python - 查找满足阈值关系的组合

Question

给定 、 、 和 的值phi，theta我n_1需要n_2找到满足以下条件的所有可能对 ( N_1, )：N_2

0 <= N_1 <= n_1
0 <= N_2 <= n_2
N_1 - phi * N_2 >= theta

在 Python 中执行此操作的最有效方法是什么？显然，我可以使用两个for循环——迭代N_1和N_2（从前两个标准）的所有可能值，并只保存满足最后一个标准的那些对——但这将是相当低效的。

Answer 1

您可以使用 numpy 和矢量化，如下所示

import numpy as np

phi = 0.5
theta = 1
n1 = 10
n2 = 20

N1 = np.random.randint(-100, 100, size=100)
N2 = np.random.randint(-100, 100, size=100)

N1 = N1[(N1 >= 0) & (N1 <= n1)]
N2 = N2[(N2 >= 0) & (N2 <= n2)]

a = N2 * theta + phi
res = N1.reshape(N1.shape[0], 1) - a.reshape(1, a.shape[0])

indices = np.argwhere(res >= 0)
pairs = zip(N1[indices[:,0]], N2[indices[:,1]])

对的示例输出

[(8, 3),
 (8, 6),
 (8, 5),
 (8, 1),
 (3, 1),
 (9, 3),
 (9, 8),
 (9, 8),
 (9, 6),
 (9, 5),
 (9, 6),
 (9, 6),
 (9, 5),
 (9, 8),
 (9, 1)]

根据@dbliss 请求，这是模块化版本及其测试

import numpy as np


def calc_combination(N1, N2, n1, n2, theta, phi):
    N1 = N1[(N1 >= 0) & (N1 <= n1)]
    N2 = N2[(N2 >= 0) & (N2 <= n2)]

    a = N2 * theta + phi
    res = N1.reshape(N1.shape[0], 1) - a.reshape(1, a.shape[0])

    indices = np.argwhere(res >= 0)
    pairs = zip(N1[indices[:,0]], N2[indices[:,1]])
    return pairs


def test_case():
    n1 = 5
    n2 = 1
    theta = 2
    phi = 2

    N1 = np.arange(n1 + 1)
    N2 = np.arange(n2 + 1)

    assert (calc_combination(N1, N2, n1, n2, theta, phi) ==
            [(2, 0), (3, 0), (4, 0), (4, 1), (5, 0), (5, 1)])

test_case()

Answer 2

这有效：

import itertools

def _get_N_1_and_N_2(n_1, n_2, phi, theta):
    """Get the (N_1, N_2) pairs as defined in Griffith (1963).

    See Equation 3.

    Parameters
    ----------
    n_1 : integer
      Number of excitatory inputs.

    n_2 : integer
      Number of inhibitory inputs.

    phi : number
      Factor that captures the difference between excitatory and
      inhibitory synaptic efficiencies.

    theta : number
      Spike threshold.

    """
    N_1 = range(n_1 + 1)
    N_2 = range(n_2 + 1)
    possible_N_1_N_2 = itertools.product(N_1, N_2)
    N_1_N_2 = []
    for N_1, N_2 in possible_N_1_N_2:
        if N_1 - phi * N_2 >= theta:
            N_1_N_2.append((N_1, N_2))
    return N_1_N_2

我想我可以用这个语句使这个for循环成为if一个非常混乱的list理解。但 。. . 呐。

这是测试：

import nose.tools as nt

def test__get_N_1_and_N_2():

    # Figure 3A in Griffith, 1963, Biophy J.
    n_1 = 4
    n_2 = 0
    theta = 2
    phi = 1
    desired = [(2, 0), (3, 0), (4, 0)]
    actual = _get_N_1_and_N_2(n_1, n_2, phi, theta)
    nt.assert_list_equal(desired, actual)

    # Figure 3B.
    n_1 = 5
    n_2 = 1
    theta = 2
    phi = 2
    desired = [(2, 0), (3, 0), (4, 0), (4, 1), (5, 0), (5, 1)]
    actual = _get_N_1_and_N_2(n_1, n_2, phi, theta)
    nt.assert_list_equal(desired, actual)