給定一個列表(或Java,我試圖在兩種語言中都這樣做),我怎麼能得到所有不同的方式來分割一個列表到不同的分組,因爲每個分組必須至少有一定的大小?我認爲獲得拆分地點的組合是最好的方式。如何獲得至少一定大小的列表分組的所有組合?
列表和最小尺寸的示例輸入是
[1,2,3], 2
和相應的輸出應是
[[1,2], [1,3], [2,3], [1,2,3]]
給定一個列表(或Java,我試圖在兩種語言中都這樣做),我怎麼能得到所有不同的方式來分割一個列表到不同的分組,因爲每個分組必須至少有一定的大小?我認爲獲得拆分地點的組合是最好的方式。如何獲得至少一定大小的列表分組的所有組合?
列表和最小尺寸的示例輸入是
[1,2,3], 2
和相應的輸出應是
[[1,2], [1,3], [2,3], [1,2,3]]
編輯
基於OP的新的輸入/輸出要求,它只是幾行:
def groupings(lst, min_size):
results = [tuple(lst)]
for i in range(min_size, len(lst)):
results.extend(combinations(lst, i))
return results
ORIGINAL
(基於不正確的假設,OP通緝給定最小分區大小的所有可能的分區)。
所以itertools.combinations()應該是一個起點。例如,
>>> list(combinations('ABCD', 2))
[('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D')]
這樣就給出了一個答案。您的分組與最小大小設置爲2「ABCD」的輸出是:
[['A', 'B', 'C', 'D']]
[['A', 'D'], ['B', 'C']]
[['A', 'C'], ['B', 'D']]
[['A', 'B'], ['C', 'D']]
所以高層次的過程應該是大致爲:
results = []
remaining = [([], list)] # Start without any groups and the full list
while remaining not empty:
groups, list = remaining.pop()
if len(list) >= min_size:
results.append(groups + list)
for size in min_size to len(list) - 1:
for combo in combinations:
new <- (groups + combo, list - combo)
if new will not cause duplicates:
remaining.append(new)
下面是一些代碼,似乎工作。爲了避免重複,並處理原始列表可能有重複的情況,我修改了itertools.combinations中的代碼,而不僅僅是使用方法。
def groupings(lst, min_size):
lst = list(lst)
# List for storing our final groupings
results = []
# Unfinished grouping, tuple storing the group and remaining sub-list
# Initialize with the empty group and original list
remaining = [([], lst)]
# Continue as long as we have unfinished groupings
while len(remaining):
# Get an unfinished grouping
current_group, current_lst = remaining.pop()
n = len(current_lst)
# If the last part of the list is big enough,
# then record the grouping
if n >= min_size:
results.append(current_group + [current_lst])
# Otherwise, discard it
else:
continue
# Helper set for getting remainder below
all_indices = set(range(n))
# Iterate the group length from min_size to the length of our current list
for r in range(min_size, n - 1):
# This is a modified version of itertools.combinations()
# http://docs.python.org/3.3/library/itertools.html#itertools.combinations
# Add the first combination to our remaining list
indices = list(range(r))
remainder = current_lst[r:]
group = current_group + [current_lst[:r]]
remaining.append((group, remainder))
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
break
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
# If one of the remaining indexes is less than the minimum used index,
# then a different iteration will handle it, so discard this one
min_index = min(indices)
remainder = []
for i in all_indices.difference(indices):
remainder.append(current_lst[i])
if i < min_index:
break
else:
# Add this combination to our remaining list
group = current_group + [[current_lst[i] for i in indices]]
remaining.append((group, remainder))
return results
結果:
>>> groupings('ABCDE', 2)
[['A', 'B', 'C', 'D', 'E']]
[['A', 'D', 'E'], ['B', 'C']]
[['A', 'C', 'E'], ['B', 'D']]
[['A', 'C', 'D'], ['B', 'E']]
[['A', 'B', 'E'], ['C', 'D']]
[['A', 'B', 'D'], ['C', 'E']]
[['A', 'B', 'C'], ['D', 'E']]
[['A', 'E'], ['B', 'C', 'D']]
[['A', 'D'], ['B', 'C', 'E']]
[['A', 'C'], ['B', 'D', 'E']]
[['A', 'B'], ['C', 'D', 'E']]
在Python,可以做到這一點遞歸:
def partition(lst, minsize=1):
yield [lst]
for n in range(minsize, len(lst)-minsize+1):
for p in partition(lst[n:], minsize):
yield [lst[:n]] + [l for l in p]
例如:
>>> lst = [1, 2, 3, 4, 5, 6, 7]
>>> partition(lst, 3)
[[[1, 2, 3, 4, 5, 6, 7]],
[[1, 2, 3], [4, 5, 6, 7]],
[[1, 2, 3, 4], [5, 6, 7]]]
>>> list(partition(lst, 2))
[[[1, 2, 3, 4, 5, 6, 7]], [[1, 2], [3, 4, 5, 6, 7]],
[[1, 2], [3, 4], [5, 6, 7]], [[1, 2], [3, 4, 5], [6, 7]],
[[1, 2, 3], [4, 5, 6, 7]], [[1, 2, 3], [4, 5], [6, 7]],
[[1, 2, 3, 4], [5, 6, 7]], [[1, 2, 3, 4, 5], [6, 7]]]
請看我上面的評論,爲什麼這不是我想要完成的。我的問題還不夠清楚。 – bab
那麼你能否編輯這個問題來準確地描述你想要達到的目標,例如樣本輸入和輸出。 – jonrsharpe
你能對你的要求更具體的每個 「分組」?例如,給定大小爲[1,2,3,4,5,6,7]的[[1,2,5,6],[3,4,7]]是有效的?還是每個組都需要它的元素在原始列表中彼此相鄰? –
我不好意思,我甚至沒有意識到接受的答案沒有完成這個。是的,我試圖找出如何獲得所有可能的組合組合,因此每個組不需要其元素彼此相鄰。 – bab
您正試圖枚舉給定集合的至少給定大小的所有子集。我在談論子集,因爲在這裏命令顯然是不重要的,我假設你不能有重複(你可以,但可能會是一個multiset,使事情變得複雜)。這是一個典型的遞歸問題。由於閾值的下限,因此您可以從完整集合開始,一次刪除一個元素,然後在小的子集上進行遞歸,而大小足夠大。 –