scipy lecture notes, create array example

Question

IN the scipy lecture notes for indexing

There is an example problem, to create this array.

[[0., 0., 0., 0., 0.],
 [2., 0., 0., 0., 0.],
 [0., 3., 0., 0., 0.],
 [0., 0., 4., 0., 0.],
 [0., 0., 0., 5., 0.],
 [0., 0., 0., 0., 6.]]

The problem for me is getting the extra blank row at the top. How do I acheive the example?

This is my current code.

d =np.zeros([5,],dtype=int) + np.diag(arange(2,7,1))

d
Out[66]: 
array([[2, 0, 0, 0, 0],
       [0, 3, 0, 0, 0],
       [0, 0, 4, 0, 0],
       [0, 0, 0, 5, 0],
       [0, 0, 0, 0, 6]])

Mark · Accepted Answer · 2019-05-12 06:17:04Z

4

You can do this with diag and indexing alone using the k argument to diag:

np.diag(np.arange(2,7), k = -1)

gives:

array([[0, 0, 0, 0, 0, 0],
       [2, 0, 0, 0, 0, 0],
       [0, 3, 0, 0, 0, 0],
       [0, 0, 4, 0, 0, 0],
       [0, 0, 0, 5, 0, 0],
       [0, 0, 0, 0, 6, 0]])

That's almost right. You just need to lose the last column which you can do with an slice:

np.diag(np.arange(2,7), k = -1)[:, :-1]

which gives the desired result:

array([[0, 0, 0, 0, 0],
       [2, 0, 0, 0, 0],
       [0, 3, 0, 0, 0],
       [0, 0, 4, 0, 0],
       [0, 0, 0, 5, 0],
       [0, 0, 0, 0, 6]])

answered May 12, 2019 at 6:17

Mark

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1 Comment

hpaulj Over a year ago

Or make it too large on the other end: np.diag(np.arange(1,7))[:,1:]

burgerhex · Accepted Answer · 2019-05-12 05:54:23Z

2

Use np.append:

>>> zero_row = np.zeros((1,5))
>>> matrix   = np.diag(np.arange(2,7,1))
>>> np.append(zero_row, matrix, axis=0)
<<< array([[0., 0., 0., 0., 0.],
           [2., 0., 0., 0., 0.],
           [0., 3., 0., 0., 0.],
           [0., 0., 4., 0., 0.],
           [0., 0., 0., 5., 0.],
           [0., 0., 0., 0., 6.]])

answered May 12, 2019 at 5:54

burgerhex

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Comments

hpaulj · Accepted Answer · 2019-05-12 06:36:50Z

2

Here's a way of generating the array without diag. Instead I index a diagonal set of elements in a zeros array:

In [167]: x = np.zeros((6,5))                                                     
In [168]: x[np.arange(1,6), np.arange(5)] = np.arange(2,7)                        
In [169]: x                                                                     
Out[169]: 
array([[0., 0., 0., 0., 0.],
       [2., 0., 0., 0., 0.],
       [0., 3., 0., 0., 0.],
       [0., 0., 4., 0., 0.],
       [0., 0., 0., 5., 0.],
       [0., 0., 0., 0., 6.]])

answered May 12, 2019 at 6:36

hpaulj

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Comments

Wang Liang · Accepted Answer · 2019-05-12 05:56:39Z

1

import numpy as np

m = np.zeros([5,])
n = np.diag(np.arange(2,7,1))

m = np.vstack((m,n))
print(m)

We can use vstack

answered May 12, 2019 at 5:56

Wang Liang

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Comments

Paul Panzer · Accepted Answer · 2019-05-12 10:09:22Z

You can also use reshape like so:

out = np.zeros((6, 5))                                                              
out.reshape(5, 6)[:, 5] = np.arange(2, 7)                                            
out                                                                                                         
# array([[0., 0., 0., 0., 0.],
#        [2., 0., 0., 0., 0.],
#        [0., 3., 0., 0., 0.],
#        [0., 0., 4., 0., 0.],
#        [0., 0., 0., 5., 0.],
#        [0., 0., 0., 0., 6.]])

or very similarly:

out = np.zeros((6, 5))
out.reshape(-1)[5::6] = np.arange(2, 7)
out
# array([[0., 0., 0., 0., 0.],
#        [2., 0., 0., 0., 0.],
#        [0., 3., 0., 0., 0.],
#        [0., 0., 4., 0., 0.],
#        [0., 0., 0., 5., 0.],
#        [0., 0., 0., 0., 6.]])

Both these methods are faster than everything posted so far:

import numpy as np
from timeit import timeit

def od_hpj():
    out = np.zeros((6, 5))
    out[np.arange(1,6), np.arange(5)] = np.arange(2,7)                        
    return out

def od_mm():
    return np.diag(np.arange(2,7), k = -1)[:, :-1]

def od_ks():
    m = np.zeros([5,])
    n = np.diag(np.arange(2,7,1))
    return np.vstack((m,n))

def od_as():
    zero_row = np.zeros((1,5))
    matrix   = np.diag(np.arange(2,7,1))
    return np.append(zero_row, matrix, axis=0)

def od_pp1():
    out = np.zeros((6, 5))
    out.reshape(5, 6)[:, 5] = np.arange(2, 7)
    return out

def od_pp2():
    out = np.zeros((6, 5))
    out.reshape(-1)[5::6] = np.arange(2, 7)
    return out

for n, o in list(globals().items()):
    if n.startswith("od_"):
        print(f"{n.replace('od_', ''):3s}: {timeit(o):.3f} us")

Sample run:

hpj: 3.379 us
mm : 2.952 us
ks : 7.804 us
as : 5.222 us
pp1: 1.735 us
pp2: 2.418 us

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