问题描述
我带来了作为一个形状（28,28,60000）矩阵60000 train_images。它是一个numpy.ndarray。我想将其转换为1二维图像阵列，这意味着每个图像重新psented为数字单行/阵列$ P $，我想60000阵列。换句话说，我想从（28，28，60000）去（60000，28 * 28）。在Python中，这将是：
I have 60000 train_images brought in as a shape (28,28,60000) matrix. It is a numpy.ndarray. I want to convert it to an array of 1 dimensional images, meaning each image is represented as a single line/array of numbers, and I want 60000 arrays. In other words, I want to go from (28, 28, 60000) to (60000, 28*28). In python, it would be:
images_features = []
for image in images:
imageLine = []
for y in range(len(image)):
for x in range(len(image[0])):
imageLine.append(image[y][x])
images_features.append(imageLine)
我怎样才能做到这一点？我怀疑我需要使用重塑，但我无法弄清楚我究竟是如何能做到这一点。
How can I do this? I suspect that I need to use reshape but I couldn't figure out how exactly I can do this.
这是我应得的图像：
data = scipy.io.loadmat('train.mat')
images = data["train_images"]
所以，图像是我说的数组。
So the "images" is the array I'm talking about.
有人向我建议：
你可能需要改变的轴或将它们组合起来做得到你想要的功能。我建议在情况下的图像横盘结束绘制它们。确保你的勤奋与轴，以避免出现更多的问题。
"You may need to change axes or combine them do get the functionality you want. I recommend plotting them as well in case an image ends up sideways. Make sure you are diligent with your axes to avoid further problems there."
我不知道什么叫轴正在这里所指的，以及如何采取什么上面说的考虑。
I have no idea what "axes" is being referred to here and how to take what's said above into account.
有人能解释什么，我需要做的，为什么呢？ （做什么）
推荐答案
由于这是通过 loadmat 即将到来，的形状（28,28 ，60000）是有道理的 - 从最后指数MATLAB进行迭代，
Since this is coming via loadmat, a shape of (28,28,60000) makes sense - MATLAB iterates starting with the last index.
images.transpose() # or images.T
重新排序的轴，所以结果是（60000,28,28）。最后两个尺寸可以与重塑组合
reorders the axes, so the result is (60000,28,28). The last two dimensions can combined with a reshape
images.T.reshape(60000,28*28)
images.T.reshape(60000,-1) # short hand
您很多需要调换28x28的图像，例如
You many need to transpose the 28x28 images, e.g.
images.transpose([2,0,1]) # instead of the default [2,1,0]
.T 是一样的MATLAB （或。）。
.T is the same as the MATLAB ' (or .').
图片也可能是为了='F'。
octave:38> images=reshape(1:30,2,3,5);
octave:39> save test.mat -v7 images
octave:40> images
images =
ans(:,:,1) =
1 3 5
2 4 6
ans(:,:,2) =
7 9 11
8 10 12
....
我选择测试的尺寸要小，并可以很容易地分辨出不同的轴。
I chose test dimensions to be small, and to make it easy to distinguish the different axes.
在一个会话IPython的：
In a Ipython session:
In [15]: data=io.loadmat('test.mat')
In [16]: data
Out[16]:
{'__globals__': [],
'__header__': 'MATLAB 5.0 MAT-file, written by Octave 3.8.2, 2016-02-10 05:19:18 UTC',
'__version__': '1.0',
'images': array([[[ 1., 7., 13., 19., 25.],
[ 3., 9., 15., 21., 27.],
[ 5., 11., 17., 23., 29.]],
[[ 2., 8., 14., 20., 26.],
[ 4., 10., 16., 22., 28.],
[ 6., 12., 18., 24., 30.]]])}
In [18]: data['images'].T
Out[18]:
array([[[ 1., 2.],
[ 3., 4.],
[ 5., 6.]],
[[ 7., 8.],
[ 9., 10.],
[ 11., 12.]],
....
In [19]: data['images'].transpose([2,0,1])
Out[19]:
array([[[ 1., 3., 5.],
[ 2., 4., 6.]],
[[ 7., 9., 11.],
[ 8., 10., 12.]],
....
In [22]: data['images'].transpose([2,1,0]).reshape(5,-1)
Out[22]:
array([[ 1., 2., 3., 4., 5., 6.],
[ 7., 8., 9., 10., 11., 12.],
...
这篇关于重塑图像阵列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！