python 如何用零填充 numpy 数组
python how to pad numpy array with zeros
我想知道如何使用带有 numpy 版本 1.5.0 的 python 2.6.6 用零填充 2D numpy 数组。但这些是我的局限性。因此我不能使用 np.pad。例如，我想用零填充 a，使其形状与 b 匹配。我之所以要这样做，是因为我可以做到：
b-a
这样
>>> a
array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
>>> b
array([[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.]])
>>> c
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]])
我能想到的唯一方法是追加，但这看起来很难看。是否有可能使用 b.shape 的更清洁的解决方案？
编辑，
感谢 MSeifert 的回答。我不得不稍微清理一下，这就是我得到的：
def pad(array, reference_shape, offsets):
"""
array: Array to be padded
reference_shape: tuple of size of ndarray to create
offsets: list of offsets (number of elements must be equal to the dimension of the array)
will throw a ValueError if offsets is too big and the reference_shape cannot handle the offsets
"""
# 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(offsets[dim], offsets[dim] + array.shape[dim]) for dim in range(array.ndim)]
# Insert the array in the result at the specified offsets
result[insertHere] = array
return result
非常简单，您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，你填充了它：
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.]])
如果您定义应该插入左上角元素的位置，还可以使它更通用一些
result = np.zeros_like(b)
x_offset = 1 # 0 would be what you wanted
y_offset = 1 # 0 in your case
result[x_offset:a.shape[0]+x_offset,y_offset:a.shape[1]+y_offset] = a
result
array([[ 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1.],
[ 0., 1., 1., 1., 1., 1.],
[ 0., 1., 1., 1., 1., 1.]])
但请注意不要让偏移量大于允许值。例如，对于 x_offset = 2，这将失败。
如果你有任意数量的维度，你可以定义一个切片列表来插入原始数组。我发现玩一下很有趣，并创建了一个填充函数，只要数组和引用具有相同的维数并且偏移量不太大，它就可以填充（带偏移量）任意形状的数组。
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 Dimension
a = np.ones(2)
b = np.ones(5)
offset = [3]
pad(a, b, offset)
# 3 Dimensions
a = np.ones((3,3,3))
b = np.ones((5,4,3))
offset = [1,0,0]
pad(a, b, offset)
我知道你的主要问题是你需要计算 d=b-a 但你的数组有不同的大小。不需要中间填充c
不用填充也能解决这个问题：
import numpy as np
a = np.array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
b = np.array([[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.]])
d = b.copy()
d[:a.shape[0],:a.shape[1]] -= a
print d
输出：
[[ 2. 2. 2. 2. 2. 3.]
[ 2. 2. 2. 2. 2. 3.]
[ 2. 2. 2. 2. 2. 3.]
[ 3. 3. 3. 3. 3. 3.]]
NumPy 1.7.0（当 numpy.pad was added) is pretty old now (it was released in 2013) so even though the question asked for a way without using that function I thought it could be useful to know how that could be achieved using numpy.pad.
其实很简单：
>>> import numpy as np
>>> a = np.array([[ 1., 1., 1., 1., 1.],
... [ 1., 1., 1., 1., 1.],
... [ 1., 1., 1., 1., 1.]])
>>> np.pad(a, [(0, 1), (0, 1)], mode='constant')
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.]])
在这种情况下，我使用 0 是 mode='constant' 的默认值。但它也可以通过显式传递来指定：
>>> np.pad(a, [(0, 1), (0, 1)], mode='constant', constant_values=0)
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.]])
以防万一第二个参数 ([(0, 1), (0, 1)]) 看起来令人困惑：每个列表项（在本例中为元组）对应一个维度，其中的项表示 before[=45= 的填充]（第一个元素）和 after（第二个元素）。所以：
[(0, 1), (0, 1)]
^^^^^^------ padding for second dimension
^^^^^^-------------- padding for first dimension
^------------------ no padding at the beginning of the first axis
^--------------- pad with one "value" at the end of the first axis.
在这种情况下，第一轴和第二轴的填充是相同的，因此也可以只传入二元组：
>>> np.pad(a, (0, 1), mode='constant')
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.]])
如果前后填充相同，甚至可以省略元组（虽然在这种情况下不适用）：
>>> np.pad(a, 1, mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 0., 0., 0., 0., 0., 0.]])
或者如果轴前后的填充相同但不同，你也可以省略内部元组中的第二个参数：
>>> np.pad(a, [(1, ), (2, )], mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0., 0.]])
但是我倾向于始终使用显式的，因为它很容易出错（当 NumPys 的期望与您的意图不同时）：
>>> np.pad(a, [1, 2], mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0.]])
