在Python中反转排列

Fin*_*fin 4 python permutation

我是编程的新手,我正在尝试使用以下代码编写一个Python函数来查找{1,2,3,...,n}上的置换的逆矩阵:

def inv(str):
    result = []
    i = list(str).index(min(list(str)))
    while min(list(str)) < len(list(str)) + 1:
        list(str)[i : i + 1] = [len(list(str)) + 1]
        result.append(i + 1)
    return result
Run Code Online (Sandbox Code Playgroud)

但是,当我尝试使用该函数时,inv('<mypermutation>')返回[].我错过了什么吗?Python是否因为某些语法原因跳过我的while循环而我不明白?我的google和stackoverflow都没有搜索我认为有用的主题.

d.b*_*d.b 13

其他答案是正确的,但就其价值而言,使用 numpy 有一个性能更高的替代方案:

inverse_perm = np.argsort(permutation)
Run Code Online (Sandbox Code Playgroud)

编辑:下面的第四个函数甚至更快。

计时码:

def invert_permutation_list_scan(p):
    return [p.index(l) for l in range(len(p))]

def invert_permutation_list_comp(permutation):
    return [i for i, j in sorted(enumerate(permutation), key=lambda i_j: i_j[1])]

def invert_permutation_numpy(permutation):
    return np.argsort(permutation)

def invert_permutation_numpy2(permutation):
    inv = np.empty_like(permutation)
    inv[permutation] = np.arange(len(inv), dtype=inv.dtype)
    return inv

x = np.random.randn(1000)
perm = np.argsort(x)
permlist = list(perm)
assert np.array_equal(invert_permutation_list_scan(permlist), invert_permutation_numpy(perm))
assert np.array_equal(invert_permutation_list_comp(perm), invert_permutation_numpy(perm))
assert np.array_equal(invert_permutation_list_comp(perm), invert_permutation_numpy2(perm))
%timeit invert_permutation_list_scan(permlist)
%timeit invert_permutation_list_comp(perm)
%timeit invert_permutation_numpy(perm)
%timeit invert_permutation_numpy2(perm)
Run Code Online (Sandbox Code Playgroud)

结果:

82.2 ms ± 7.28 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
479 µs ± 9.19 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
18 µs ± 1.17 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
4.22 µs ± 388 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Run Code Online (Sandbox Code Playgroud)

  • 好吧,如果您追求速度,那么您可以做更少的工作: 。``` def invert_permutation_numpy2(permutation): x = np.empty_like(permutation) x[permutation] = np.arange(len(permutation), dtype=permutation.dtype) return x ``` 。在迷你基准测试中,它比 numpy 排序高出 10-20 倍 (2认同)

Rei*_*ica 7

如果您只想要反向排列,则可以使用

def inv(perm):
    inverse = [0] * len(perm)
    for i, p in enumerate(perm):
        inverse[p] = i
    return inverse

perm = [3, 0, 2, 1]
print(inv(perm))
for i in perm:
    print(inv(perm)[i])

[1, 3, 2, 0]
0
1
2
3
Run Code Online (Sandbox Code Playgroud)