在列表中生成字符串元素的组合，将它们限制为特定的字符数量

Question

我想生成以下电影的 5 种组合，将它们限制为特定的字符数量。

films = ['Pulp Fiction','The Lion King','Reservoir Dogs','The Wolf of Wall Street','Jackie Brown','The Shawshank Redemption','Django Unchained','The Godfather','Gone Girl','The Dark Knight']

我打算让字符数量可变（本例假设为 50 个字符）。

预期结果：

for i in film_combinations_limited:
    i[0] = ['The Shawshank Redemption, The Wolf of Wall Street'] (49 characters inc comma)
    i[1] = ['Pulp Fiction, Gone Girl, The Wolf of Wall Street'] (48 characters inc comma)
    i[2] = ['Reservoir Dogs, Pulp Fiction, The Dark Knight'] (45 characters inc comma)
    i[3] = ['Jackie Brown, Django Unchained, Pulp Fiction'] (44 characters inc comma)
    i[4] = ['The Wolf of Wall Street, The Lion King'] (38 characters inc comma)
    i[5] = ['Pulp Fiction, The Shawshank Redemption'] (38 characters inc comma)

希望充分利用字数限制，逗号和空格也需要考虑字数限制。

当前代码：

import itertools

x_raw=[el.split(' ') for el in films] 
x=[el for sublist in x_raw for el in sublist] #Not sure if I understood, what do you mean by "substring" - these 2 lines will produce substring ~ word

n=50 # character limit

res=[]
for i in range(len(x)):
   for obj in itertools.combinations(x, i+1):
      res_temp = " ".join(obj)
      #to ensure total number of characters <25 but it's high enough, that no other word from lorem/x will fit
      if((len(res_temp) < n) and (n-len(res_temp)<=min([len(el) for el in [el_x for el_x in x if el_x not in obj]] or [100]))): res.append(res_temp)

这会生成一个不包含逗号或空格的组合实例。我正在尝试实现尽可能多地填充字符限制的输出。

此代码的输出无关紧要，可以从列表更改。

如需更多信息/说明，请询问。

谢谢

Answer 1

我认为您的解决方案使事情变得过于复杂。无需用空格等分隔电影名称。

import itertools

films = ['Pulp Fiction','The Lion King','Reservoir Dogs',
         'The Wolf of Wall Street','Jackie Brown','The Shawshank Redemption',
         'Django Unchained','The Godfather','Gone Girl','The Dark Knight']

def each_short_combination(films, max_length=50):
  for i in range(len(films)):
    yielded_something = False
    for combination in itertools.combinations(films, i):
      output = ', '.join(combination)
      if len(output) < max_length:
        yield output
        yielded_something = True
    if not yielded_something:  # nothing yielded with i movie names?
      break  # no need to try longer combinations then

answers = list(each_short_combination(films))
answers.sort(key=lambda x: len(x), reverse=True)
answers = answers[:5]

for answer in answers:
  print(answer, len(answer))

打印出来：

The Wolf of Wall Street, The Shawshank Redemption 49
Pulp Fiction, The Shawshank Redemption, Gone Girl 49
The Lion King, The Wolf of Wall Street, Gone Girl 49
Reservoir Dogs, Django Unchained, The Dark Knight 49
The Wolf of Wall Street, The Godfather, Gone Girl 49

Answer 2

这是一种方法，假设您希望选择最长的：

from itertools import chain, combinations

# Itertools recipe
def powerset(iterable):
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

def get_longest_combinations(names, num, max_length):
    # All combinations with their total length
    g = ((sum(map(len, c)) + 2 * len(c) - 1, c) for c in powerset(names))
    # Filter by total length and negate length for sorting
    g = ((s, c) for s, c in g if s <= max_length)
    # Pick longest ones
    longest = sorted(g)[:-num-1:-1]
    # Format
    return [', '.join(c) for _, c in longest]

films = ['Pulp Fiction', 'The Lion King', 'Reservoir Dogs', 'The Wolf of Wall Street',
         'Jackie Brown', 'The Shawshank Redemption', 'Django Unchained', 'The Godfather',
         'Gone Girl', 'The Dark Knight']
n = 50
m = 5
result = get_longest_combinations(films, m, n)
print(*result, sep='\n')
# The Wolf of Wall Street, The Shawshank Redemption
# The Wolf of Wall Street, The Godfather, Gone Girl
# The Lion King, The Wolf of Wall Street, Gone Girl
# Reservoir Dogs, Django Unchained, The Dark Knight
# Pulp Fiction, The Shawshank Redemption, Gone Girl

你可以这样写：

longest = sorted(g)[-num:]

如果你使用堆来挑选最长的元素会更快：

import heapq

def max_n(it, n):
    it = iter(it)
    h = [next(it)]
    for elem in it:
        if len(h) < n:
            heapq.heappush(h, elem)
        elif elem > h[0]:
            heapq.heappop(h)
            heapq.heappush(h, elem)
    return sorted(h, reverse=True)

# ...
longest = max_n(g, num)

---

如果名称个数的大小很长，则幂集的大小（2ⁿ）会太大。如果您想获得最佳组合，则无法真正“解决”该问题，但如果您停止探索无法成功的部分组合，则可以稍微减少搜索空间。你可以用这样的递归算法来做到这一点：

import heapq

def get_longest_combinations(names, num, max_length):
    h = []
    _get_longest_combinations_rec(names, num, max_length, h, [], -2, 0)
    return [', '.join(c) for _, c in sorted(h, reverse=True)]

def _get_longest_combinations_rec(names, num, max_length, h, cur, cur_size, name_idx):
    if h and cur_size > h[0][0]:
        heapq.heappop(h)
    if len(h) < num:
        heapq.heappush(h, (cur_size, tuple(cur)))
    cur_size += 2
    for i in range(name_idx, len(names)):
        name = names[i]
        cur.append(name)
        cur_size += len(name)
        if cur_size < max_length:
            _get_longest_combinations_rec(
                names, num, max_length, h, cur, cur_size, i + 1)
        cur_size -= len(name)
        cur.pop()

films = ['Pulp Fiction', 'The Lion King', 'Reservoir Dogs', 'The Wolf of Wall Street',
         'Jackie Brown', 'The Shawshank Redemption', 'Django Unchained', 'The Godfather',
         'Gone Girl', 'The Dark Knight']
n = 50
m = 5
result = get_longest_combinations(films, m, n)
print(*result, sep='\n')
# The Wolf of Wall Street, The Shawshank Redemption
# The Wolf of Wall Street, The Godfather, Gone Girl
# The Lion King, The Wolf of Wall Street, Gone Girl
# Reservoir Dogs, Django Unchained, The Dark Knight
# Pulp Fiction, The Shawshank Redemption, Gone Girl

Answer 3

您已经在使用combinations，这是一个完美的工具。您可以将所有内容简化为列表理解：

from itertools import combinations

def combo(l):
    result = sorted([", ".join(y) for i in range(len(l)) for y in combinations(l,i) if len(", ".join(y))<50],
                    key=lambda s: len(s), reverse=True)
    for i in result[:5]: #get the best 5 result only
        print (i, len(i))

combo(films)

#The Wolf of Wall Street, The Shawshank Redemption 49
#Pulp Fiction, The Shawshank Redemption, Gone Girl 49
#The Lion King, The Wolf of Wall Street, Gone Girl 49
#Reservoir Dogs, Django Unchained, The Dark Knight 49
#The Wolf of Wall Street, The Godfather, Gone Girl 49