这里 NumPy 认为您想用每个轴前 1 个元素和后 2 个元素填充所有轴！即使您打算在轴 1 中填充 1 个元素，在轴 2 中填充 2 个元素。
我使用元组列表作为填充，请注意这只是 "my convention"，您还可以使用列表列表或元组元组，甚至数组元组。 NumPy 只检查参数的长度（或者如果它没有长度）和每个项目的长度（或者如果它有长度）！
如果您需要将 1 的栅栏添加到数组中：
>>> mat = np.zeros((4,4), np.int32)
>>> mat
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]])
>>> mat[0,:] = mat[:,0] = mat[:,-1] = mat[-1,:] = 1
>>> mat
array([[1, 1, 1, 1],
[1, 0, 0, 1],
[1, 0, 0, 1],
[1, 1, 1, 1]])
我知道我有点晚了，但如果您想执行相对填充（也称为边缘填充），可以按照以下方法实现它。请注意，赋值的第一个实例导致 zero-padding，因此您可以将其用于 zero-padding 和相对填充（这是您将原始数组的边缘值复制到填充数组的位置）。
def replicate_padding(arr):
"""Perform replicate padding on a numpy array."""
new_pad_shape = tuple(np.array(arr.shape) + 2) # 2 indicates the width + height to change, a (512, 512) image --> (514, 514) padded image.
padded_array = np.zeros(new_pad_shape) #create an array of zeros with new dimensions
# perform replication
padded_array[1:-1,1:-1] = arr # result will be zero-pad
padded_array[0,1:-1] = arr[0] # perform edge pad for top row
padded_array[-1, 1:-1] = arr[-1] # edge pad for bottom row
padded_array.T[0, 1:-1] = arr.T[0] # edge pad for first column
padded_array.T[-1, 1:-1] = arr.T[-1] # edge pad for last column
#at this point, all values except for the 4 corners should have been replicated
padded_array[0][0] = arr[0][0] # top left corner
padded_array[-1][0] = arr[-1][0] # bottom left corner
padded_array[0][-1] = arr[0][-1] # top right corner
padded_array[-1][-1] = arr[-1][-1] # bottom right corner
return padded_array
复杂度分析：
最佳解决方案是numpy的pad方法。
平均运行 5 次后，具有相对填充的 np.pad 仅比上面定义的函数好 8%。这表明这是相对和 zero-padding 填充的最佳方法。
#My method, replicate_padding
start = time.time()
padded = replicate_padding(input_image)
end = time.time()
delta0 = end - start
#np.pad with edge padding
start = time.time()
padded = np.pad(input_image, 1, mode='edge')
end = time.time()
delta = end - start
print(delta0) # np Output: 0.0008790493011474609
print(delta) # My Output: 0.0008130073547363281
print(100*((delta0-delta)/delta)) # Percent difference: 8.12316715542522%
Tensorflow 还为 resizing/padding 图像 tf.image.pad tf.pad.
实现了函数
padded_image = tf.image.pad_to_bounding_box(image, top_padding, left_padding, target_height, target_width)
padded_image = tf.pad(image, paddings, "CONSTANT")
这些函数的工作方式与 tensorflow 的其他输入管道功能一样，对于机器学习应用程序来说效果会更好。
TL;DR
def pad_n_cols_left_of_2d_matrix(arr, n):
"""Adds n columns of zeros to left of 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of columns that are added to the left of the matrix.
"""
padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
padded_array[:, n:] = arr
return padded_array
def pad_n_cols_right_of_2d_matrix(arr, n):
"""Adds n columns of zeros to right of 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of columns that are added to the right of the matrix.
"""
padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
padded_array[:, : arr.shape[1]] = arr
return padded_array
def pad_n_rows_above_2d_matrix(arr, n):
"""Adds n rows of zeros above 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of rows that are added above the matrix.
"""
padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
padded_array[n:, :] = arr
return padded_array
def pad_n_rows_below_2d_matrix(arr, n):
"""Adds n rows of zeros below 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of rows that are added below the matrix.
"""
padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
padded_array[: arr.shape[0], :] = arr
return padded_array
输出
Original array:
[[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad left:
[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad right:
[[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]]
Pad above:
[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad below:
[[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]]
动机
我来这里搜索如何填充数组并找到了很多文本，尽管我的 objective 问题很简单：用 n 行或列填充二维数组.
我认为解释更好，因为它们有助于建立理解。但是，为了让一些人可以节省一些时间，如果他们愿意的话，这里有一个 copy-pastable 功能。
代码说明
每个函数向传入数组添加 n 行或列。代码假定传入数组是 2D numpy 数组。填充的方向根据函数definition/names。有关更详细的解释，我想参考 reader MSeifert 的回答。
我想知道如何使用带有 numpy 版本 1.5.0 的 python 2.6.6 用零填充 2D numpy 数组。但这些是我的局限性。因此我不能使用 np.pad。例如，我想用零填充 a，使其形状与 b 匹配。我之所以要这样做，是因为我可以做到：
b-a
这样
>>> a
array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
>>> b
array([[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.]])
