查找矩阵中对角线的单词字母

我有一个5*6的矩阵，所有的值都是英文字母。我必须在矩阵中从左到右、从右到左、从上到下、从上到对角线找到特定的单词。其实这是一个字谜。

我能找到“左右”、“右左”、“上下”和“下上”这些词。当涉及到对角线查找单词时，事情会变得更加混乱。我提供了从左到右和从右到左搜索的示例代码。

#the word I am looking for
word_list = ["CRAM", "ROTLQ", "TDML", "COOI"]

# create the matrix array or puzzle
letter_array = np.matrix([['C','R','A','M','B','A' ],
 ['R','O','T','L','Q','C' ],
 ['E','O','O','A','U','A'],
 ['I','T','E','I','A','L' ],
 ['I','A','L','M','D','T']])

#itenarate through word list
for word in word_list:

    #flatten the array, convert it to list, 
    #finally join everythig to make a long string
    left_right = ''.join(letter_array.flatten().tolist()[0])

    # flip the string to search from right to left
    right_left = left_right[::-1]


    # if the word is there, it gives the index otherwise it gives -1
    if left_right.find(word)>-1:
        row, col = divmod(left_right.find(word), 6)
        print ("The word is in left to right combination, row and column = ", row, col)

    # look from right to left
    elif right_left.find(word)>-1:
        row, col = divmod(right_left.find(word), 6)
        print ("The word is in right to left combination, row and column = ", row, col)

    else:
        print ("The word is in up down or diagonally located")

Results

The word is in left to right combination, row and column =  0 0
The word is in left to right combination, row and column =  1 0
The word is in right to left combination, row and column =  0 0
The word is in up down or diagonally located

然而，通过对矩阵进行转置，也可以完成上下和上下的搜索。但是，我不知道如何对角搜索。有什么方法可以对角搜索吗？或者，还有其他简单的解决方案可以解决整个问题吗？

正如您已经知道如何搜索字符串中的单词（但请阅读我的评论），我在这里提供了一种如何提取对角线的方法。

def diagelem(ll, a, b):
    try:
        if a < 0 or b < 0:
            raise IndexError
        return ll[a, b]
    except IndexError:
        return None

我需要定义这个函数来提取矩阵的单个对角元素。为了避免矩阵周围出现圆形，我添加了一个raise IndexError如果索引小于 0，则声明。我认为您不希望这样，但如果我错了，请删除if函数内的语句应该赋予您循环性。

dd = []
for j in range(-letter_array.shape[0]+1, letter_array.shape[0]):
    dd.append([diagelem(letter_array, i, i+j) for i in range(letter_array.shape[1])])

for j in range(0, 2*letter_array.shape[0]):
    dd.append([diagelem(letter_array, i, -i+j) for i in range(letter_array.shape[1])])

diagonals = []
for diag in dd:
    diagword = ''.join([letter for letter in diag if letter is not None])
    if len(diagword) > 0:
        diagonals.append(diagword)

print(diagonals)

前两个循环从对角线构建单词。第一个循环从左上到右下对角线构建单词，第二个循环从左下到右上构建单词。dd这是一个列表列表，其中有几个None里面。第三个循环将构建单词的内部列表连接在一起。到底diagonals是从对角线构建的单词列表。您可以搜索是否包含以下单词word_list位于任何单词内diagonal使用您发布的相同逻辑。