python如何用零填充numpy数组 [英] python how to pad numpy array with zeros

问题描述

我想知道如何使用python 2.6.6和numpy版本1.5.0用零填充2D numpy数组.对不起！但是这些是我的局限性.因此，我不能使用np.pad.例如，我想用零填充a，以使其形状匹配b.我想这样做的原因是我可以这样做:

I want to know how I can pad a 2D numpy array with zeros using python 2.6.6 with numpy version 1.5.0. Sorry! But these are my limitations. Therefore I cannot use np.pad. For example, I want to pad a with zeros such that its shape matches b. The reason why I want to do this is so I can do:

b-a

如此

>>> 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，但这看起来很丑.是否有可能使用b.shape的清洁解决方案?

The only way I can think of doing this is appending, however this seems pretty ugly. is there a cleaner solution possibly using b.shape?

编辑， 谢谢MSeiferts的回答.我不得不清理一下，这就是我得到的:

Edit, Thank you to MSeiferts answer. I had to clean it up a bit, and this is what I got:

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

推荐答案

非常简单，您可以使用参考形状创建一个包含零的数组:

Very simple, you create an array containing zeros using the reference shape:

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

，然后将数组插入所需的位置:

and then insert the array where you need it:

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角插入元素的位置，也可以使其更通用

---

You can also make it a bit more general if you define where your upper left element should be inserted

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.]])

，但是请注意偏移量不要超过允许的范围.例如，对于x_offset = 2，此操作将失败.

but then be careful that you don't have offsets bigger than allowed. For x_offset = 2 for example this will fail.

如果有任意数量的维，则可以定义切片列表以插入原始数组.我发现有趣的是，可以玩一下并创建一个填充函数，该函数可以填充(偏移)任意形状的数组，只要该数组和参考的维数相同且偏移量不太大即可.

If you have an arbitary number of dimensions you can define a list of slices to insert the original array. I've found it interesting to play around a bit and created a padding function that can pad (with offset) an arbitary shaped array as long as the array and reference have the same number of dimensions and the offsets are not too big.

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)

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