finding consecutive numbers in a matrix with python numpy-CodePudding

I am practicing some exercises and have been going in circles trying to figure it out. The first part is to create a 5x5 matrix with NumPy giving it random values, and I’ve already solved it.

Now I need to see if the matrix, either horizontally or vertically, has consecutive numbers (for example: The matrix of the attached image does not have consecutive numbers). Here is there are 4 consecutive numbers in the first column: [[50 92 78 84 36] [51 33 94 73 32] [52 94 35 47 9] [53 5 60 55 67] [83 51 56 99 18]]` Here are 4 consecutive numbers in the last row [[50 92 78 84 36] [41 33 94 73 32] [72 94 35 47 9] [55 5 60 55 67] [84 85 86 87 18]]"

The last step is to continue randomizing the array until you find those consecutive numbers.

CodePudding user response：

All things you need to do is to check whether a position in a mtrix minus its previous position (both horizontally and vertically) equals to 1 or -1.

You can use np.diff method, this calculates the difference of a matrix along a given axis, for example:

M = np.random.randint(1, 100, size=(5,5))
print(M)
print(np.diff(M))  # Horizontally
print(np.diff(M.T))  # Vertically

[[ 4 69 99 11 61]
 [81 36 12 27  3]
 [92 86 68 40 49]
 [11 58 49  6 32]
 [74 52 87 95  1]]
[[ 65  30 -88  50]
 [-45 -24  15 -24]
 [ -6 -18 -28   9]
 [ 47  -9 -43  26]
 [-22  35   8 -94]]
[[ 77  11 -81  63]
 [-33  50 -28  -6]
 [-87  56 -19  38]
 [ 16  13 -34  89]
 [-58  46 -17 -31]]

Then you can use np.any together with np.abs to find whether these difference matrices contain 1 or -1:

def has_consecutive_number(M):
    return np.any(np.abs(np.diff(M)) == 1) or np.any(np.abs(np.diff(M.T)) == 1)

Testing:

M = np.random.randint(1, 100, size=(5,5))
print(M)
print(has_consecutive_number(M))

[[73 72 23 16 43]
 [ 4 43 14 42 98]
 [76 89 82 13 18]
 [43 93 73 17 29]
 [85 49 77 42 78]]
True

[[ 8 96  2 91  2]
 [37 55 37 74 94]
 [51 97  6 94 15]
 [74 95 52 46 82]
 [ 3 95 42 72 23]]
False