How to create matrix in python of repeating number?

Question

I want to:
Create a vector list from 0 to 4, i.e. [0, 1, 2, 3, 4] and from that
Create a matrix containing a "tiered list" from 0 to 4, 3 times over, once for each dimension. The matrix has 4^3 = 64 rows, so for example

This is what I have so far:

n=5;
ind=list(range(0,n))
print(ind)

I am just getting started with Python so any help would be greatly appreciated!

Kingsley · Accepted Answer · 2018-12-11 05:15:27Z

4

The python itertools module product() function can do this:

for code in itertools.product( range(5), repeat=3 ):
    print(code)

Giving the result:

(0, 0, 0)
(0, 0, 1)
(0, 0, 2)
(0, 0, 3)
...
(4, 4, 2)
(4, 4, 3)
(4, 4, 4)

So to make this into a matrix:

import itertools
matrix = []
for code in itertools.product( range(5), repeat=3 ):
     matrix.append( list(code) )

edited Dec 11, 2018 at 5:15

answered Dec 11, 2018 at 5:10

Kingsley

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2 Comments

Kingsley Over a year ago

@Tomothy32 - I had to go lookup the syntax, you beat me to it. I don't like the "pythonic" way, it's unreadable, especially for a beginner.

iz_ Over a year ago

Also, matrix = [list(code) for code in itertools.product(range(5), repeat=3)] is easier and more Pythonic.

phil · Accepted Answer · 2018-12-11 05:10:20Z

0

 list_ = []
 for a in range(5):
     for b in range(5):
        for c in range(5):
              list_ += [a ,b ,c ]

 print(list_)

answered Dec 11, 2018 at 5:10

phil

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2 Comments

Mike Lane Over a year ago

This is incorrect. First, it makes a list that is one dimensional instead of having many rows of 3. Second, what would be the rows are backwards compared to what is asked for.

phil Over a year ago

The OP mentions matrix in their question but offers up a python list T = [0 0 0 etc with no commas just newlines and spaces

Mike Lane · Accepted Answer · 2018-12-11 05:28:45Z

0

Note, you really want the matrix to have 5^3 = 125 rows. The basic answer is to just iterate in nested for loops:

T = []
for a in range(5):
    for b in range(5):
        for c in range(5):
            T.append([a, b, c])

There are other, probably faster, ways of doing this, but for sheer get 'er done velocity, it's hard to beat this.

edited Dec 11, 2018 at 5:28

answered Dec 11, 2018 at 5:15

Mike Lane

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6 Comments

iz_ Over a year ago

Wouldn't it be T.append([a, b, c])?

Mike Lane Over a year ago

That would make the order backwards compared to what OP is asking for.

iz_ Over a year ago

Are you sure? Running it on my machine:

[[0, 0, 0], [1, 0, 0], [2, 0, 0], [3, 0, 0], [4, 0, 0], [0, 1, 0], [1, 1, 0], [2, 1, 0], [3, 1, 0], [4, 1, 0], [0, 2, 0], [1, 2, 0], [2, 2, 0], [3, 2, 0], [4, 2, 0], [0, 3, 0], [1, 3, 0], [2, 3, 0], [3, 3, 0], [4, 3, 0], [0, 4, 0], [1, 4, 0], [2, 4, 0], [3, 4, 0], [4, 4, 0], [0, 0, 1], [1, 0, 1], [2, 0, 1], [3, 0, 1], [4, 0, 1], [0, 1, 1], [1, 1, 1], [2, 1, 1], [3, 1, 1], [4, 1, 1], [0, 2, 1], [1, 2, 1], [2, 2, 1], [3, 2, 1], [4, 2, 1], [0, 3, 1], [1, 3, 1], [2, 3, 1], [3, 3, 1], [4, 3, 1], [0, 4, 1], [1, 4, 1], [2, 4, 1], [3, 4, 1] ...

Mike Lane Over a year ago

Compare your result carefully with what OP asked for. It was [[0,0,0], [0,0,1], [0,0,2], ... ]

iz_ Over a year ago

Exactly, running your solution outputs the wrong result.

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