How to make matrices in Python?

Question

I've googled it and searched StackOverflow and YouTube.. I just can't get matrices in Python to click in my head. Can someone please help me? I'm just trying to create a basic 5x5 box that displays:

A A A A A
B B B B B
C C C C C
D D D D D
E E E E E

I got

a b c d e
a b c d e
a b c d e
a b c d e
a b c d e

To display but I couldn't even get them to break lines that, instead all they would appear as

[['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E']]

And if I try to add \n to them or print "" etc it just doesn't work.. \n will display as 'A\n' and print will display before the matrix.

Please someone help me, even if you direct me to somewhere that should be really obvious and make me look like an idiot, I just want to learn this.

Answer 1

If you don't want to use numpy, you could use the list of lists concept. To create any 2D array, just use the following syntax:

  mat = [[input() for i in range (col)] for j in range (row)]

and then enter the values you want.

Answer 2

I got a simple fix to this by casting the lists into strings and performing string operations to get the proper print out of the matrix.

Creating the function

By creating a function, it saves you the trouble of writing the for loop every time you want to print out a matrix.

def print_matrix(matrix):
    for row in matrix:
        new_row = str(row)
        new_row = new_row.replace(',','')
        new_row = new_row.replace('[','')
        new_row = new_row.replace(']','')
        print(new_row)

Examples

Example of a 5x5 matrix with 0 as every entry:

>>> test_matrix = [[0] * 5 for i in range(5)]
>>> print_matrix(test_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 

Example of a 2x3 matrix with 0 as every entry:

>>> test_matrix = [[0] * 3 for i in range(2)]
>>> print_matrix(test_matrix)
0 0 0
0 0 0

---

EDIT

If you want to make it print:

A A A A A
B B B B B
C C C C C
D D D D D 
E E E E E

I suggest you just change the way you enter your data into your lists within lists. In my method, each list within the larger list represents a line in the matrix, not columns.

Answer 3

you can do it short like this:

matrix = [["A, B, C, D, E"]*5]
print(matrix)


[['A, B, C, D, E', 'A, B, C, D, E', 'A, B, C, D, E', 'A, B, C, D, E', 'A, B, C, D, E']]

Answer 4

you can also use append function

b = [ ]

for x in range(0, 5):
    b.append(["O"] * 5)

def print_b(b):
    for row in b:
        print " ".join(row)

Answer 5

Looping helps:

for row in matrix:
    print ' '.join(row)

or use nested str.join() calls:

print '\n'.join([' '.join(row) for row in matrix])

Demo:

>>> matrix = [['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E']]
>>> for row in matrix:
...     print ' '.join(row)
... 
A B C D E
A B C D E
A B C D E
A B C D E
A B C D E
>>> print '\n'.join([' '.join(row) for row in matrix])
A B C D E
A B C D E
A B C D E
A B C D E
A B C D E

If you wanted to show the rows and columns transposed, transpose the matrix by using the zip() function; if you pass each row as a separate argument to the function, zip() recombines these value by value as tuples of columns instead. The *args syntax lets you apply a whole sequence of rows as separate arguments:

>>> for cols in zip(*matrix):  # transposed
...     print ' '.join(cols)
... 
A A A A A
B B B B B
C C C C C
D D D D D
E E E E E

Answer 6

The answer to your question depends on what your learning goals are. If you are trying to get matrices to "click" so you can use them later, I would suggest looking at a Numpy array instead of a list of lists. This will let you slice out rows and columns and subsets easily. Just try to get a column from a list of lists and you will be frustrated.

Using a list of lists as a matrix...

Let's take your list of lists for example:

L = [list("ABCDE") for i in range(5)]

It is easy to get sub-elements for any row:

>>> L[1][0:3]
['A', 'B', 'C']

Or an entire row:

>>> L[1][:]
['A', 'B', 'C', 'D', 'E']

But try to flip that around to get the same elements in column format, and it won't work...

>>> L[0:3][1]
['A', 'B', 'C', 'D', 'E']

>>> L[:][1]
['A', 'B', 'C', 'D', 'E']

You would have to use something like list comprehension to get all the 1th elements....

>>> [x[1] for x in L]
['B', 'B', 'B', 'B', 'B']

Enter matrices

If you use an array instead, you will get the slicing and indexing that you expect from MATLAB or R, (or most other languages, for that matter):

>>> import numpy as np
>>> Y = np.array(list("ABCDE"*5)).reshape(5,5)
>>> print Y
[['A' 'B' 'C' 'D' 'E']
 ['A' 'B' 'C' 'D' 'E']
 ['A' 'B' 'C' 'D' 'E']
 ['A' 'B' 'C' 'D' 'E']
 ['A' 'B' 'C' 'D' 'E']]
>>> print Y.transpose()
[['A' 'A' 'A' 'A' 'A']
 ['B' 'B' 'B' 'B' 'B']
 ['C' 'C' 'C' 'C' 'C']
 ['D' 'D' 'D' 'D' 'D']
 ['E' 'E' 'E' 'E' 'E']]

Grab row 1 (as with lists):

>>> Y[1,:]
array(['A', 'B', 'C', 'D', 'E'], 
      dtype='|S1')

Grab column 1 (new!):

>>> Y[:,1]
array(['B', 'B', 'B', 'B', 'B'], 
      dtype='|S1')

So now to generate your printed matrix:

for mycol in Y.transpose():
    print " ".join(mycol)


A A A A A
B B B B B
C C C C C
D D D D D
E E E E E