Python ValueError：所有输入数组必须具有相同的维数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/38848759/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
ValueError: all the input arrays must have same number of dimensions
提问by odo22
I'm having a problem with np.append.
我有问题np.append。
I'm trying to duplicate the last column of 20x361 matrix n_list_convertedby using the code below:
我正在尝试n_list_converted使用以下代码复制 20x361 矩阵的最后一列：
n_last = []
n_last = n_list_converted[:, -1]
n_lists = np.append(n_list_converted, n_last, axis=1)
But I get error:
但我得到错误：
ValueError: all the input arrays must have same number of dimensions
ValueError：所有输入数组必须具有相同的维数
However, I've checked the matrix dimensions by doing
但是，我通过执行检查了矩阵维度
print(n_last.shape, type(n_last), n_list_converted.shape, type(n_list_converted))
and I get
我得到
(20L,) (20L, 361L)
(20L,) (20L, 361L)
so the dimensions match? Where is the mistake?
所以尺寸匹配？错误在哪里？
采纳答案by hpaulj
If I start with a 3x4 array, and concatenate a 3x1 array, with axis 1, I get a 3x5 array:
如果我从一个 3x4 数组开始，并用轴 1 连接一个 3x1 数组，我会得到一个 3x5 数组：
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))
Note that both inputs to concatenate have 2 dimensions.
请注意，要连接的两个输入都有 2 个维度。
Omit the :, and x[:,-1]is (3,) shape - it is 1d, and hence the error:
省略:, 和x[:,-1](3,) 形状 - 它是 1d，因此错误：
In [914]: np.concatenate([x,x[:,-1]], axis=1)
...
ValueError: all the input arrays must have same number of dimensions
The code for np.appendis (in this case where axis is specified)
的代码np.append是（在这种情况下指定轴）
return concatenate((arr, values), axis=axis)
So with a slight change of syntax appendworks. Instead of a list it takes 2 arguments. It imitates the list appendis syntax, but should not be confused with that list method.
所以稍微改变语法就行了append。它需要 2 个参数而不是列表。它模仿 list appendis 语法，但不应与该 list 方法混淆。
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.hstackfirst makes sure all inputs are atleast_1d, and then does concatenate:
np.hstack首先确保所有输入都是atleast_1d，然后进行连接：
return np.concatenate([np.atleast_1d(a) for a in arrs], 1)
So it requires the same x[:,-1:]input. Essentially the same action.
所以它需要相同的x[:,-1:]输入。基本相同的动作。
np.column_stackalso does a concatenate on axis 1. But first it passes 1d inputs through
np.column_stack也在轴 1 上进行连接。但首先它通过 1d 输入
array(arr, copy=False, subok=True, ndmin=2).T
This is a general way of turning that (3,) array into a (3,1) array.
这是将 (3,) 数组转换为 (3,1) 数组的一般方法。
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]])
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.
所有这些“堆栈”都很方便，但从长远来看，了解维度和基数很重要np.concatenate。还知道如何查找此类函数的代码。我经常使用ipython??魔法。
And in time tests, the np.concatenateis noticeably faster - with a small array like this the extra layers of function calls makes a big time difference.
在时间测试中，np.concatenate速度明显更快——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。
回答by Aguy
(n,) and (n,1) are not the same shape. Try casting the vector to an array by using the [:, None]notation:
(n,) 和 (n,1) 不是相同的形状。尝试使用[:, None]符号将向量转换为数组：
n_lists = np.append(n_list_converted, n_last[:, None], axis=1)
Alternatively, when extracting n_lastyou can use
或者，在提取时，n_last您可以使用
n_last = n_list_converted[:, -1:]
to get a (20, 1)array.
得到一个(20, 1)数组。
回答by RuRo
The reason why you get your error is because a "1 by n" matrix is different from an array of length n.
出现错误的原因是“1 x n”矩阵与长度为 n 的数组不同。
I recommend using hstack()and vstack()instead.
Like this:
我建议使用hstack()andvstack()代替。像这样：
import numpy as np
a = np.arange(32).reshape(4,8) # 4 rows 8 columns matrix.
b = a[:,-1:] # last column of that matrix.
result = np.hstack((a,b)) # stack them horizontally like this:
#array([[ 0, 1, 2, 3, 4, 5, 6, 7, 7],
# [ 8, 9, 10, 11, 12, 13, 14, 15, 15],
# [16, 17, 18, 19, 20, 21, 22, 23, 23],
# [24, 25, 26, 27, 28, 29, 30, 31, 31]])
Notice the repeated "7, 15, 23, 31" column.
Also, notice that I used a[:,-1:]instead of a[:,-1]. My version generates a column:
请注意重复的“7、15、23、31”列。另外，请注意我使用的a[:,-1:]不是a[:,-1]. 我的版本生成一列：
array([[7],
[15],
[23],
[31]])
Instead of a row array([7,15,23,31])
而不是一行 array([7,15,23,31])
Edit: append()is muchslower. Read this answer.
编辑：append()是多慢。阅读这个答案。
回答by ZZZ
You can also cast (n,) to (n,1) by enclosing within brackets [ ].
您还可以通过将 (n,) 括在方括号 [ ] 中来将 (n,) 转换为 (n,1)。
e.g. Instead of np.append(b,a,axis=0)use np.append(b,[a],axis=0)
例如，而不是np.append(b,a,axis=0)使用np.append(b,[a],axis=0)
a=[1,2]
b=[[5,6],[7,8]]
np.append(b,[a],axis=0)
returns
返回
array([[5, 6],
[7, 8],
[1, 2]])
