在列表中的列表数据 [英] Data in a list within a list
本文介绍了在列表中的列表数据的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
由于:
g=[[], [], [0, 2], [1, 5], [0, 2, 3, 7], [4, 6], [1, 4, 5, 6], [], [], [3, 7]]
我如何能够在摹比较每个列表,以便对列表共享任何常见的数字可以合并到一组?
How can I compare each list within g so that for lists sharing anyone common number can merge to a set?
例如,
0
中存在政[2]
和政[4]
因此他们合并为一组 {0,2,3,7}
e.g.
0
exists in g[2]
and g[4]
so they merge to a set {0,2,3,7}
我曾尝试以下,但它不工作:
I have tried the following but it doesn't work:
for i in g:
for j in g:
if k in i == l in j:
m=set(i+j)
我希望让尽可能多。
I want to make the largest possible set.
推荐答案
作为更快的方式你可以先创建一组项目的列表,len个以上的(<$ C $ ç>取值)。然后通过你的清单,更新到位与 工会
功能!
As a much faster way You can first create a list of the set of items with len more than one (s
) . then go through your list and update in place with union
function !
s=map(set,g)
def find_intersection(m_list):
for i,v in enumerate(m_list) :
for j,k in enumerate(m_list[i+1:],i+1):
if v & k:
m_list[i]=v.union(m_list.pop(j))
return find_intersection(m_list)
return m_list
演示:
g=[[], [], [0, 2], [1, 5], [0, 2, 3, 7], [4, 6], [1, 4, 5, 6], [], [], [3, 7]]
s=map(set,g)
print find_intersection(s)
[set([0, 2, 3, 7]), set([1, 4, 5, 6])]
g=[[1,2,3],[3,4,5],[5,6],[6,7],[9,10],[10,11]]
s=map(set,g)
print find_intersection(s)
[set([1, 2, 3, 4, 5, 6, 7]), set([9, 10, 11])]
g=[[], [1], [0,2], [1, 5], [0, 2, 3, 7], [4, 6], [1, 4, 5, 6], [], [], [3, 7]]
s=map(set,g)
print find_intersection(s)
[set([1, 4, 5, 6]), set([0, 2, 3, 7])]
基准与@马克的回答是:
Benchmark with @Mark's answer :
from timeit import timeit
s1="""g=[[], [], [0, 2], [1, 5], [0, 2, 3, 7], [4, 6], [1, 4, 5, 6], [], [], [3, 7]]
sets = [set(i+j) for i in g for j in g if i!=j and (set(i) & set(j))]
"""
s2="""g=[[], [], [0, 2], [1, 5], [0, 2, 3, 7], [4, 6], [1, 4, 5, 6], [], [], [3, 7]]
s=map(set,g)
def find_intersection(m_list):
for i,v in enumerate(m_list) :
for j,k in enumerate(m_list[i+1:],i+1):
if v & k:
s[i]=v.union(m_list.pop(j))
return find_intersection(m_list)
return m_list
"""
print ' first: ' ,timeit(stmt=s1, number=100000)
print 'second : ',timeit(stmt=s2, number=100000)
first: 3.8284008503
second : 0.213887929916
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