很简单，使用参考形状创建一个包含零的数组：
result = np.zeros(b.shape)# actually you can also use result = np.zeros_like(b) # but that also copies the dtype not only the shape
然后在需要的地方插入数组：
result[:a.shape[0],:a.shape[1]] = a
瞧，您已经填充了它：
print(result)array([[ 1., 1., 1., 1., 1., 0.], [ 1., 1., 1., 1., 1., 0.], [ 1., 1., 1., 1., 1., 0.], [ 0., 0., 0., 0., 0., 0.]])
如果您定义应该在左上方插入元素的位置，也可以使其更通用
result = np.zeros_like(b)x_offset = 1 # 0 would be what you wantedy_offset = 1 # 0 in your caseresult[x_offset:a.shape[0]+x_offset,y_offset:a.shape[1]+y_offset] = aresultarray([[ 0., 0., 0., 0., 0., 0.], [ 0., 1., 1., 1., 1., 1.], [ 0., 1., 1., 1., 1., 1.], [ 0., 1., 1., 1., 1., 1.]])
但请注意，偏移量不要超过允许的范围。例如
x_offset = 2，这将失败。
如果您有任意数量的维，则可以定义切片列表以插入原始数组。我发现有趣的是可以玩一下，并创建了一个填充函数，该函数可以填充（偏移）任意形状的数组，只要数组和引用的维数相同且偏移量不太大即可。
def pad(array, reference, offsets): """ array: Array to be padded reference: Reference array with the desired shape offsets: list of offsets (number of elements must be equal to the dimension of the array) """ # Create an array of zeros with the reference shape result = np.zeros(reference.shape) # Create a list of slices from offset to offset + shape in each dimension insertHere = [slice(offset[dim], offset[dim] + array.shape[dim]) for dim in range(a.ndim)] # Insert the array in the result at the specified offsets result[insertHere] = a return result
和一些测试用例：
import numpy as np# 1 Dimensiona = np.ones(2)b = np.ones(5)offset = [3]pad(a, b, offset)# 3 Dimensionsa = np.ones((3,3,3))b = np.ones((5,4,3))offset = [1,0,0]pad(a, b, offset)
