Question

我正在尝试查看字符串中的字符是否在键盘上相邻。

此代码将键盘初始化为3d数组“Array3”：

KeyboardRow1 = ["`", "1", "2", "3", "4", "5", "6", "7", "8", "9", "0", "-", "=", ""]
KeyboardRow2 = ["", "q", "w", "e", "r", "t", "y", "u", "i", "o", "p", "[", "]", ""] 
KeyboardRow3 = ["", "a", "s", "d", "f", "g", "h", "j", "k", "l", ";", "", "", ""] 
KeyboardRow4 = ["", "z", "x", "c", "v", "b", "n", "m", ",", ".", "/", "", "", ""]
KeyboardRow1S = ["~", "!", "@", "#", "$", "%", "^", "&", "*", "(", ")", "_", "+", ""]
KeyboardRow2S = ["", "Q", "W", "E", "R", "T", "Y", "U", "I", "O", "P", "{", "}", "|"]
KeyboardRow3S = ["","A", "S", "D", "F", "G", "H", "J", "K", "L", ":", "", "", ""] 
KeyboardRow4S = ["", "Z", "X", "C", "V", "B", "N", "M", "<", ">", "?", "", "", ""]
Array2R = [KeyboardRow1, KeyboardRow2, KeyboardRow3, KeyboardRow4]
Array2S = [KeyboardRow1S, KeyboardRow2S, KeyboardRow3S, KeyboardRow4S]
Array3 = [Array2R, Array2S]

此代码将输入密码的每个字符转换为数组中的位置坐标，并将这些坐标存储在KeyboardPositions中。

KeyboardPositions = []
for z in range(0,PasswordLength):
    for i in range(2):
        for j in range(4):
            for k in range (14):
                if Password[z] == str(Array3[i][j][k]):
                    KeyboardPositions.append((i,j,k))

这段代码，我遇到麻烦，检查邻接。

  for x in range(PasswordLength-1):
        for y in range(3):
            if KeyboardPositions[x][y] == (KeyboardPositions[x+1][y]-1) or KeyboardPositions[x][y] == (KeyboardPositions[x+1][y]+1):
                Adjacency = Adjacency + 1

print("There are " + str(Adjacency) + " adjacent characters")

最后一段代码试图查看行，列或移位（如果保持移位，1或0）坐标是否为彼此的+ - 1。但是，它会将“t”和“a”等相邻的内容计算在一起，因为它们只相隔一行。我怎样才能解决这个问题？感谢

Answer 1

为什么不为键盘中的每个字符分配一个数字，并通过减去它们来检查它们是否是邻居？这样的事情：
首先定义一个字典来存储数字：

n = {}

n['q'] = 1
n['w'] = 2
n['e'] = 3
... n['p'] = 10

对于第二行，从20开始计数，以避免a和p成为邻居：

n['a'] = 20
n['s'] = 21
... n['l'] = 28

然后，您可以使用此功能：

def check_neighbourhood(a,b):
    return n[a] == n[b] or abs(n[a] - n[b]) == 1 or abs(n[a] - n[b]) == 19

n[a] - n[q] = 19，所以我们应该处理这个。

Answer 2

我这样做（假设代码中存在KeyboardRowN）：

DESCRIBE <http://pl.dbpedia.org/resource/Robin_Wright>

代码评论很好，所以希望你不需要额外的解释。

您可以按照以下方式检查（我已注释掉public static double[][] deepCopy(double[][] array) { double[][] d = new double[array.length][]; for (int i = 0; i < array.length; i++) d[i] = Arrays.copyOf(array[i], array[i].length); return d; }，以便您了解我们获得这些结果的原因）：

from itertools import combinations

Rows = [KeyboardRow1, KeyboardRow2, KeyboardRow3, KeyboardRow4, \
        KeyboardRow1S, KeyboardRow2S, KeyboardRow3S, KeyboardRow4S]

row_len = len(KeyboardRow1)

def pass2index(passwd):
    """Produce a list of pairs (coordinates) for each character of a given passphrase. 
    Each pair encodes row and column in a list of keyboard keys."""
    coordinates = []
    for c in passwd:
        for i, r in enumerate(Rows):
            # check if password's char is in the current row
            try:
                # modulo 4 used for shifted keys,
                # so that 'k' and 'K' have the same row index
                coordinates.append((i % 4, r.index(c)))
            except:
                pass
    return coordinates

def adjacency(coords):
    """Produce a number of adjacency for a list of coordinates."""
    adjacent = 0
    # produce combinations of coordinates to compare each two letters
    # i.e. combinations of two coordinates
    for pairs in combinations(coords, 2):
        # two coordinates are adjacent if their rows and columns
        # differ by +/- 1
        if (abs(pairs[0][0] - pairs[1][0]) < 2) \
           and (abs(pairs[0][1] - pairs[1][1]) < 2):
            # print(pairs) # used for examples below
            adjacent += 1
    return adjacent

检查相邻键盘字符的字符串

2 个答案: