预分配NumPy数组的首选方法是什么? [英] What is the preferred way to preallocate NumPy arrays?

问题描述

我是NumPy/SciPy的新手.从文档中看，预分配似乎更有效 单个数组，而不是调用append/insert/concatenate.

I am new to NumPy/SciPy. From the documentation, it seems more efficient to preallocate a single array rather than call append/insert/concatenate.

例如，要将1的列添加到数组中，我认为是这样的:

For example, to add a column of 1's to an array, i think that this:

ar0 = np.linspace(10, 20, 16).reshape(4, 4)
ar0[:,-1] = np.ones_like(ar0[:,0])

首选:

ar0 = np.linspace(10, 20, 12).reshape(4, 3)
ar0 = np.insert(ar0, ar0.shape[1], np.ones_like(ar0[:,0]), axis=1)

我的第一个问题是这是否正确(第一个更好)，而我的第二个问题是，目前，我只是像这样预先分配我的数组(我在SciPy上的一些Cookbook示例中注意到了这一点)网站):

my first question is whether this is correct (that the first is better), and my second question is, at the moment, I am just preallocating my arrays like this (which I noticed in several of the Cookbook examples on the SciPy Site):

np.zeros((8,5))

首选NumPy"的方式是什么?

what is the 'NumPy-preferred' way to do this?

推荐答案

预分配m在一次调用中分配您需要的所有内存，而调整数组大小(通过调用append，insert，concatenate或resize的大小)可能需要将数组复制到更大的内存块.因此，您是对的，预分配比调整大小更可取(并且应该比调整大小更快).

Preallocation mallocs all the memory you need in one call, while resizing the array (through calls to append,insert,concatenate or resize) may require copying the array to a larger block of memory. So you are correct, preallocation is preferred over (and should be faster than) resizing.

根据您要创建的内容，有许多首选"方法可以预分配numpy数组.有np.zeros，np.ones，np.empty，np.zeros_like，np.ones_like和np.empty_like，还有许多其他创建有用的数组的文件，例如np.linspace和np.arange.

There are a number of "preferred" ways to preallocate numpy arrays depending on what you want to create. There is np.zeros, np.ones, np.empty, np.zeros_like, np.ones_like, and np.empty_like, and many others that create useful arrays such as np.linspace, and np.arange.

所以

ar0 = np.linspace(10, 20, 16).reshape(4, 4)

如果

最接近您想要的ar0，就很好.

但是，要使最后一栏全为1，我认为首选的方法是说

However, to make the last column all 1's, I think the preferred way would be to just say

ar0[:,-1]=1

由于ar0[:,-1]的形状为(4,)，因此1为广播以匹配此形状

Since the shape of ar0[:,-1] is (4,), the 1 is broadcasted to match this shape.

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