How to define a two-dimensional array?

Question

I want to define a two-dimensional array without an initialized length like this:

Matrix = [][]

But this gives an error:

IndexError: list index out of range

Answer 1

You're technically trying to index an uninitialized array. You have to first initialize the outer list with lists before adding items; Python calls this "list comprehension".

# Creates a list containing 5 lists, each of 8 items, all set to 0
w, h = 8, 5
Matrix = [[0 for x in range(w)] for y in range(h)] 

#You can now add items to the list:

Matrix[0][0] = 1
Matrix[6][0] = 3 # error! range... 
Matrix[0][6] = 3 # valid

Note that the matrix is "y" address major, in other words, the "y index" comes before the "x index".

print Matrix[0][0] # prints 1
x, y = 0, 6 
print Matrix[x][y] # prints 3; be careful with indexing! 

Although you can name them as you wish, I look at it this way to avoid some confusion that could arise with the indexing, if you use "x" for both the inner and outer lists, and want a non-square Matrix.

Answer 2

If you really want a matrix, you might be better off using numpy. Matrix operations in numpy most often use an array type with two dimensions. There are many ways to create a new array; one of the most useful is the zeros function, which takes a shape parameter and returns an array of the given shape, with the values initialized to zero:

>>> import numpy
>>> numpy.zeros((5, 5))
array([[ 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.]])

Here are some other ways to create 2-d arrays and matrices (with output removed for compactness):

numpy.arange(25).reshape((5, 5))         # create a 1-d range and reshape
numpy.array(range(25)).reshape((5, 5))   # pass a Python range and reshape
numpy.array([5] * 25).reshape((5, 5))    # pass a Python list and reshape
numpy.empty((5, 5))                      # allocate, but don't initialize
numpy.ones((5, 5))                       # initialize with ones

numpy provides a matrix type as well, but it is no longer recommended for any use, and may be removed from numpy in the future.

Answer 3

Here is a shorter notation for initializing a list of lists:

matrix = [[0]*5 for i in range(5)]

Unfortunately shortening this to something like 5*[5*[0]] doesn't really work because you end up with 5 copies of the same list, so when you modify one of them they all change, for example:

>>> matrix = 5*[5*[0]]
>>> 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]]
>>> matrix[4][4] = 2
>>> matrix
[[0, 0, 0, 0, 2], [0, 0, 0, 0, 2], [0, 0, 0, 0, 2], [0, 0, 0, 0, 2], [0, 0, 0, 0, 2]]

Answer 4

If you want to create an empty matrix, the correct syntax is

matrix = [[]]

And if you want to generate a matrix of size 5 filled with 0,

matrix = [[0 for i in xrange(5)] for i in xrange(5)]

Answer 5

If all you want is a two dimensional container to hold some elements, you could conveniently use a dictionary instead:

Matrix = {}

Then you can do:

Matrix[1,2] = 15
print Matrix[1,2]

This works because 1,2 is a tuple, and you're using it as a key to index the dictionary. The result is similar to a dumb sparse matrix.

As indicated by osa and Josap Valls, you can also use Matrix = collections.defaultdict(lambda:0) so that the missing elements have a default value of 0.

Vatsal further points that this method is probably not very efficient for large matrices and should only be used in non performance-critical parts of the code.

Answer 6

In Python you will be creating a list of lists. You do not have to declare the dimensions ahead of time, but you can. For example:

matrix = []
matrix.append([])
matrix.append([])
matrix[0].append(2)
matrix[1].append(3)

Now matrix[0][0] == 2 and matrix[1][0] == 3. You can also use the list comprehension syntax. This example uses it twice over to build a "two-dimensional list":

from itertools import count, takewhile
matrix = [[i for i in takewhile(lambda j: j < (k+1) * 10, count(k*10))] for k in range(10)]

Answer 7

Here's the code for a beginner whose coming from C, CPP and Java background

rows = int(input())
cols = int(input())

matrix = []
for i in range(rows):
  row = []
  for j in range(cols):
    row.append(0)
  matrix.append(row)

print(matrix)

Why such a long code, that too in Python you ask?

