ValueError:所有输入数组必须具有相同的维数 [英] ValueError: all the input arrays must have same number of dimensions

问题描述

我在使用 np.append 时遇到问题.

I'm having a problem with np.append.

我正在尝试使用以下代码复制 20x361 矩阵 n_list_converted 的最后一列:

I'm trying to duplicate the last column of 20x361 matrix n_list_converted by using the code below:

n_last = []
n_last = n_list_converted[:, -1]
n_lists = np.append(n_list_converted, n_last, axis=1)

但我得到错误:

ValueError: 所有输入数组的维数必须相同

ValueError: all the input arrays must have same number of dimensions

但是，我已经通过执行检查了矩阵维度

However, I've checked the matrix dimensions by doing

 print(n_last.shape, type(n_last), n_list_converted.shape, type(n_list_converted))

我得到

(20L,) (20L, 361L)

(20L,) (20L, 361L)

所以尺寸匹配?错在哪里?

so the dimensions match? Where is the mistake?

推荐答案

如果我从一个 3x4 数组开始，然后将一个 3x1 数组与轴 1 连接起来，我会得到一个 3x5 数组:

If I start with a 3x4 array, and concatenate a 3x1 array, with axis 1, I get a 3x5 array:

In [911]: x = np.arange(12).reshape(3,4)
In [912]: np.concatenate([x,x[:,-1:]], axis=1)
Out[912]: 
array([[ 0,  1,  2,  3,  3],
       [ 4,  5,  6,  7,  7],
       [ 8,  9, 10, 11, 11]])
In [913]: x.shape,x[:,-1:].shape
Out[913]: ((3, 4), (3, 1))

请注意，要连接的两个输入都有 2 个维度.

Note that both inputs to concatenate have 2 dimensions.

省略 :，x[:,-1] 是 (3,) 形状 - 它是 1d，因此错误:

Omit the :, and x[:,-1] is (3,) shape - it is 1d, and hence the error:

In [914]: np.concatenate([x,x[:,-1]], axis=1)
...
ValueError: all the input arrays must have same number of dimensions

np.append 的代码是(在这种情况下指定轴)

The code for np.append is (in this case where axis is specified)

return concatenate((arr, values), axis=axis)

所以稍微改变语法 append 就可以了.它需要 2 个参数而不是列表.它模仿列表append 的语法，但不应与列表方法混淆.

So with a slight change of syntax append works. Instead of a list it takes 2 arguments. It imitates the list append is syntax, but should not be confused with that list method.

In [916]: np.append(x, x[:,-1:], axis=1)
Out[916]: 
array([[ 0,  1,  2,  3,  3],
       [ 4,  5,  6,  7,  7],
       [ 8,  9, 10, 11, 11]])

np.hstack 首先确保所有输入都是 atleast_1d，然后进行连接:

np.hstack first makes sure all inputs are atleast_1d, and then does concatenate:

return np.concatenate([np.atleast_1d(a) for a in arrs], 1)

所以它需要相同的 x[:,-1:] 输入.基本相同的操作.

So it requires the same x[:,-1:] input. Essentially the same action.

np.column_stack 也在轴 1 上进行连接.但首先它通过 1d 输入通过

np.column_stack also does a concatenate on axis 1. But first it passes 1d inputs through

array(arr, copy=False, subok=True, ndmin=2).T

这是将 (3,) 数组转换为 (3,1) 数组的一般方法.

This is a general way of turning that (3,) array into a (3,1) array.

In [922]: np.array(x[:,-1], copy=False, subok=True, ndmin=2).T
Out[922]: 
array([[ 3],
       [ 7],
       [11]])
In [923]: np.column_stack([x,x[:,-1]])
Out[923]: 
array([[ 0,  1,  2,  3,  3],
       [ 4,  5,  6,  7,  7],
       [ 8,  9, 10, 11, 11]])

所有这些堆栈"都可能很方便，但从长远来看，了解维度和基础 np.concatenate 很重要.还知道如何查找此类函数的代码.我经常使用 ipython ?? 魔法.

All these 'stacks' can be convenient, but in the long run, it's important to understand dimensions and the base np.concatenate. Also know how to look up the code for functions like this. I use the ipython ?? magic a lot.

在时间测试中，np.concatenate 明显更快 - 对于像这样的小数组，额外的函数调用层会产生很大的时间差异.

And in time tests, the np.concatenate is noticeably faster - with a small array like this the extra layers of function calls makes a big time difference.

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