将位置处的布尔掩码应用于 2D numpy 数组

Question

我已经四处寻找了一段时间，但找不到具体问题的答案。

假设我有这个 dtype Object 的空 2D numpy 数组和一个布尔掩码：

class Object(object):
    def __repr__(self):
        return "aObj"

a = np.empty((7, 10), dtype=Object)
mask = np.array([[False, True, False], [True, False, True], [True, True, True]])

我想应用此掩码将记录分配给 Object 的实例，给定必须应用此掩码的左上位置：

top_left = (2, 3)
obj = Object()
result = apply_mask(a, mask, top_left, obj)
result

array([[None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, aObj, None, None, None, None, None],
       [None, None, None, aObj, None, aObj, None, None, None, None],
       [None, None, None, aObj, aObj, aObj, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None]],
      dtype=Object)

如您所见，行索引 2 的第四列设置为 aObject，因为第一个掩码行是 [False, True False]，只有第二个索引应该设置为 obj。同样，行索引 3 的第 3 列和第 5 列在掩码的第二行之后设置为 obj：[True, False, True]。

我一直在查看 numpy.ma 模块，但找不到我一直在寻找的东西。到目前为止，我一直在搞乱np.indeces：

yd, xd = np.indices(a.shape)
result = a
result[(xd >= top_left[1]) &
       (xd < top_left[1] + mask.shape[1]) &
       (yd >= top_left[0]) &
       (yd < top_left[0] + mask.shape[0])] = obj

但显然这只会产生obj的矩形，我还没有真正应用蒙版，只是形状。

有什么建议吗？

---

编辑：一个重要的细节是，如果另一个 Object 实例已经存在于掩码中 False 覆盖的记录之一中，则该实例应该持续存在。请参阅以下示例：

a = np.empty((7, 10), dtype=Object)
a[2 ,3] = Object()
mask = np.array([[False, True, False], [True, False, True], [True, True, True]])
top_left = (2, 3)
result = apply_mask(a, mask, top_left, Object())
result

array([[None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, aObj, aObj, None, None, None, None, None],
       [None, None, None, aObj, None, aObj, None, None, None, None],
       [None, None, None, aObj, aObj, aObj, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None]],
      dtype=Object)

这确实让我相信我需要 np.ma 模块，因为那些掩码数组实际上支持 empty 记录。

---

编辑：

使用切片，我可以在我想要的确切位置分配一个对象数组

result[top_left[1]:top_left[1] + mask.shape[1],
       top_left[0]:top_left[0] + mask.shape[0]] = obj_array

但是这不满足之前编辑中提到的约束。

所以我正在查看掩码数组，我希望使用以下代码它会起作用，但是唉：

obj_mask = np.ma.masked_where(~mask, np.full(mask.shape, obj))
result = a
result[top_left[1]:top_left[1] + mask.shape[1],
       top_left[0]:top_left[0] + mask.shape[0]] = obj_mask
result

array([[None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, aObj, aObj, aObj, None, None, None, None, None],
       [None, None, aObj, aObj, aObj, None, None, None, None, None],
       [None, None, aObj, aObj, aObj, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None]],
      dtype=object)

撇开非一错误不谈，为什么将掩码数据分配给数组时突然可用？

Answer 1

我找到了答案！在编辑中我已经非常接近了。

off-by-one 错误是由索引被交换的事实引起的，我必须使用 np.ma.compressed 方法应用对象掩码。

全文：

class Object(object):
    def __repr__(self):
        return "aObj"

a = np.empty((7, 10), dtype=Object)
a[2, 3] = Object()
mask = np.array([[False, True, False], [True, False, True], [True, True, True]])
top_left = (2, 3)
obj = Object()
a

array([[None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, aObj, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None]],
      dtype=object)

我添加了一个对象以表明它在应用蒙版后仍然存在。

obj_mask = np.ma.masked_where(~mask, np.full(mask.shape, obj))
obj_mask

masked_array(
  data=[[--, aObj, --],
        [aObj, --, aObj],
        [aObj, aObj, aObj]],
  mask=[[ True, False,  True],
        [False,  True, False],
        [False, False, False]],
  fill_value='?',
  dtype=object)

蒙版的创建方式与编辑相同。

result = a
result[top_left[0]:top_left[0] + mask.shape[0],
       top_left[1]:top_left[1] + mask.shape[1]][~obj_mask.mask] = np.ma.compressed(obj_mask)
result

array([[None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, aObj, aObj, None, None, None, None, None],
       [None, None, None, aObj, None, aObj, None, None, None, None],
       [None, None, None, aObj, aObj, aObj, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None],
       [None, None, None, None, None, None, None, None, None, None]],
      dtype=object)

如您所见，所有值都已应用。

最后：

>>> a[2, 3] == a[2, 4]
False

>>> a[2, 4] == a[3, 3]
True

正如预期的那样。