我的问题与this question完全相同.我有字符的数组(列表).我想从该列表中获取所有可能的序列组合,但具有字符限制(例如：最多2个字符).此外,在排列行中不能重复单个字符：
chars = ['a','b','c','d']
# output
output = [['a','d'],['ab',['a','bc','cd'],['abc',# this one will be exempted
['a','bcd'],# this one will be exempted
['abcd']] # this one will be exempted
我知道我可以在生成和构建序列时检查条件以省略超限字符组合.但它会增加运行时间.我的目的是减少现有的执行时间.
没有字符数限制,组合将生成为2 ^(N-1).如果列表超过15个字符,则执行程序将花费很长时间.因此,我想减少字符数限制的组合数.
优先级是性能.我已经研究并试了两天没有任何成功.
一种方法是迭代输入列表并逐步构建组合.在每个步骤中,从输入列表中取出下一个字符并将其添加到先前生成的组合中.
from collections import defaultdict
def make_combinations(seq,maxlen):
# memo is a dict of {length_of_last_word: list_of_combinations}
memo = defaultdict(list)
memo[1] = [[seq[0]]] # put the first character into the memo
seq_iter = iter(seq)
next(seq_iter) # skip the first character
for char in seq_iter:
new_memo = defaultdict(list)
# iterate over the memo and expand it
for wordlen,combos in memo.items():
# add the current character as a separate word
new_memo[1].extend(combo + [char] for combo in combos)
# if the maximum word length isn't reached yet,add a character to the last word
if wordlen < maxlen:
word = combos[0][-1] + char
new_memo[wordlen+1] = newcombos = []
for combo in combos:
combo[-1] = word # overwrite the last word with a longer one
newcombos.append(combo)
memo = new_memo
# flatten the memo into a list and return it
return [combo for combos in memo.values() for combo in combos]
输出：
[['a','cd']]
对于短输入,此实现比天真的itertools.product方法慢：
input: a b c d
maxlen: 2
iterations: 10000
itertools.product: 0.11653625800136069 seconds
make_combinations: 0.1657387from collections import defaultdict
def make_combinations(seq,add a character to the last word
if wordlen < maxlen:
word = combos[0][-1] + char
new_memo[wordlen+1] = newcombos = []
for combo in combos:
combo[-1] = word # overwrite the last word with a longer one
newcombos.append(combo)
memo = new_memo
# flatten the memo into a list and return it
return [combo for combos in memo.values() for combo in combos]
41118 seconds
但是当输入列表更长时,它会快速恢复：
input: a b c d e f g h i j k
maxlen: 2
iterations: 10000
itertools.product: 6.9087735799985240 seconds
make_combinations: 1.2037671390007745 seconds
