如何在 Python 中将列或行矩阵转换为对角矩阵？
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How to convert a column or row matrix to a diagonal matrix in Python?
提问by Tom Kurushingal
I have a row vector A, A = [a1 a2 a3 ..... an] and I would like to create a diagonal matrix, B = diag(a1, a2, a3, ....., an) with the elements of this row vector. How can this be done in Python?
我有一个行向量 A, A = [a1 a2 a3 ..... an] 并且我想创建一个对角矩阵 B = diag(a1, a2, a3, ....., an) 与此行向量的元素。这如何在 Python 中完成？
UPDATE
更新
This is the code to illustrate the problem:
这是说明问题的代码：
import numpy as np
a = np.matrix([1,2,3,4])
d = np.diag(a)
print (d)
the output of this code is [1], but my desired output is:
这段代码的输出是 [1]，但我想要的输出是：
[[1 0 0 0]
[0 2 0 0]
[0 0 3 0]
[0 0 0 4]]
采纳答案by Marcin
You can use diagmethod:
您可以使用diag方法：
import numpy as np
a = np.array([1,2,3,4])
d = np.diag(a)
# or simpler: d = np.diag([1,2,3,4])
print(d)
Results in:
结果是：
[[1 0 0 0]
[0 2 0 0]
[0 0 3 0]
[0 0 0 4]]
If you have a row vector, you can do this:
如果你有一个行向量，你可以这样做：
a = np.array([[1, 2, 3, 4]])
d = np.diag(a[0])
Results in:
结果是：
[[1 0 0 0]
[0 2 0 0]
[0 0 3 0]
[0 0 0 4]]
For the given matrix in the question:
对于问题中的给定矩阵：
import numpy as np
a = np.matrix([1,2,3,4])
d = np.diag(a.A1)
print (d)
Result is again:
结果又是：
[[1 0 0 0]
[0 2 0 0]
[0 0 3 0]
[0 0 0 4]]
回答by deadcode
Assuming you are working in numpy based on your tags, this will do it:
假设您根据标签在 numpy 中工作，则可以这样做：
import numpy
def make_diag( A ):
my_diag = numpy.zeroes( ( 2, 2 ) )
for i, a in enumerate( A ):
my_diag[i,i] = a
return my_diag
enumerate( LIST ) creates an iterator over the list that returns tuples like:
enumerate( LIST ) 在列表上创建一个迭代器，返回元组，如：
( 0, 1st element), ( 1, 2nd element), ... ( N-1, Nth element )
( 0, 1st element), ( 1, 2nd element), ... ( N-1, Nth element )
回答by Bokee
回答by jeannej
Another solution could be:
另一种解决方案可能是：
import numpy as np
a = np.array([1,2,3,4])
d = a * np.identity(len(a))
As for performances for the various answers here, I get with timeiton 100000 repetitions:
至于这里各种答案的表现，我得到了timeit100000 次重复：
np.arrayandnp.diag(Marcin's answer): 2.18E-02 snp.arrayandnp.identity(this answer): 6.12E-01 snp.matrixandnp.diagflat(Bokee's answer): 1.00E-00 s
np.array和np.diag（Marcin 的回答）：2.18E-02 snp.array和np.identity（这个答案）：6.12E-01 snp.matrix和np.diagflat（Bokee 的回答）：1.00E-00 秒
