如何在 python 的二維數組中找到元素的行和列？

Question

我有這個二維數組：

   import numpy as np

R = int(input("Enter the number of rows:")) //4
C = int(input("Enter the number of columns:")) //5

randnums= np.random.randint(1,100, size=(R,C))

print(randnums)
    [[98 25 33  9 41]
     [67 32 67 27 85]
     [38 79 52 40 58]
     [84 76 44  9  2]]

現在，我想要搜索一個元素，output 將是它的列和行示例。

輸入要搜索的數字：40

在第 3 行第 4 列中找到第 40 號

輸入數字：100

找不到號碼

像這樣的東西？ 提前致謝

Answer 1

l = [[98, 25, 33,  9, 41],
[67, 32, 67, 27, 85],
[38, 79, 52, 40, 58],
[84, 76, 44,  9,  2]]

def fnd(l,value):
    for i,v in enumerate(l):
        if value in v:
            return {'row':i+1,'col':v.index(value)+1}
    return {'row':-1,'col':-1}
print(fnd(l,40))

{'row': 3, 'col': 4}

Answer 2

如果列數如示例所示是恆定的，則可以使用以下代碼進行搜索。

a = [[98, 25, 33, 9, 41],
     [67, 32, 67, 27, 85],
     [38, 79, 52, 40, 58],
     [84, 76, 44, 9, 2]]
a_ind = [p[x] for p in a for x in range(len(p))] # Construct 1d array as index for a to search efficiently.


def find(x):
    return a_ind.index(x) // 5 + 1, a_ind.index(x) % 5 + 1 # Here 5 is the number of columns


print(find(98), find(58), find(40))

#Output
(1, 1) (3, 5) (3, 4)

Answer 3

您可以使用numpy.where function。

r = np.random.randint(1,10, size=(5,5))

# array([[6, 5, 3, 1, 8],
#        [3, 9, 7, 5, 6],
#        [6, 2, 5, 5, 8],
#        [1, 5, 1, 1, 1],
#        [1, 6, 5, 8, 6]])

s = np.where(r == 8)

# the first array is row indices, second is column indices
# (array([0, 2, 4], dtype=int64), array([4, 4, 3], dtype=int64))

s = np.array(np.where(r == 8)).T

# transpose to get 2d array of indices
# array([[0, 4],
#        [2, 4],
#        [4, 3]], dtype=int64)