I am having an issue with Ipython - Numpy. I want to do the following operation:
我对Ipython - Numpy有意见。我想做以下操作:
x^T.x
with x belonging to R^n, and x^T the transpose operation on vector x. x is extracted from a txt file with the instruction:
属于R ^ n x,x ^ T向量x x上转置操作提取从txt文件指令:
x = np.loadtxt('myfile.txt')
The problem is that if i use the transpose function
问题是如果我用转置函数
np.transpose(x)
and uses the shape function to know the size of x, I get the same dimensions for x and x^T. Numpy gives the size with a L uppercase indice after each dimensions. e.g.
并使用大小的形状函数知道x,我得到同样的尺寸x,x ^ T。Numpy在每个维度之后都给出一个L大写标记的大小。如。
print x.shape
print np.transpose(x).shape
(3L, 5L)
(3L, 5L)
Does anybody know how to solve this, and compute x^T.x as a matrix product?
有人知道如何解决这个问题,并计算x ^ T。x作为矩阵乘积?
Thank you!
谢谢你！
6 个解决方案
#1
6
As explained by others, transposition won't "work" like you want it to for 1D arrays. You might want to use np.atleast_2d to have a consistent scalar product definition:
正如其他人所解释的，转置不会像你希望的那样在一维数组中“工作”。你可能想用np。至少有一个一致的标量积定义:
def vprod(x):
y = np.atleast_2d(x)
return np.dot(y.T, y)
#2
27
What np.transpose does is reverse the shape tuple, i.e. you feed it an array of shape (m, n), it returns an array of shape (n, m), you feed it an array of shape (n,)... and it returns you the same array with shape(n,).
np。转置做的是反转形状元组，即你给它提供一个形状数组(m, n)，它返回一个形状数组(n, m)，你给它提供一个形状数组(n，)…它返回相同的数组(n)
What you are implicitly expecting is for numpy to take your 1D vector as a 2D array of shape (1, n), that will get transposed into a (n, 1) vector. Numpy will not do that on its own, but you can tell it that's what you want, e.g.:
你所隐含期望的是numpy将你的1D向量作为一个二维的形状数组(1,n)它会被转置成一个(n, 1)向量。Numpy自己不会这么做，但是你可以告诉它那是你想要的，例如:
>>> a = np.arange(4)
>>> a
array([0, 1, 2, 3])
>>> a.T
array([0, 1, 2, 3])
>>> a[np.newaxis, :].T
array([[0],
[1],
[2],
[3]])
#3
1
For starters L just means that the type is a long int. This shouldn't be an issue. You'll have to give additional information about your problem though since I cannot reproduce it with a simple test case:
对于初学者来说，我只是说这个类型是一个很长的int类型，这应该不是问题。你将不得不提供更多关于你的问题的信息，因为我不能用一个简单的测试用例来重现它:
In [1]: import numpy as np
In [2]: a = np.arange(12).reshape((4,3))
In [3]: a
Out[3]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
In [4]: a.T #same as np.transpose(a)
Out[4]:
array([[ 0, 3, 6, 9],
[ 1, 4, 7, 10],
[ 2, 5, 8, 11]])
In [5]: a.shape
Out[5]: (4, 3)
In [6]: np.transpose(a).shape
Out[6]: (3, 4)
There is likely something subtle going on with your particular case which is causing problems. Can you post the contents of the file that you're reading into x?
在你的特殊情况下，可能有一些微妙的事情正在发生，它正在引起问题。你能把读到的文件的内容贴在x上吗?
#4
1
I had the same problem, I used numpy matrix to solve it:
我也有同样的问题，我用numpy矩阵来求解:
# assuming x is a list or a numpy 1d-array
>>> x = [1,2,3,4,5]
# convert it to a numpy matrix
>>> x = np.matrix(x)
>>> x
matrix([[1, 2, 3, 4, 5]])
# take the transpose of x
>>> x.T
matrix([[1],
[2],
[3],
[4],
[5]])
# use * for the matrix product
>>> x*x.T
matrix([[55]])
>>> (x*x.T)[0,0]
55
>>> x.T*x
matrix([[ 1, 2, 3, 4, 5],
[ 2, 4, 6, 8, 10],
[ 3, 6, 9, 12, 15],
[ 4, 8, 12, 16, 20],
[ 5, 10, 15, 20, 25]])
While using numpy matrices may not be the best way to represent your data from a coding perspective, it's pretty good if you are going to do a lot of matrix operations!
虽然使用numpy矩阵可能不是从编码的角度表示数据的最佳方式，但是如果您要进行大量的矩阵操作，那么它是相当不错的!
#5
1
This is either the inner or outer product of the two vectors, depending on the orientation you assign to them. Here is how to calculate either without changing x.
这是两个向量的内积或外积，取决于你给它们的方向。这是如何在不改变x的情况下计算。
import numpy
x = numpy.array([1, 2, 3])
inner = x.dot(x)
outer = numpy.outer(x, x)
#6
0
The file 'myfile.txt' contain lines such as
文件的myfile。txt包含诸如。
5.100000 3.500000 1.400000 0.200000 1
4.900000 3.000000 1.400000 0.200000 1
Here is the code I run:
下面是我运行的代码:
import numpy as np
data = np.loadtxt('iris.txt')
x = data[1,:]
print x.shape
print np.transpose(x).shape
print x*np.transpose(x)
print np.transpose(x)*x
And I get as a result
结果是
(5L,)
(5L,)
[ 24.01 9. 1.96 0.04 1. ]
[ 24.01 9. 1.96 0.04 1. ]
I would be expecting one of the two last result to be a scalar instead of a vector, because x^T.x (or x.x^T) should give a scalar.
