How to convert a column or row matrix to a diagonal matrix in Python?

Question

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?

UPDATE

This is the code to illustrate the problem:

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:

[[1 0 0 0]
 [0 2 0 0]
 [0 0 3 0]
 [0 0 0 4]]

Solution 1

You can use diag method:

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:

[[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:

a = np.array([[1, 2, 3, 4]])
d = np.diag(a[0])

Results in:

[[1 0 0 0]
 [0 2 0 0]
 [0 0 3 0]
 [0 0 0 4]]

For the given matrix in the question:

import numpy as np
a = np.matrix([1,2,3,4])
d = np.diag(a.A1)
print (d)

Result is again:

[[1 0 0 0]
 [0 2 0 0]
 [0 0 3 0]
 [0 0 0 4]]

Solution 2

I suppose you could also use diagflat:

import numpy
a = np.matrix([1,2,3,4])
d = np.diagflat(a)
print (d)

Which like the diag method results in

[[1 0 0 0]
 [0 2 0 0]
 [0 0 3 0]
 [0 0 0 4]]

but there's no need for flattening with .A1

Solution 3

Another solution could be:

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 timeit on 100000 repetitions:

1. np.array and np.diag (Marcin's answer): 2.18E-02 s
2. np.array and np.identity (this answer): 6.12E-01 s
3. np.matrix and np.diagflat (Bokee's answer): 1.00E-00 s