>>> c
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]])
我能想到的唯一方法是追加，但这看起来很难看。是否有可能使用 b.shape 的更清洁的解决方案？
编辑， 感谢 MSeifert 的回答。我不得不稍微清理一下，这就是我得到的：
def pad(array, reference_shape, offsets):
"""
array: Array to be padded
reference_shape: tuple of size of ndarray to create
offsets: list of offsets (number of elements must be equal to the dimension of the array)
will throw a ValueError if offsets is too big and the reference_shape cannot handle the offsets
"""
# 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(offsets[dim], offsets[dim] + array.shape[dim]) for dim in range(array.ndim)]
# Insert the array in the result at the specified offsets
result[insertHere] = array
return result
非常简单，您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，你填充了它：
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.]])
如果您定义应该插入左上角元素的位置，还可以使它更通用一些
result = np.zeros_like(b)
x_offset = 1 # 0 would be what you wanted
y_offset = 1 # 0 in your case
result[x_offset:a.shape[0]+x_offset,y_offset:a.shape[1]+y_offset] = a
result
array([[ 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1.],
[ 0., 1., 1., 1., 1., 1.],
[ 0., 1., 1., 1., 1., 1.]])
但请注意不要让偏移量大于允许值。例如，对于 x_offset = 2，这将失败。
如果你有任意数量的维度，你可以定义一个切片列表来插入原始数组。我发现玩一下很有趣，并创建了一个填充函数，只要数组和引用具有相同的维数并且偏移量不太大，它就可以填充（带偏移量）任意形状的数组。
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 Dimension
a = np.ones(2)
b = np.ones(5)
offset = [3]
pad(a, b, offset)
# 3 Dimensions
a = np.ones((3,3,3))
b = np.ones((5,4,3))
offset = [1,0,0]
pad(a, b, offset)
我知道你的主要问题是你需要计算 d=b-a 但你的数组有不同的大小。不需要中间填充c
不用填充也能解决这个问题：
import numpy as np
a = np.array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
b = np.array([[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3.]])
d = b.copy()
d[:a.shape[0],:a.shape[1]] -= a
print d
输出：
[[ 2. 2. 2. 2. 2. 3.]
[ 2. 2. 2. 2. 2. 3.]
[ 2. 2. 2. 2. 2. 3.]
[ 3. 3. 3. 3. 3. 3.]]
NumPy 1.7.0（当 numpy.pad was added) is pretty old now (it was released in 2013) so even though the question asked for a way without using that function I thought it could be useful to know how that could be achieved using numpy.pad.
其实很简单：
>>> import numpy as np
>>> a = np.array([[ 1., 1., 1., 1., 1.],
... [ 1., 1., 1., 1., 1.],
... [ 1., 1., 1., 1., 1.]])
>>> np.pad(a, [(0, 1), (0, 1)], mode='constant')
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.]])
在这种情况下，我使用 0 是 mode='constant' 的默认值。但它也可以通过显式传递来指定：
>>> np.pad(a, [(0, 1), (0, 1)], mode='constant', constant_values=0)
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.]])
以防万一第二个参数 ([(0, 1), (0, 1)]) 看起来令人困惑：每个列表项（在本例中为元组）对应一个维度，其中的项表示 before[=45= 的填充]（第一个元素）和 after（第二个元素）。所以：
[(0, 1), (0, 1)]
^^^^^^------ padding for second dimension
^^^^^^-------------- padding for first dimension
^------------------ no padding at the beginning of the first axis
^--------------- pad with one "value" at the end of the first axis.
在这种情况下，第一轴和第二轴的填充是相同的，因此也可以只传入二元组：
>>> np.pad(a, (0, 1), mode='constant')
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.]])
如果前后填充相同，甚至可以省略元组（虽然在这种情况下不适用）：
>>> np.pad(a, 1, mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 1., 1., 1., 1., 1., 0.],
[ 0., 0., 0., 0., 0., 0., 0.]])
或者如果轴前后的填充相同但不同，你也可以省略内部元组中的第二个参数：
>>> np.pad(a, [(1, ), (2, )], mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0., 0.]])
但是我倾向于始终使用显式的，因为它很容易出错（当 NumPys 的期望与您的意图不同时）：
>>> np.pad(a, [1, 2], mode='constant')
array([[ 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 1., 1., 1., 1., 1., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0.]])