我将期待的两个最后的结果是一个标量,而不是一个矢量,因为x ^ T。x(或x.x ^ T)应该给一个标量。
#1
6
As explained by others, transposition won't "work" like you want it to for 1D arrays. You might want to use np.atleast_2d to have a consistent scalar product definition:
正如其他人所解释的，转置不会像你希望的那样在一维数组中“工作”。你可能想用np。至少有一个一致的标量积定义:
def vprod(x):
y = np.atleast_2d(x)
return np.dot(y.T, y)
#2
27
What np.transpose does is reverse the shape tuple, i.e. you feed it an array of shape (m, n), it returns an array of shape (n, m), you feed it an array of shape (n,)... and it returns you the same array with shape(n,).
np。转置做的是反转形状元组，即你给它提供一个形状数组(m, n)，它返回一个形状数组(n, m)，你给它提供一个形状数组(n，)…它返回相同的数组(n)
What you are implicitly expecting is for numpy to take your 1D vector as a 2D array of shape (1, n), that will get transposed into a (n, 1) vector. Numpy will not do that on its own, but you can tell it that's what you want, e.g.:
你所隐含期望的是numpy将你的1D向量作为一个二维的形状数组(1,n)它会被转置成一个(n, 1)向量。Numpy自己不会这么做，但是你可以告诉它那是你想要的，例如:
>>> a = np.arange(4)
>>> a
array([0, 1, 2, 3])
>>> a.T
array([0, 1, 2, 3])
>>> a[np.newaxis, :].T
array([[0],
[1],
[2],
[3]])
#3
1
For starters L just means that the type is a long int. This shouldn't be an issue. You'll have to give additional information about your problem though since I cannot reproduce it with a simple test case:
对于初学者来说，我只是说这个类型是一个很长的int类型，这应该不是问题。你将不得不提供更多关于你的问题的信息，因为我不能用一个简单的测试用例来重现它:
In [1]: import numpy as np
In [2]: a = np.arange(12).reshape((4,3))
In [3]: a
Out[3]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
In [4]: a.T #same as np.transpose(a)
Out[4]:
array([[ 0, 3, 6, 9],
[ 1, 4, 7, 10],
[ 2, 5, 8, 11]])
In [5]: a.shape
Out[5]: (4, 3)
In [6]: np.transpose(a).shape
Out[6]: (3, 4)
There is likely something subtle going on with your particular case which is causing problems. Can you post the contents of the file that you're reading into x?
在你的特殊情况下，可能有一些微妙的事情正在发生，它正在引起问题。你能把读到的文件的内容贴在x上吗?
#4
1
I had the same problem, I used numpy matrix to solve it:
我也有同样的问题，我用numpy矩阵来求解:
# assuming x is a list or a numpy 1d-array
>>> x = [1,2,3,4,5]
# convert it to a numpy matrix
>>> x = np.matrix(x)
>>> x
matrix([[1, 2, 3, 4, 5]])
# take the transpose of x
>>> x.T
matrix([[1],
[2],
[3],
[4],
[5]])
# use * for the matrix product
>>> x*x.T
matrix([[55]])
>>> (x*x.T)[0,0]
55
>>> x.T*x
matrix([[ 1, 2, 3, 4, 5],
[ 2, 4, 6, 8, 10],
[ 3, 6, 9, 12, 15],
[ 4, 8, 12, 16, 20],
[ 5, 10, 15, 20, 25]])
While using numpy matrices may not be the best way to represent your data from a coding perspective, it's pretty good if you are going to do a lot of matrix operations!
虽然使用numpy矩阵可能不是从编码的角度表示数据的最佳方式，但是如果您要进行大量的矩阵操作，那么它是相当不错的!
#5
1
This is either the inner or outer product of the two vectors, depending on the orientation you assign to them. Here is how to calculate either without changing x.
这是两个向量的内积或外积，取决于你给它们的方向。这是如何在不改变x的情况下计算。
import numpy
x = numpy.array([1, 2, 3])
inner = x.dot(x)
outer = numpy.outer(x, x)
#6
0
The file 'myfile.txt' contain lines such as
文件的myfile。txt包含诸如。
5.100000 3.500000 1.400000 0.200000 1
4.900000 3.000000 1.400000 0.200000 1
Here is the code I run:
下面是我运行的代码:
import numpy as np
data = np.loadtxt('iris.txt')
x = data[1,:]
print x.shape
print np.transpose(x).shape
print x*np.transpose(x)
print np.transpose(x)*x
And I get as a result
结果是
(5L,)
(5L,)
[ 24.01 9. 1.96 0.04 1. ]
[ 24.01 9. 1.96 0.04 1. ]
I would be expecting one of the two last result to be a scalar instead of a vector, because x^T.x (or x.x^T) should give a scalar.
我将期待的两个最后的结果是一个标量,而不是一个矢量,因为x ^ T。x(或x.x ^ T)应该给一个标量。