Question

单调性百分比是指数组按升序或降序排序的方式。我需要一个python（最好是numpy）方法来实现这个考虑下面的例子

$slug = str_slug($request->company_name, '-');

$validator = Validator::make(['company_name' => $slug], [
    'company_name' => 'required|unique:admin_users,company_name,slug|max:191'
]);

if (!$validator->fails()) {
    $db_filed->company_name = $slug;
    $db_filled->save();
}

我想在AI启发式游戏中运行它，速度非常重要，感谢任何帮助

EDITED 这就是我现在所拥有的它只是返回一个bool而且它是1D

   array([[2,3,4,6,5] # 60% sorted
         [1,2,3,4,5] # 100% sorted
         [0,2,4,8,10]# 100% sorted
         [0,2,4,8,10]]# 100% sorted
         /  | |  \  \
        /   | |   \  \
    100% 100% 80% 80% 100%

Monotonicity percentage is average(80,100,100,100,100,100,80,80,100)

我是PYTHON的新手

Answer 1

您可以使用zip()和list comprehension执行此类操作：

def get_pourcent(a, order='ascending'):
    if order == 'ascending':
        # Check if every element is inferior than his next in the list
        b = [1 if j < v else 0 for j, v in zip(a, a[1:])]
        # Get how much zeros in b
        zeros = len(b) - sum(b)
        percent = (1 - (float(zeros)/len(a)))*100

    elif order == 'descending':
        b = [1 if j > v else 0 for j, v in zip(a, a[1:])]
        zeros = sum(b)
        percent = (float(zeros)/len(a))*100

    else:
        return None

    return '"%s": %.2f%% sorted' % (order, percent)


# Test
tests = [('ascending', [2,3,4,6,5]), ('ascending', [1,2,3,4,5]),
 ('ascending', [0,2,4,8,10]), ('descending', [2,3,4,6,5]), ('descending', [0,2,4,8,10])]

for k, v in tests:
    print v, get_pourcent(v, order=k)

输出：

[2, 3, 4, 6, 5] "ascending": 80.00% sorted
[1, 2, 3, 4, 5] "ascending": 100.00% sorted
[0, 2, 4, 8, 10] "ascending": 100.00% sorted
[2, 3, 4, 6, 5] "descending": 20.00% sorted
[0, 2, 4, 8, 10] "descending": 0.00% sorted

修改：

tests = [[ 2, 4, 0, 8], [ 4, 24, 0, 16], [ 16, 2, 16, 32], [ 16, 2, 16, 128]]

for k in tests:
    print get_pourcent(k)

将输出：

"ascending": 75.00% sorted
"ascending": 75.00% sorted
"ascending": 75.00% sorted
"ascending": 75.00% sorted

Answer 2

您想沿轴（0或1）获取差异，然后查看沿此轴的正值比例：

(np.diff(a, axis=1) > 0).mean(axis=1)

对于diff和mean中axis=0的列，也可以这样做。

我不确定为什么你想要第一行80％而不是75％，但如果你想要，你可以这样做：

(np.diff(a, axis=1) > 0).sum(axis=1) / a.shape[1]

和列：

(np.diff(a, axis=0) > 0).sum(axis=0) / a.shape[0]

Answer 3

这是我构建并按预期工作的功能

def merge(mat):
  monotone = 0
  matrix = numpy.copy(mat)
  for i in range(4):
    m_vertical = 4 if numpy.size(numpy.where(numpy.diff(matrix[:, i]) < 0)[0]) == 0 else numpy.where(numpy.diff(matrix[:, i]) < 0)[0][0]+1
    m_horizontal = 4 if numpy.size(numpy.where(numpy.diff(matrix[i]) < 0)[0]) == 0 else numpy.where(numpy.diff(matrix[i]) < 0)[0][0]+1
    monotone += (m_vertical + m_horizontal)*3.125
  return monotone

PYTHON：numpy 2D阵列的排序等级百分比

3 个答案: