[英]Finding common elements from two lists of lists
我有兩個包含一些3D坐標的列表列表(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”中有任何坐標,那么快速查找的方法是什么? 例如,在給定的示例中,“a [2]”和“b [2] [1]”是相同的,因此我希望程序返回“true”。
轉換的最里面列出b
為一組( s
),然后遍歷a
檢查是否有項目在a
中不存在s
與否。
tot_items_b = sum(1 for x in b for y in x) #total items in b
集提供O(1)
查找,因此整體復雜性將是:
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
幫助:
>>> 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來獲取所有常見元素:
>>> 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)])
使用itertools.chain.from_iterable
首先展平列表,然后簡單地遍歷它:
>>> 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]
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
試試這段代碼:
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,但我不確定它是否快速。
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