这里 NumPy 认为您想用每个轴前 1 个元素和后 2 个元素填充所有轴！即使您打算在轴 1 中填充 1 个元素，在轴 2 中填充 2 个元素。
我使用元组列表作为填充，请注意这只是 "my convention"，您还可以使用列表列表或元组元组，甚至数组元组。 NumPy 只检查参数的长度（或者如果它没有长度）和每个项目的长度（或者如果它有长度）！
如果您需要将 1 的栅栏添加到数组中：
>>> mat = np.zeros((4,4), np.int32)
>>> mat
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]])
>>> mat[0,:] = mat[:,0] = mat[:,-1] = mat[-1,:] = 1
>>> mat
array([[1, 1, 1, 1],
[1, 0, 0, 1],
[1, 0, 0, 1],
[1, 1, 1, 1]])
我知道我有点晚了，但如果您想执行相对填充（也称为边缘填充），可以按照以下方法实现它。请注意，赋值的第一个实例导致 zero-padding，因此您可以将其用于 zero-padding 和相对填充（这是您将原始数组的边缘值复制到填充数组的位置）。
def replicate_padding(arr):
"""Perform replicate padding on a numpy array."""
new_pad_shape = tuple(np.array(arr.shape) + 2) # 2 indicates the width + height to change, a (512, 512) image --> (514, 514) padded image.
padded_array = np.zeros(new_pad_shape) #create an array of zeros with new dimensions
# perform replication
padded_array[1:-1,1:-1] = arr # result will be zero-pad
padded_array[0,1:-1] = arr[0] # perform edge pad for top row
padded_array[-1, 1:-1] = arr[-1] # edge pad for bottom row
padded_array.T[0, 1:-1] = arr.T[0] # edge pad for first column
padded_array.T[-1, 1:-1] = arr.T[-1] # edge pad for last column
#at this point, all values except for the 4 corners should have been replicated
padded_array[0][0] = arr[0][0] # top left corner
padded_array[-1][0] = arr[-1][0] # bottom left corner
padded_array[0][-1] = arr[0][-1] # top right corner
padded_array[-1][-1] = arr[-1][-1] # bottom right corner
return padded_array
复杂度分析：
最佳解决方案是numpy的pad方法。
平均运行 5 次后，具有相对填充的 np.pad 仅比上面定义的函数好 8%。这表明这是相对和 zero-padding 填充的最佳方法。
#My method, replicate_padding
start = time.time()
padded = replicate_padding(input_image)
end = time.time()
delta0 = end - start
#np.pad with edge padding
start = time.time()
padded = np.pad(input_image, 1, mode='edge')
end = time.time()
delta = end - start
print(delta0) # np Output: 0.0008790493011474609
print(delta) # My Output: 0.0008130073547363281
print(100*((delta0-delta)/delta)) # Percent difference: 8.12316715542522%
Tensorflow 还为 resizing/padding 图像 tf.image.pad tf.pad.
实现了函数padded_image = tf.image.pad_to_bounding_box(image, top_padding, left_padding, target_height, target_width)
padded_image = tf.pad(image, paddings, "CONSTANT")
这些函数的工作方式与 tensorflow 的其他输入管道功能一样，对于机器学习应用程序来说效果会更好。
TL;DR
def pad_n_cols_left_of_2d_matrix(arr, n):
"""Adds n columns of zeros to left of 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of columns that are added to the left of the matrix.
"""
padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
padded_array[:, n:] = arr
return padded_array
def pad_n_cols_right_of_2d_matrix(arr, n):
"""Adds n columns of zeros to right of 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of columns that are added to the right of the matrix.
"""
padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
padded_array[:, : arr.shape[1]] = arr
return padded_array
def pad_n_rows_above_2d_matrix(arr, n):
"""Adds n rows of zeros above 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of rows that are added above the matrix.
"""
padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
padded_array[n:, :] = arr
return padded_array
def pad_n_rows_below_2d_matrix(arr, n):
"""Adds n rows of zeros below 2D numpy array matrix.
:param arr: A two dimensional numpy array that is padded.
:param n: the number of rows that are added below the matrix.
"""
padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
padded_array[: arr.shape[0], :] = arr
return padded_array
输出
Original array:
[[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad left:
[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad right:
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[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]]
Pad above:
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[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]]
Pad below:
[[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.5 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.3 0.7 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.5 0.8 1. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.2 0.4 0.6 0.8 1. 0. 0. 0. 0. 0. 0. ]
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[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]]
动机
我来这里搜索如何填充数组并找到了很多文本，尽管我的 objective 问题很简单：用 n 行或列填充二维数组.
我认为解释更好，因为它们有助于建立理解。但是，为了让一些人可以节省一些时间，如果他们愿意的话，这里有一个 copy-pastable 功能。
代码说明
每个函数向传入数组添加 n 行或列。代码假定传入数组是 2D numpy 数组。填充的方向根据函数definition/names。有关更详细的解释，我想参考 reader MSeifert 的回答。