仅当其他任何单词中第一个单词的所有索引都匹配时，如何将一个单词的索引与列表中所有其他单词的索引匹配？

Question

我正在尝试用 python 解决一个难题，而这个难题说：

Two words “interlock” if taking alternating letters from each forms a 
new word. For example, “shoe” and “cold” interlock to form “schooled. 
interlocks word's characters can be random position ex : "iam" can be
 "mia" now if "iam" and  "boy” ==> combined word  cab be yobmia

所以如果我有一个列表：

a=["shoe","cold","schooled","any","boy","hair","iahroby","i","yani"]

在此列表中：

schooled => shoe + cold 的组合

iahroby => hair+ boy的组合

yani => any + i 的组合

所以我写的论文算法是：

根据该算法，我可以执行到第 2 步，但在第 2 步之后，我如何将一个单词的每个索引与单词的其他索引匹配，我可以使用 zip 模块、finditer 或嵌套循环来做到这一点，但问题是我不必匹配一个相同的位置索引，如 "iam"=="iam" ，索引可以在随机位置，如 "iam" 将是 "mia" ，那么我该如何解决这个问题？

我尝试的是错误的代码，还没有工作，我还在分享

a=["shoe","cold","schooled","any","s","pre","our","uorpre"]




    def hello(ae):
        for i in ae:
            yield i

    ar=hello(a)

    aq=list(ar.__next__())



    def check(tr):
        for i in tr:
            for j in i:
                if j in aq:
                    ??

    check(a)

由于模式是重复的，所以我试图实现递归方法，但这不起作用。

Answer 1

订单被忽略

看起来你不关心字母位置。在这种情况下，您只需要检查来自word1 和word2 的字母是否与来自goal 的字母相加。由于您不关心位置，因此对字母进行排序就足够了：

words = ["shoe","cold","schooled","any","s","pre","our","uorpre"]

for goal in words:
    lg = len(goal)
    goal_letters = sorted(goal)
    for word1 in words:
        l1 = len(word1)
        if lg > l1:
            for word2 in words:
                if word2 > word1 and len(word2) == lg - l1:
                    if sorted(word1 + word2) == goal_letters:
                        print('%s + %s = %s' % (word1, word2, goal))

它输出：

cold + shoe = schooled
boy + hair = iahroby
any + i = yani

它效率不高，但它适用于您的示例。

考虑顺序

如果考虑到顺序，您可以将单词与::2 和1::2 分开并检查它们是否是已知单词（感谢@StefanPochmann）：

words = set(["shoe", "cold", "schooled", "any", "boy", "hair", "iahroby", "i", "yani"])

for goal in words:
    word1 = goal[::2]
    word2 = goal[1::2]
    if word1 in words and word2 in words:
      print('%s * %s = %s' % (word1, word2, goal))
# shoe * cold = schooled

它输出：

shoe * cold = schooled