Question

可能重复：
Power set and Cartesian Product of a set python

划伤旧问题。我想出了一切。现在我有一个更疯狂的问题。这是我应该得到的：

输入：scoreList（[“a”，“s”，“m”，“t”，“p”]）

输出：[['a'，1]，['am'，4]，['at'，2]，['spam'，8]]

这个I / O工作得很好，但如果我添加这样的第6个元素：

输入：scoreList（[“a”，“s”，“m”，“t”，“p”，“e”]）

程序像疯了一样漏掉了。请告诉我如何解决这个问题。感谢任何帮助

我的代码：

from itertools import chain, combinations

def ind(e,L):
    if L==[] or L=="":
        return 0
    elif L[0]==e:
        return 0
    else:
        return ind(e,L[1:])+1

def letterScore(letter, scorelist):
    if scorelist[0][0] == letter:
        return scorelist[0][1]
    elif (len(scorelist) == 1) and (scorelist[0][0] != letter):
        return 'lol. stop trying to crash my program'
    else:
        return letterScore(letter, scorelist[1:])

scorelist = [ ["a", 1], ["b", 3], ["c", 3], ["d", 2], ["e", 1], ["f", 4], ["g", 2], ["h", 4], ["i", 1], ["j", 8], ["k", 5], ["l", 1], ["m", 3], ["n", 1], ["o", 1], ["p", 3], ["q", 10], ["r", 1], ["s", 1], ["t", 1], ["u", 1], ["v", 4], ["w", 4], ["x", 8], ["y", 4], ["z", 10] ]

def wordScore(S, scorelist):
    if (len(S) == 1):
        return letterScore(S[0],scorelist)
    elif (letterScore(S[0],scorelist) == 'lol. stop trying to crash my program'):
        return 'you really want to crash me, dont you'
    else:
        return letterScore(S[0],scorelist) + wordScore(S[1:], scorelist)


def perm(l):
    sz = len(l)
    if sz <= 1:
        return [l]
    return [p[:i]+[l[0]]+p[i:]
        for i in xrange(sz) for p in perm(l[1:])]


from itertools import combinations, permutations

def findall(my_input):
    return [''.join(p) for x in range(len(my_input)) for c in combinations(my_input, x+1)
            for p in permutations(c)]

d = ["a", "am", "cab", "apple", "at", "bat", "bar", "babble", "can", "foo", "spam", "spammy", "zzyzva"]

def match(lis): 
    return match2(findall(lis))

def match2(lis): 
    if lis == []:
        return []
    elif(len(d) != ind(lis[0],d)):
        return [lis[0]] + match2(lis[1:])
    else:
        return match2(lis[1:])

def scoreList(lis):
    return match3(match(lis))

def match3(lis):
    if (lis == []):
        return []
    else:
        return [[lis[0],wordScore(lis[0],scorelist)]] + match3(lis[1:])

Answer 1

可能不是最具可读性的，但这是另一种解决方案，使用itertools和此answer：

>>> from itertools import permutations
>>> inpt = ['a', 'b', 'c']
>>> sum([map(''.join, list(permutations(inpt, l + 1))) for l in xrange(len(inpt))], [])
['a', 'b', 'c', 'ab', 'ac', 'ba', 'bc', 'ca', 'cb', 'abc', 'acb', 'bac', 'bca', 'cab', 'cba']

Answer 2

之前的答案显示了itertools包的使用情况，但是如果您不想使用它（功课就是您的唯一理由），我发现this算法是最简单的算法。实施

Python查找列表的所有组合

2 个答案: