如何将多个额外的列添加到NumPy数组

问题描述：

假设我有两个NumPy数组，分别是 a 和 b :

Let’s say I have two NumPy arrays, a and b:

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

b = np.array([8,9])

我想将相同的数组 b 附加到每一行(即添加多列)以获得一个数组， c :

And I would like to append the same array b to every row (ie. adding multiple columns) to get an array, c:

b = np.array([
    [1, 2, 3, 8, 9],
    [2, 3, 4, 8, 9]
    ])

如何在NumPy中轻松高效地做到这一点?

How can I do this easily and efficiently in NumPy?

我特别担心它在大型数据集(其中 a 比 b 大)的行为，是否有办法创建多个副本(即 b 的> a.shape [0] )?

I am especially concerned about its behaviour with big datasets (where a is much bigger than b), is there any way around creating many copies (ie. a.shape[0]) of b?

与此问题有关，但具有多个值.

Related to this question, but with multiple values.

答

concatenate 方法的一种替代方法是制作一个收件人数组，并将值复制到其中:

An alternative to concatenate approach is to make a recipient array, and copy values to it:

In [483]: a = np.arange(300).reshape(100,3)
In [484]: b=np.array([8,9])
In [485]: res = np.zeros((100,5),int)
In [486]: res[:,:3]=a
In [487]: res[:,3:]=b

采样时间

sample timings

In [488]: %%timeit
     ...: res = np.zeros((100,5),int)
     ...: res[:,:3]=a
     ...: res[:,3:]=b
     ...: 
     ...: 
6.11 µs ± 20.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [491]: timeit np.concatenate((a, b.repeat(100).reshape(2,-1).T),1)
7.74 µs ± 15.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [164]: timeit np.concatenate([a, np.ones([a.shape[0],1], dtype=int).dot(np.array([b]))], axis=1) 
8.58 µs ± 160 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

如何将多个额外的列添加到NumPy数组

相关推荐