如何在python中填充二维数组？

Question

What's the pythonic way to rewrite the following C code?

int a[16][4];
int s[16] = {1,0,2,3,0,1,1,3,3,2,0,2,0,3,2,1};

for (int i = 0; i < 16; ++i) {
    for (int j = 0; j < 16; ++j) {
        int diff = i ^ j;
        int val = s[i] ^ s[j];
        ++a[diff][val];
    }
}

Answer 1

Here is some equivalent Python code:

a = [[0]*4 for i in range(16)]
s = [1,0,2,3,0,1,1,3,3,2,0,2,0,3,2,1]
for i in range(16):
    for j in range(16):
        diff = i ^ j
        val = s[i] ^ s[j]
        a[diff][val] += 1

Answer 2

The array is initialized by

In [1]: a=[[0]*4]*16

In [2]: a
Out[2]: 
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

Then the second array (list):

In [5]: s  = [1,0,2,3,0,1,1,3,3,2,0,2,0,3,2,1]

Bitwise operators in Python are similar to C. As FJ already posted it's the same.

Answer 3

Implementation

a = [[0]*4 for _ in range(16)]
s = [1,0,2,3,0,1,1,3,3,2,0,2,0,3,2,1]
from itertools import product
for diff, val in ((i ^ j, s[i] ^ s[j]) 
                  for i, j in product(range(16), repeat = 2):
    a[diff][val] += 1