[英]combining sets within a list
嗨,我正在嘗試執行以下操作,但是有點卡住了。 假設我有一組清單:
A = [set([1,2]), set([3,4]), set([1,6]), set([1,5])]
我想創建一個新列表,如下所示:
B = [ set([1,2,5,6]), set([3,4]) ]
即創建一個集合列表,如果集合重疊,則將它們合並在一起。 這可能很簡單,但是今天早上我不太明白。
這也有效並且很短:
import itertools
groups = [{'1', '2'}, {'3', '2'}, {'2', '4'}, {'5', '6'}, {'7', '8'}, {'7','9'}]
while True:
for s1, s2 in itertools.combinations(groups, 2):
if s1.intersection(s2):
break
else:
break
groups.remove(s1)
groups.remove(s2)
groups.append(s1.union(s2))
groups
這給出以下輸出:
[{'5', '6'}, {'1', '2', '3', '4'}, {'7', '8', '9'}]
while True
確實對我來說有點危險,但有人有什么想法嗎?
怎么樣:
from collections import defaultdict
def sortOverlap(listOfTuples):
# The locations of the values
locations = defaultdict(lambda: [])
# 'Sorted' list to return
sortedList = []
# For each tuple in the original list
for i, a in enumerate(listOfTuples):
for k, element in enumerate(a):
locations[element].append(i)
# Now construct the sorted list
coveredElements = set()
for element, tupleIndices in locations.iteritems():
# If we've seen this element already then skip it
if element in coveredElements:
continue
# Combine the lists
temp = []
for index in tupleIndices:
temp += listOfTuples[index]
# Add to the list of sorted tuples
sortedList.append(list(set(temp)))
# Record that we've covered this element
for element in sortedList[-1]:
coveredElements.add(element)
return sortedList
# Run the example (with tuples)
print sortOverlap([(1,2), (3,4), (1,5), (1,6)])
# Run the example (with sets)
print sortOverlap([set([1,2]), set([3,4]), set([1,5]), set([1,6])])
您可以在for循環中使用intersection()和union():
A = [set([1,2]), set([3,4]), set([1,6]), set([1,5])]
intersecting = []
for someSet in A:
for anotherSet in A:
if someSet.intersection(anotherSet) and someSet != anotherSet:
intersecting.append(someSet.union(anotherSet))
A.pop(A.index(anotherSet))
A.pop(A.index(someSet))
finalSet = set([])
for someSet in intersecting:
finalSet = finalSet.union(someSet)
A.append(finalSet)
print A
輸出: [set([3, 4]), set([1, 2, 5, 6])]
一個更直接的解決方案,
def overlaps(sets):
overlapping = []
for a in sets:
match = False
for b in overlapping:
if a.intersection(b):
b.update(a)
match = True
break
if not match:
overlapping.append(a)
return overlapping
例子
>>> overlaps([set([1,2]), set([1,3]), set([1,6]), set([3,5])])
[{1, 2, 3, 5, 6}]
>>> overlaps([set([1,2]), set([3,4]), set([1,6]), set([1,5])])
[{1, 2, 5, 6}, {3, 4}]
for set_ in A:
new_set = set(set_)
for other_set in A:
if other_set == new_set:
continue
for item in other_set:
if item in set_:
new_set = new_set.union(other_set)
break
if new_set not in B:
B.append(new_set)
輸入輸出:
A = [set([1,2]), set([3,4]), set([2,3]) ]
B = [set([1, 2, 3]), set([2, 3, 4]), set([1, 2, 3, 4])]
A = [set([1,2]), set([3,4]), set([1,6]), set([1,5])]
B = [set([1, 2, 5, 6]), set([3, 4])]
A = [set([1,2]), set([1,3]), set([1,6]), set([3,5])]
B = [set([1, 2, 3, 6]), set([1, 2, 3, 5, 6]), set([1, 3, 5])]
此功能將完成工作,而無需觸摸輸入:
from copy import deepcopy
def remove_overlapped(input_list):
input = deepcopy(input_list)
output = []
index = 1
while input:
head = input[0]
try:
next_item = input[index]
except IndexError:
output.append(head)
input.remove(head)
index = 1
continue
if head & next_item:
head.update(next_item)
input.remove(next_item)
index = 1
else:
index += 1
return output
這是執行您想要的功能。 可能不是最Python的,但能完成工作,最有可能改進很多。
from sets import Set
A = [set([1,2]), set([3,4]), set([2,3]) ]
merges = any( a&b for a in A for b in A if a!=b)
while(merges):
B = [A[0]]
for a in A[1:] :
merged = False
for i,b in enumerate(B):
if a&b :
B[i]=b | a
merged =True
break
if not merged:
B.append(a)
A = B
merges = any( a&b for a in A for b in A if a!=b)
print B
發生了以下情況,我們循環了A中的所有集合(除了第一個,因為我們已經將其添加到了B中。我們檢查與B中所有集合的相交,如果相交結果不是False(又稱空集合) ),我們在集合上執行並集並開始下一個迭代,關於集合操作,請檢查以下頁面: https : //docs.python.org/2/library/sets.html&是交集運算符|是並集運算符
您可能可以使用any()等使用更多pythonic,但是wuold需要更多處理,因此我避免了
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.