python - 通过 if 条件在二维数组之间进行元素明智的比较

Question

我有两个二维 numpy 数组 a 和 b，大小为 200*200我想根据条件对这两个数组进行元素明智的比较:

如果 a[x][y] 和 b[x][y] 是 0 打印 0

如果 a[x][y] 和 b[x][y] 中的任何一个是 0 打印 non-零元素

如果a[x][y]和b[x][y]是非零打印小元素/大元素

我试过 np.all 但它只与一个条件进行比较

我想要这样的 C 代码:

for(i=0;i<200;i++)

  for(j=0:j<200;j++)


     if( a[i][j]==0 and b[i][j]==0)
       ....
       ....

比较后我想将结果打印到另一个二维数组。我如何使用 Python 编码来实现？

Answer 1

取决于你的应用程序，但 numpy 的好处是你不必循环(因此它更快)

import numpy as np

np.random.seed(42)
a = np.random.randint(0,3,size=(5,5))
b = np.random.randint(0,3,size=(5,5))
print(a)
print(b)


where_a_eq_zero = a==0
where_b_eq_zero = b==0


# conditino: a and b is zero
con1 = where_a_eq_zero & where_b_eq_zero
print(con1)

# conditino: a or b is zero
con2 = np.logical_or(where_a_eq_zero,where_b_eq_zero)
print(con2)

# conditino: none of both is a zero element
con3 = ~where_a_eq_zero & ~where_b_eq_zero
print(con3)

# a
array([[2, 0, 2, 2, 0],
       [0, 2, 1, 2, 2],
       [2, 2, 0, 2, 1],
       [0, 1, 1, 1, 1],
       [0, 0, 1, 1, 0]])

# b
array([[0, 0, 2, 2, 2],
       [1, 2, 1, 1, 2],
       [1, 2, 2, 0, 2],
       [0, 2, 2, 0, 0],
       [2, 1, 0, 1, 1]])

# conditino: a and b is zero
array([[False,  True, False, False, False],
       [False, False, False, False, False],
       [False, False, False, False, False],
       [ True, False, False, False, False],
       [False, False, False, False, False]])

# conditino: a or b is zero
array([[ True,  True, False, False,  True],
       [ True, False, False, False, False],
       [False, False,  True,  True, False],
       [ True, False, False,  True,  True],
       [ True,  True,  True, False,  True]])

# conditino: none of both is a zero element
array([[False, False,  True,  True, False],
       [False,  True,  True,  True,  True],
       [ True,  True, False, False,  True],
       [False,  True,  True, False, False],
       [False, False, False,  True, False]])

现在如果你想进一步使用它(根据你的评论)你可以像这样使用它:

# fill up the array with np.nan because zero could be a possible 
# outcome and we can not differentiate if we missed that item or we wrote a 
# zero to this position. Later we can check if all np.nan´s are gone.
skor = np.full(a.shape,np.nan)

skor[con2] = -np.abs(a[con2]-b[con2])

my_min = np.minimum(a,b)
my_max = np.maximum(a,b)
skor[con3] = my_min[con3]/my_max[con3]

skor[con1] = 0

assert not np.any(skor==np.nan)

skor
>>> array([[-2. ,  0. ,  1. ,  1. , -2. ],
           [-1. ,  1. ,  1. ,  0.5,  1. ],
           [ 0.5,  1. , -2. , -2. ,  0.5],
           [ 0. ,  0.5,  0.5, -1. , -1. ],
           [-2. , -1. , -1. ,  1. , -1. ]])