bit_length()
    x = a_xor_b >> n
    return a % (b * (x | 1))

def mod_c7(a, b):
    a_xor_b = a ^ b
    n = a_xor_b.bit_length()
    x = a_xor_b >> n
    return a % ((-b & x) | (b & ~x))

def mod_c8(a, b):
    q, r = divmod(a, b)
    if (a >= 0) != (b >= 0) and r:
        q += 1
    return a - q * b

def mod_c9(a, b):
    if a >= 0:
        if b >= 0:
            return a % b
        else:
            return a % -b
    else:
        if b >= 0:
            return -(-a % b)
        else:
            return a % b

which all work as expected, e.g.:

print(mod_c0(31, -3))
# 1

Essentially, mod_c0() implements an optimized version of mod_c1() and mod_c2(), which are identical except that in mod_c1() the call to (relatively expensive) call to abs() is replaced by a ternary conditional operator with the same semantic. Instead, mod_c3() and mod_c4() try to directly fix the a % b value for the cases where it is needed. The difference between the two is in how they detect opposite signs of the arguments: (a < 0) != (b != 0) versus a * b < 0. The mod_c5() approach is inspired by @ArborealAnole's answer, and essentially uses the bit-wise xor to handle the cases correctly, while mod_c6() and mod_c7() are the same as @ArborealAnole's answer but using adaptive right shift with int.bit_length(). The mod_c8() approach uses a corrected definition of integer division to fix up the modulus value. The mod_c9() method is inspired by @NeverGoodEnough's answer, and essentially goes full conditional.

---

Covering all sign cases:

vals = (3, -3, 31, -31)
s = '{:<{n}}' * 4
n = 14
print(s.format('a', 'b', 'mod(a, b)', 'mod_c(a, b)', n=n))
print(s.format(*(('-' * (n - 1),) * 4), n=n))
for a, b in itertools.product(vals, repeat=2):
    print(s.format(a, b, mod(a, b), mod_c0(a, b), n=n))

a             b             mod(a, b)     mod_c(a, b)   
------------- ------------- ------------- ------------- 
3             3             0             0             
3             -3            0             0             
3             31            3             3             
3             -31           -28           3             
-3            3             0             0             
-3            -3            0             0             
-3            31            28            -3            
-3            -31           -3            -3            
31            3             1             1             
31            -3            -2            1             
31            31            0             0             
31            -31           0             0             
-31           3             2             -1            
-31           -3            -1            -1            
-31           31            0             0             
-31           -31           0             0             

---

A bit more tests and benchmarks:

n = 100
k = 1
l = [x for x in range(-n, n + k, k)]
ll = [(a, b) for a, b in itertools.product(l, repeat=2) if b]

funcs = mod_c0, mod_c1, mod_c2, mod_c3, mod_c4, mod_c5, mod_c6, mod_c7, mod_c8, mod_c9
for func in funcs:
    correct = all(func(a, b) == funcs[0](a, b) for a, b in ll)
    print(func.__name__, 'correct:', all_equal)
    %timeit [func(a, b) for a, b in ll]
    print()

mod_c0 correct: True
100 loops, best of 3: 6.6 ms per loop

mod_c1 correct: True
100 loops, best of 3: 7.86 ms per loop

mod_c2 correct: True
100 loops, best of 3: 8.49 ms per loop

mod_c3 correct: True
100 loops, best of 3: 7.56 ms per loop

mod_c4 correct: True
100 loops, best of 3: 7.5 ms per loop

mod_c5 correct: True
100 loops, best of 3: 7.94 ms per loop

mod_c6 correct: True
100 loops, best of 3: 13.4 ms per loop

mod_c7 correct: True
100 loops, best of 3: 16.8 ms per loop

mod_c8 correct: True
100 loops, best of 3: 12.4 ms per loop

mod_c9 correct: True
100 loops, best of 3: 6.48 ms per loop

---

Perhaps there are better (shorter?, faster?) ways, given that the implementation of Python's % using C's % seems much simpler:

((a % b) + b) % b

---

To get some feeling on how the C-style % computation (mod_c*() functions from above) stands against the usual % or the operations required to get Python-style % from C:

def mod_py(a, b):
    return a % b

def mod_c2py(a, b):
    return ((a % b) + b) % b


%timeit [mod_py(a, b) for a, b in ll]
# 100 loops, best of 3: 5.85 ms per loop
%timeit [mod_c2py(a, b) for a, b in ll]
# 100 loops, best of 3: 7.84 ms per loop

Note of course that mod_c2py() is only useful to get a feeling of what performances we could expect from a mod_c() function.

---

(EDITED to fix some of the proposed methods and include some timings)

(EDITED-2 to add the mod_c5() solution)

(EDITED-3 to add the mod_c6() to mod_c9() solutions)

Answer 2

I am following up the very comprehensive answer of @norok2. I have tried the super-naive approach with branches, and it appears to be slightly but consistently faster (~2-4%).

def mod_naive(x,y):
  if y < 0:
    if x < 0:
      return x%y
    else:
      return (x%-y)
  else:
    if x < 0:
      return -(-x%y)
    else:
      return x%y

or with a lambda (does not affect speed, only coolness):

mod_naive = lambda x,y: (x%y if x < 0 else x%-y) if y < 0 else (-(-x%y) if x < 0 else x%y)

Compared to @norok2's fastest solution (mod_c0):

mod_c0 correct: True
100 loops, best of 3: 6.86 ms per loop

mod_naive correct: True
100 loops, best of 3: 6.58 ms per loop

My (naive) guess on the reason why is that the branch prediction algorithms will eventually produce less operations overall.

Answer 3

For 64-bit integers, either of these should work:

def mod_c_AA0(a,b):
    x=(a^b)>>63
    return a % (b*(x|1))

def mod_c_AA1(a,b):
    x=(a^b)>>63
    return a % ((-b & x)|(b & ~x))

using two's complement binary. As norok2 suggests, substitute a_xor_b=a^b; x=a_xor_b>>a_xor_b.bit_length(); for the first line to have optimal specificity on bit shifting depending on the magnitude of a and b.