I have two lists of lists:
a = [[1,2],[5,3],[7,9]]
b = [[2,4], [6,7]]
I would like to concatenate the lists to
[[1,2,4],[5,3],[6,7,9]]
The goal is that if there are elements in the list that are the same, they will be concatenated.
2 Answers
This should work for the more general cases:
def connected_components(list_of_lists):
""" based on Howard's answer https://stackoverflow.com/a/4842897/10693596 """
temp_list_copy = list_of_lists.copy()
result = []
while len(temp_list_copy)>0:
first, *rest = temp_list_copy
first = set(first)
lf = -1
while len(first)>lf:
lf = len(first)
rest2 = []
for r in rest:
if len(first.intersection(set(r)))>0:
first |= set(r)
else:
rest2.append(r)
rest = rest2
result.append(list(first))
temp_list_copy = rest
return result
a = [[1,2],[5,3],[7,9]]
b = [[2,4], [6,7]]
a = connected_components(a)
b = connected_components(b)
for n, i in enumerate(a):
combined_list = a[n]+ [jj for j in b if set(j).intersection(set(i)) for jj in j]
a[n] = sorted(list(set(combined_list)))
print(a)
Or perhaps the following is a more pythonic version:
result = [
sorted(
set([
k
for j in b
for k in (set(i)|set(j) if set(i)&set(j) else set(i))
])
)
for i in a
]
print(result)
7 Comments
Malo
This does not seems to work for second example in the comments :( It seems better to put "a" and "b" in only one list to work with.
SultanOrazbayev
I see, in general that's a much harder question, because you are essentially trying to find connected components in your original list
a, so there are several solutions here: stackoverflow.com/questions/4842613/…SultanOrazbayev
I incorporated a function from the link to the answer.
Malo
Good! I updated my answer, with a shorter way to do it also. I will test yours.
Moch. Chamdani M
both of those codes do not works if the cases are
a = [[1,2],[5,3],[7,9]] and b = [[2,4], [6,7], [9,3]] |
Here is a solution working with sets: It works with OP example and also example given in the comments.
a = [[1,2], [2,3],[7,9]]
b = [[2,4], [6,7]]
a = a+b
b = []
while a != []:
i = a.pop()
for j in range(len(b)):
if set(b[j]).intersection(set(i)) != set():
b[j] = list(set(b[j]).union(set(i)))
break
else:
if i != []:
b.append(i)
print(b)
## [[9, 6, 7], [1, 2, 3, 4]]
Other tests:
a = [[8, 9], [1,2],[5,3],[7,9], [5, 6]]
b = [[2,4], [6,7]]
## [[3, 5, 6, 7, 8, 9], [1, 2, 4]]
a = [[1,2], [2,3],[7,9]]
b = [[2,4], [6,7]]
## [[9, 6, 7], [1, 2, 3, 4]]
[[1,2], [2,3],[7,9]]and[[2,4], [6,7]][[1,2,3,4],[6,7,9]]