我有一个列表列表，如我所附的代码所示。如果有任何共同值，我想链接每个子列表。然后我想用列表的精简列表替换列表的列表。例子：如果我有一个清单[[1,2,3],[3,4]] I want [1,2,3,4]。如果我有[[4,3],[1,2,3]] I want [4,3,1,2]。如果我有[[1,2,3],[a,b],[3,4],[b,c]] I want [[1,2,3,4],[a,b,c]] or [[a,b,c],[1,2,3,4]]我不在乎是哪一个。
我快到了...
我的问题是当我遇到这样的情况时[[1,2,3],[10,5],[3,8,5]] I want [1,2,3,10,5,8]但用我当前的代码我得到[1,2,3,8,10,5]
这是我的代码：
import itertools
a = [1,2,3]
b = [3,4]
i = [21,22]
c = [88,7,8]
e = [5,4]
d = [3, 50]
f = [8,9]
g= [9,10]
h = [20,21]
lst = [a,b,c,i,e,d,f,g,h,a,c,i]*1000
#I have a lot of list but not very many different lists
def any_overlap(a, b):
sb = set(b)
return any(itertools.imap(sb.__contains__, a))
def find_uniq(lst):
''' return the uniq parts of lst'''
seen = set()
seen_add = seen.add
return [ x for x in lst if x not in seen and not seen_add(x)]
def overlap_inlist(o_lst, lstoflst):
'''
Search for overlap, using "any_overlap", of a list( o_lst) in a list of lists (lstoflst).
If there is overlap add the uniq part of the found list to the search list, and keep
track of where that list was found
'''
used_lst =[ ]
n_lst =[ ]
for lst_num, each_lst in enumerate(lstoflst):
if any_overlap(o_lst,each_lst):
n_lst.extend(each_lst)
used_lst.append(lst_num)
n_lst= find_uniq(n_lst)
return n_lst, used_lst
def comb_list(lst):
'''
For each list in a list of list find all the overlaps using 'ovelap_inlist'.
Update the list each time to delete the found lists. Return the final combined list.
'''
for updated_lst in lst:
n_lst, used_lst = overlap_inlist(updated_lst,lst)
lst[:] = [x for i,x in enumerate(lst) if i not in used_lst]
lst.insert(0,n_lst)
return lst
comb_lst = comb_list(lst)
print comb_lst
该脚本的输出是：
[[88, 7, 8, 9, 10], [1, 2, 3, 4, 50, 5], [21, 22, 20]]
我想要它，所以密钥按原始顺序排列，例如：
[[88, 7, 8, 9, 10], [1, 2, 3, 4, 5, 50,], [21, 22, 20]]
The 5 和 50 互换在新的列表[2]中
我对 python 有点陌生。我将不胜感激任何问题的解决方案或对我当前代码的评论。我不是计算机科学家，我想可能有某种算法可以快速做到这一点（也许来自集合论？）。如果有这样的算法，请指出我正确的方向。
我可能会让这种方式变得更加复杂......
谢谢你！！
