嵌套numpy的阵列和使用类似分裂它们的方法

问题描述

我是新来numpy的，我试图用它在我的code的一些表。

I am new to NumPy and am trying to use it in my code for some tables.

我有坐标列表看起来像这样的：

I have a list of coordinates that looks like this:

coordinates = [["2 0"], ["0 1"], ["3 4"]]

和希望它这样写：

coordinatesNumpy = np.array([[2, 0], [0, 1], [3, 4]])

在常规的Python这很容易做到，但你怎么用numpy的呢？我应该只是做定期的Python功能表列表，然后将二维表转换为 np.array 或不numpy的有分裂和东西的方法呢？

In regular Python that's easy to do but how do you do it with NumPy? Should I just make the table with regular Python functions for lists and then convert the 2d table to np.array or does NumPy have methods for splitting and stuff?

我尝试了一些东西，但他们都让我一个错误。我尝试最新的东西：

I tried some things but they all give me an error. The latest thing I tried:

flowers = np.array([np.array([int(coordinate[0]), int(coordinate[2])]) for coordinate in coordinates])

我怎么会做这样的事情与numpy的？

How could I do something like this with NumPy?

推荐答案

本作品：

>>> flowers = np.array([[int(x)  for x in coordinate[0].split()] 
                        for coordinate in coordinates])
>>> flowers
array([[2, 0],
       [0, 1],
       [3, 4]])

我不知道有任何numpy的功能，将做到这一步的。

I am not aware of any NumPy function that would do this in one step.

让我们来看看事情的速度有多快。

Let's check how fast things are.

有关您的数据。例如，纯Python版本是最快的：

For your example data, the pure Python version is the fastest:

%timeit np.array([np.fromstring(i, dtype=int, sep=' ') for j in coordinates for i in j])
100000 loops, best of 3: 18.4 µs per loop

%timeit np.array([np.fromstring(item[0], dtype=int, sep=' ').tolist() for item in coordinates])
10000 loops, best of 3: 19 µs per loop

%timeit np.array([[int(x)  for x in coordinate[0].split()] for coordinate in coordinates])
100000 loops, best of 3: 12.1 µs per loop

请数据更大的：

long_coords = coordinates * 1000

但尽管如此，纯Python版本是最快的：

But still, the pure Python version is the fastest:

%timeit np.array([np.fromstring(i, dtype=int, sep=' ') for j in long_coords for i in j])
100 loops, best of 3: 12.2 ms per loop

%timeit np.array([np.fromstring(item[0], dtype=int, sep=' ').tolist() for item in long_coords])
100 loops, best of 3: 14.2 ms per loop

%timeit np.array([[int(x)  for x in coordinate[0].split()] for coordinate in long_coords])
100 loops, best of 3: 7.54 ms per loop

对于更大的数据一致的结果：

Consistent results for even larger data:

very_long_coords = coordinates * 10000

%timeit np.array([np.fromstring(i, dtype=int, sep=' ') for j in very_long_coords for i in j])
10 loops, best of 3: 125 ms per loop

%timeit np.array([np.fromstring(item[0], dtype=int, sep=' ').tolist() for item in very_long_coords])
10 loops, best of 3: 140 ms per loop

%timeit np.array([[int(x)  for x in coordinate[0].split()] for coordinate in very_long_coords])
10 loops, best of 3: 73.5 ms per loop

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