给定一个字符串列表，打印字符串的所有组合，从每个组合中选择 1 个字符

Question

我正在尝试找到解决此问题的最有效方法。目前，我有一个解决方案，在该解决方案中，我创建了字符串与其字符串长度的映射，然后使用辅助函数将字符拼接在一起，并在执行过程中递减映射列表。当当前值减为 0 时，其左侧的数字减 1，然后该数字 + 其右侧的数字重置为其长度为 1。其实现如下所示：

def printCombinations(s):
    data_lens = []
    s = [x for x in s if x]
    for idx,val in enumerate(s): #create string length mapping list
        if len(val) == 0:
            s = s[0:idx]+s[idx+1:] #remove empty strings
            idx = idx -1
        else:
            data_lens.append(len(s[idx])-1)
    total_combos = 1
    for i in data_lens:
        total_combos = total_combos * (i+1) #total combos = lengths of the strings multiplied by each other
    current_index = len(s)-1

    while total_combos > 0:
        data_lens_copy = data_lens[:]
        if data_lens[current_index] >= 0: #if current number >= 0
            print(generateString(data_lens_copy, s))
            data_lens[current_index] -= 1
            total_combos -=1
        else:
            if current_index > 0:
                while data_lens[current_index] <= 0: #shift left while <= 0
                    current_index -= 1

                if data_lens[current_index] >= 0:
                    data_lens[current_index] -= 1
                    for i in range(current_index+1,len(s)):
                        data_lens[i] = len(s[i])-1
                    current_index = len(s)-1

def generateString(indices, strings):
    resultStr = ""
    for i in range(len(indices)-1,-1,-1):
        current_str = strings[i]
        current_index = indices[i]
        if current_str != "":
            resultStr = current_str[current_index] + resultStr
        indices[i] -= 1
    return resultStr

当这个解决方案完成工作时，它会创建一个大小相等的映射列表，并在数字达到 0 时迭代地将映射值重置到右侧。打印出包含 1 个字符的所有字符串组合的更有效方法是什么？来自每个字符串元素？

ex: ["dog","cat"] -> dc,da,dt,oc,oa,ot,gc,ga,gt

Answer 1

如果您不想使用itertools.product 并希望了解该算法，一种方法是定义递归算法。它比原始代码快（大约 16 倍），但与所涉及的 print 的比较是不确定的。

def product(item,*args):
    # termination condition: if only one string, yield the characters.
    if not args:
        yield from item
    else:
        # Recursion, yield the character for each item,
        # combined with the results of the remaining strings
        for c in item:
            for d in product(*args):
                yield c + d

print(list(product('dog','cat')))
print(list(product('ab','cd','def')))

输出：

['dc', 'da', 'dt', 'oc', 'oa', 'ot', 'gc', 'ga', 'gt']
['acd', 'ace', 'acf', 'add', 'ade', 'adf', 'bcd', 'bce', 'bcf', 'bdd', 'bde', 'bdf']