How to create a Matrix through an iteration in Python

Question

I want to create a (3n*7) Matrix in python through Iteration, I did this in VB.Net and it worked but it gives me some challenge in Python as this a new language to me. n can be the number of Iterations to build up the matrix, i.e. ; if u iterated 3 times, the matrix will be a 9*7 matrix. Here is how I did it in VB.Net. :

'Assign some values to the A matrix
A_mat(p, 0) = 1 : A_mat(1 + p, 1) = 1 : A_mat(2 + p, 2) = 1 'row/column 1,2 & 3 data
A_mat(p, 3) = c : A_mat(p, 4) = 0 : A_mat(p, 5) = a : A_mat(p, 6) = b 'row 1 data
A_mat(1 + p, 3) = g : A_mat(1 + p, 4) = d : A_mat(1 + p, 5) = t : A_mat(1 + p, 6) = f 'row 2 data
A_mat(2 + p, 3) = m : A_mat(2 + p, 4) = h : A_mat(2 + p, 5) = u : A_mat(2 + p, 6) = k 'row 3 data

this yielded: 
1     0    0   c   0   a   b 
0     1    0   g   d   t   f
0     0    1   m   h   u   k
.     .
.          .
.              .

Why not just make the matrix the right size in the first place? `mat = numpy.zeros((3,7))`. — Katriel, Jul 18 '11 at 10:54
What is the purpose of the matrix? Because if you want to use numpy arrays you cannot mix numbers and strings like this. — joris, Jul 18 '11 at 10:55
At Joris, the strings are replaced by numbers calculated before, but within the iteration, i.e. the change as u iterate. — Blessed, Jul 18 '11 at 12:43
@katrielalex, the dimensions changes...nt just 3*7, but depend on the number of iterations. — Blessed, Jul 18 '11 at 12:45

score 3 · Answer 1 · edited May 23 '17 at 11:55

One solution is to use a numpy (download here) matrix:

>>> from numpy import zeros, matrix
>>> n = 2
>>> mat = matrix(zeros([3*n, 7]))
>>> mat
matrix([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  0.,  0.,  0.]])

However, as noted above by @joris, with a numpy matrix all of your data must be of the same type.

>>> mat[0,2] = 14
>>> mat
matrix([[  0.,   0.,  14.,   0.,   0.,   0.,   0.],
        [  0.,   0.,   0.,   0.,   0.,   0.,   0.],
        [  0.,   0.,   0.,   0.,   0.,   0.,   0.],
        [  0.,   0.,   0.,   0.,   0.,   0.,   0.],
        [  0.,   0.,   0.,   0.,   0.,   0.,   0.],
        [  0.,   0.,   0.,   0.,   0.,   0.,   0.]])
>>> mat[0,1] = 's'

Traceback (most recent call last):
  File "<pyshell#139>", line 1, in <module>
    mat[0,1] = 's'
ValueError: could not convert string to float: s

You probably want to use a nested list instead, since this is by far a more flexible and generic data type, and in addition is easier to get the hang of. For that I would refer you to this question, since @Mike Graham does a right proper job of explaining them. You might want to define a method like:

def shape_zeros(n):
    # Here's where you build up your "matrix"-style nested list
    pass # The pass statement does nothing!

Here are a couple of hints for filling up the code body of the above function. First, you can construct a whole list of duplicates with the * asterisk operator:

>>> row = [0] * 7
>>> row
[0, 0, 0, 0, 0, 0, 0]

Next, you can nest lists within lists; but be careful when moving lists around, as the list's name is actually like a pointer:

>>> mat = []
>>> n = 2
>>> for i in range(n*3):
    mat.append(row)

>>> mat
[[0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0]]
>>> mat[0][0] = 1
>>> mat
[[1, 0, 0, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0]]

You can avoid the above problem by creating your sub-lists (rows) separately or using a list(old_list) constructor function to make a copy. When you build it right, you can access/manipulate an element of a nested list like this:

>>> mat[0][0] = 1
>>> mat[1][2] = 'Q_Q'
>>> mat[2][0] = 3
>>> mat[2][2] = 5
>>> mat
[[1, 0, 0, 0, 0, 0, 0],
 [0, 0, 'Q_Q', 0, 0, 0, 0],
 [3, 0, 5, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0]]

Good luck!

lprsd · Answer 2 · 2011-07-18T11:01:26.240

In [13]: [[0]*8 for el in range(8)]
Out[13]: 
[[0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0]]

Or similarly,

In [21]: [[a for a in range(el*8,(el+1)*8)] for el in range(8)]
Out[21]: 
[[0, 1, 2, 3, 4, 5, 6, 7],
 [8, 9, 10, 11, 12, 13, 14, 15],
 [16, 17, 18, 19, 20, 21, 22, 23],
 [24, 25, 26, 27, 28, 29, 30, 31],
 [32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47],
 [48, 49, 50, 51, 52, 53, 54, 55],
 [56, 57, 58, 59, 60, 61, 62, 63]]

agf · Answer 3 · 2011-07-18T12:20:54.760

Your question is tagged Numpy, so here it is in a form that lets you turn it into a Numpy matrix.

from numpy import matrix

repeat = 3
width = 7
rows_per_iteration = 3

# The 1st row is also the 4th and 7th, the 2nd is also the 5th and 8th, etc.
A = [[0] * width for Null in range(rows_per_iteration)] * repeat

A[0][0] = 1
A[1][1] = 1
A[2][2] = 1
A[0][3] = 'c'
A[0][4] = 0
A[0][5] = 'a'
A[0][6] = 'b'
A[1][3] = 'g'
A[1][4] = 'd'
A[1][5] = 't'
A[1][6] = 'f'
A[2][3] = 'm'
A[2][4] = 'h'
A[2][5] = 'u'
A[2][6] = 'k'

A_mat = matrix(A)

If you're going to use them without going to a matrix,

repeat = 3
width = 7
rows_per_iteration = 3
total_rows = repeat * rows_per_iteration

A = [[0] * width for Null in range(total_rows)]

for np in range(0, total_rows, rows_per_iteration):
    A[np][0] = 1
    A[1 + np][1] = 1
    A[2 + np][2] = 1
    A[np][3] = 'c'
    A[np][4] = 0
    A[np][5] = 'a'
    A[np][6] = 'b'
    A[1 + np][3] = 'g'
    A[1 + np][4] = 'd'
    A[1 + np][5] = 't'
    A[1 + np][6] = 'f'
    A[2 + np][3] = 'm'
    A[2 + np][4] = 'h'
    A[2 + np][5] = 'u'
    A[2 + np][6] = 'k'

Which will actually create unique lists for each row instead of reusing them every 3rd row.

How to create a Matrix through an iteration in Python

3 Answers3