Question

我有两个列表，其中包含一些像这样的3D坐标（例如）：

a = [[1, 2, 3], [4, 5, 6], [4, 2, 3]]
b[0] = [[11, 22, 3], [12, 34, 6], [41, 2, 34], [198, 213, 536], [1198, 1123, 1156]]
b[1] = [[11, 22, 3], [42, 25, 64], [43, 45, 23]]
b[2] = [[3, 532, 23], [4, 5, 6], [98, 23, 56], [918, 231, 526]]
b[n] = [[11, 22, 3], [42, 54, 36], [41, 2432, 34]]

如果“b”中的任何元素在列表“a”中有任何坐标，那么快速查找的方法是什么？例如，在给定的示例中，“a [2]”和“b [2] [1]”是相同的，因此我希望程序返回“true”。

Answer 1

将b的最里面的列表转换为集合（s），然后迭代a以检查a中是否存在s中的任何项目或不。

tot_items_b = sum(1 for x in b for y in x) #total items in b

集提供O(1)查找，因此整体复杂性将是：

O(max(len(a), tot_items_b))

def func(a, b):
   #sets can't contain mutable items, so convert lists to tuple while storing

   s = set(tuple(y) for x in b for y in x)
   #s is set([(41, 2, 34), (98, 23, 56), (42, 25, 64),...])

   return any(tuple(item) in s for item in a)

演示：

>>> a = [[1, 2, 3], [4, 5, 6], [4, 2, 3]]
>>> b = [[[11, 22, 3], [12, 34, 6], [41, 2, 34], [198, 213, 536], [1198, 1123, 1156]], [[11, 22, 3], [42, 25, 64], [43, 45, 23]], [[3, 532, 23], [4, 5, 6], [98, 23, 56], [918, 231, 526]]]
>>> func(a,b)
True

any的帮助：

>>> print any.__doc__
any(iterable) -> bool

Return True if bool(x) is True for any x in the iterable.
If the iterable is empty, return False.

使用set intersection获取所有常用元素：

>>> s_b = set(tuple(y) for x in b for y in x)
>>> s_a = set(tuple(x) for x in a)
>>> s_a & s_b
set([(4, 5, 6)])

Answer 2

首先使用itertools.chain.from_iterable展开列表，然后简单地遍历它：

>>> B = [[[11, 22, 3], [12, 34, 6], [41, 2, 34], [198, 213, 536], [1198, 1123, 1156]], [[11, 22, 3], [42, 25, 64], [43, 45, 23]], [[3, 532, 23], [4, 5, 6], [98, 23, 56], [918, 231, 526]], [[11, 22, 3], [42, 54, 36], [41, 2432, 34]]]
>>> A = [[1, 2, 3], [4, 5, 6], [4, 2, 3]]
>>> for i in itertools.chain.from_iterable(B):
...     if i in A:
...             print True, i
... 
True [4, 5, 6]

Answer 3

使用numpy：

import numpy
b=[ [], [] , [] ]
a = [[1, 2, 3], [4, 5, 6], [4, 2, 3]]
b[0] = [[11, 22, 3], [12, 34, 6], [41, 2, 34], [198, 213, 536], 
        [1198, 1123, 1156]]
b[1] = [[11, 22, 3], [42, 25, 64], [43, 45, 23]]
b[2] = [[3, 532, 23], [4, 5, 6], [98, 23, 56], [918, 231, 526]]

for p1 in a:
    for p2 in [ p for bb in b for p in bb]:
        v=numpy.linalg.norm(numpy.array(p1)-numpy.array(p2))
        if v == 0: print p1

Answer 4

试试这段代码：

a = [[1, 2, 3], [4, 5, 6], [4, 2, 3]]
b={}
b[0] = [[11, 22, 3], [12, 34, 6], [41, 2, 34], [198, 213, 536], [1198, 1123, 1156]]
b[1] = [[11, 22, 3], [42, 25, 64], [43, 45, 23]]
b[2] = [[3, 532, 23], [4, 5, 6], [98, 23, 56], [918, 231, 526]]
b[3] = [[11, 22, 3], [42, 54, 36], [41, 2432, 34]]

for i in b:
    for ii in i:
        if ii in a:
            return True

它有效，但我不确定它是否快速。

从两个列表列表中查找公共元素

4 个答案: