Python：检查列表中的每个元素是否存在特定时间

Question

我有以下问题。我有一个这样的范围列表：

parameterRanges2 = [(1,5),(1,5),(1,7),(1,7),(0,10),(1,20),(1,3),(0,1)]

我有一个像这样的 numpy 数组：

arr = np.array([[2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0],
[4.0,2.0,3.0,4.0,2.0,4.0,5.0,1.0,2.0,4.0,1.0,3.0,4.0,2.0,3.0,5.0,1.0,3.0,4.0,2.0],
[2.0,3.0,4.0,6.0,7.0,1.0,2.0,3.0,5.0,6.0,1.0,2.0,4.0,5.0,6.0,2.0,3.0,4.0,5.0,6.0],
[6.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,2.0,2.0,3.0,4.0,5.0],
[8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,7.0,8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0],
[11.0,13.0,14.0,16.0,17.0,19.0,1.0,3.0,4.0,6.0,7.0,9.0,10.0,11.0,13.0,14.0,16.0,17.0,19.0,1.0],
[1.0,1.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,3.0,3.0,3.0,3.0,1.0,1.0,1.0,1.0,1.0,2.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,0.0,0.0]])

列表中的每个 parameterRange 元组都对应于 numpy 数组中的子数组。有没有办法检查相应范围内的所有元素是否至少存在一次？例如，在 numpy 数组的第一个子列表中，所有数字 1,2,3,4,5 至少存在一次，在第二个子列表中存在一次，在第三个列表中，例如数字 1,2,3 ,4,5,6,7 存在一次，以此类推。

Answer 1

利用范围是整数，我们可以给出 O(nm) 解，nxm 是arr 的形状。该算法的工作原理如下：

• 丢弃所有非整数元素和所有超出其范围的元素
• 使用 np.add.at 有效地 (O(mn)) 为范围内的数字生成 bincounts
• 计算每行中的上述阈值箱并与范围进行比较

.

import numpy as np

parameterRanges2 = np.array([(1,5),(1,5),(1,7),(1,7),(0,10),(1,20),(1,3),(0,1)])

arr = np.array([[2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0],
[4.0,2.0,3.0,4.0,2.0,4.0,5.0,1.0,2.0,4.0,1.0,3.0,4.0,2.0,3.0,5.0,1.0,3.0,4.0,2.0],
[2.0,3.0,4.0,6.0,7.0,1.0,2.0,3.0,5.0,6.0,1.0,2.0,4.0,5.0,6.0,2.0,3.0,4.0,5.0,6.0],
[6.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,2.0,2.0,3.0,4.0,5.0],
[8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,7.0,8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0],
[11.0,13.0,14.0,16.0,17.0,19.0,1.0,3.0,4.0,6.0,7.0,9.0,10.0,11.0,13.0,14.0,16.0,17.0,19.0,1.0],
[1.0,1.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,3.0,3.0,3.0,3.0,1.0,1.0,1.0,1.0,1.0,2.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,0.0,0.0]])

min_occ = 2

dp = np.diff(parameterRanges2, axis=-1)
m = np.zeros((parameterRanges2.shape[0], np.max(dp) + 2), dtype=int)
arr = arr - parameterRanges2[:, :1]
ia = arr.astype(int)
idx = np.where((arr==ia) & (ia>=0) & (ia<=dp), ia, -1)
np.add.at(m, (np.arange(parameterRanges2.shape[0])[:, None], idx), 1)
res = (m[:, :-1] >= min_occ).sum(axis=-1) == dp.ravel() + 1
print(res)

输出：

[ True  True False  True False False  True  True]

Answer 2

仅使用 Numpy 函数可能有更有效的方法，但下面的代码有效。我想不出一个简单的 Numpy 方法来做到这一点，因为我们无法制作标准的 Numpy 数组 ranges，因为所有行的长度都不相同。

import numpy as np

arr = np.array([
    [2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0,1.0,3.0,2.0,4.0,2.0,4.0,3.0,5.0],
    [4.0,2.0,3.0,4.0,2.0,4.0,5.0,1.0,2.0,4.0,1.0,3.0,4.0,2.0,3.0,5.0,1.0,3.0,4.0,2.0],
    [2.0,3.0,4.0,6.0,7.0,1.0,2.0,3.0,5.0,6.0,1.0,2.0,4.0,5.0,6.0,2.0,3.0,4.0,5.0,6.0],
    [6.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,2.0,2.0,3.0,4.0,5.0],
    [8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0,7.0,7.0,8.0,9.0,0.0,1.0,2.0,3.0,4.0,5.0,6.0],
    [11.0,13.0,14.0,16.0,17.0,19.0,1.0,3.0,4.0,6.0,7.0,9.0,10.0,11.0,13.0,14.0,16.0,17.0,19.0,1.0],
    [1.0,1.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,3.0,3.0,3.0,3.0,1.0,1.0,1.0,1.0,1.0,2.0],
    [0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,0.0,0.0]
])

parameterRanges2 = [(1,5),(1,5),(1,7),(1,7),(0,10),(1,20),(1,3),(0,1)]
ranges = [np.arange(u, v+1, dtype='float64') for u, v in parameterRanges2]

print([np.all(np.isin(u,v)) for u, v in zip(ranges, arr)])

输出

[True, True, True, True, False, False, True, True]