从 ND 到 1D 阵列 [英] From ND to 1D arrays

问题描述

假设我有一个数组 a:

a = np.array([[1,2,3], [4,5,6]])

array([[1, 2, 3],
       [4, 5, 6]])

我想将其转换为一维数组(即列向量):

I would like to convert it to a 1D array (i.e. a column vector):

b = np.reshape(a, (1,np.product(a.shape)))

但这会返回

array([[1, 2, 3, 4, 5, 6]])

与以下不同:

array([1, 2, 3, 4, 5, 6])

我可以把这个数组的第一个元素手动转换成一维数组:

I can take the first element of this array to manually convert it to a 1D array:

b = np.reshape(a, (1,np.product(a.shape)))[0]

但这需要我知道原始数组有多少维(并在处理更高维时连接 [0])

but this requires me to know how many dimensions the original array has (and concatenate [0]'s when working with higher dimensions)

是否有一种从任意 ndarray 中获取列/行向量的与维度无关的方法?

Is there a dimensions-independent way of getting a column/row vector from an arbitrary ndarray?

推荐答案

使用 np.ravel(用于一维视图)或 np.ndarray.flatten(用于一维副本)或 np.ndarray.flat(用于一维迭代器):

Use np.ravel (for a 1D view) or np.ndarray.flatten (for a 1D copy) or np.ndarray.flat (for an 1D iterator):

In [12]: a = np.array([[1,2,3], [4,5,6]])

In [13]: b = a.ravel()

In [14]: b
Out[14]: array([1, 2, 3, 4, 5, 6])

请注意，ravel() 在可能的情况下返回 a 的 view.所以修改b 也会修改a.当一维元素在内存中是连续的时，ravel() 返回一个 view，但如果例如 a 是通过使用非单位步长(例如 a = x[::2])对另一个数组进行切片而制成的.

Note that ravel() returns a view of a when possible. So modifying b also modifies a. ravel() returns a view when the 1D elements are contiguous in memory, but would return a copy if, for example, a were made from slicing another array using a non-unit step size (e.g. a = x[::2]).

如果您想要副本而不是视图，请使用

If you want a copy rather than a view, use

In [15]: c = a.flatten()

如果你只想要一个迭代器，使用 np.ndarray.flat:

If you just want an iterator, use np.ndarray.flat:

In [20]: d = a.flat

In [21]: d
Out[21]: <numpy.flatiter object at 0x8ec2068>

In [22]: list(d)
Out[22]: [1, 2, 3, 4, 5, 6]

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