CODEWITHSUNDEEP
pythonlist
Pradeep Kumar Jha
Pradeep Kumar Jha

Reputation: 877

Finding common elements from two lists of lists

I have two lists of lists containing some 3D coordinates like this (for example):

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]]

What will be a fast way of finding if any of the elements of "b" have any coordinate in the list "a"? For example, in the given example "a[2]" and "b[2][1]" are the same and so I want the program to return "true".

Upvotes: 3

Views: 573

Answers (4)

perreal
perreal

Reputation: 98118

Using 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

Upvotes: 1

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 251196

Convert the innermost lists of b into a set(s), and then iterate over a to check whether any item in a exist in s or not.

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

Sets provide an O(1) lookup, so the overall complexity is going to be :

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)

Demo:

>>> 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

Help on any:

>>> 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.

Use set intersection to get all the common elements:

>>> 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)])

Upvotes: 4

htynkn
htynkn

Reputation: 104

try this code:

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

It works but I don't sure it's fast or not.

Upvotes: 0

TerryA
TerryA

Reputation: 60024

Use itertools.chain.from_iterable to flatten the list first, then simply iterate through it:

>>> 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]

Upvotes: 3

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