Long back when I was not comfortable with Python, I saw the single line answers for writing 2D matrix and told myself I am not going to use 2-D matrix in Python again. (Those single lines were pretty scary and It didn't give me any information on what Python was doing. Also note that I am not aware of these shorthands.)

Answer 8

The accepted answer is good and correct, but it took me a while to understand that I could also use it to create a completely empty array.

l =  [[] for _ in range(3)]

results in

[[], [], []]

Answer 9

You should make a list of lists, and the best way is to use nested comprehensions:

>>> matrix = [[0 for i in range(5)] for j in range(5)]
>>> pprint.pprint(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]]

On your [5][5] example, you are creating a list with an integer "5" inside, and try to access its 5th item, and that naturally raises an IndexError because there is no 5th item:

>>> l = [5]
>>> l[5]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

Answer 10

This is how I usually create 2D arrays in python.

col = 3
row = 4
array = [[0] * col for _ in range(row)]

I find this syntax easy to remember compared to using two for loops in a list comprehension.

Answer 11

Use:

matrix = [[0]*5 for i in range(5)]

The *5 for the first dimension works because at this level the data is immutable.

Answer 12

A rewrite for easy reading:

# 2D array/ matrix

# 5 rows, 5 cols
rows_count = 5
cols_count = 5

# create
#     creation looks reverse
#     create an array of "cols_count" cols, for each of the "rows_count" rows
#        all elements are initialized to 0
two_d_array = [[0 for j in range(cols_count)] for i in range(rows_count)]

# index is from 0 to 4
#     for both rows & cols
#     since 5 rows, 5 cols

# use
two_d_array[0][0] = 1
print two_d_array[0][0]  # prints 1   # 1st row, 1st col (top-left element of matrix)

two_d_array[1][0] = 2
print two_d_array[1][0]  # prints 2   # 2nd row, 1st col

two_d_array[1][4] = 3
print two_d_array[1][4]  # prints 3   # 2nd row, last col

two_d_array[4][4] = 4
print two_d_array[4][4]  # prints 4   # last row, last col (right, bottom element of matrix)

Answer 13

To declare a matrix of zeros (ones):

numpy.zeros((x, y))

e.g.

>>> numpy.zeros((3, 5))
    array([[ 0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.],
   [ 0.,  0.,  0.,  0.,  0.]])

or numpy.ones((x, y)) e.g.

>>> np.ones((3, 5))
array([[ 1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.],
   [ 1.,  1.,  1.,  1.,  1.]])

Even three dimensions are possible. (http://www.astro.ufl.edu/~warner/prog/python.html see --> Multi-dimensional arrays)

Answer 14

You can create an empty two dimensional list by nesting two or more square bracing or third bracket ([], separated by comma) with a square bracing, just like below:

Matrix = [[], []]

Now suppose you want to append 1 to Matrix[0][0] then you type:

Matrix[0].append(1)

Now, type Matrix and hit Enter. The output will be:

[[1], []]

If you entered the following statement instead

Matrix[1].append(1)

then the Matrix would be

[[], [1]]

Answer 15

I'm on my first Python script, and I was a little confused by the square matrix example so I hope the below example will help you save some time:

 # Creates a 2 x 5 matrix
 Matrix = [[0 for y in xrange(5)] for x in xrange(2)]

so that

Matrix[1][4] = 2 # Valid
Matrix[4][1] = 3 # IndexError: list index out of range

Answer 16

Using NumPy you can initialize empty matrix like this:

import numpy as np
mm = np.matrix([])

And later append data like this:

mm = np.append(mm, [[1,2]], axis=1)

Answer 17

I read in comma separated files like this:

data=[]
for l in infile:
    l = split(',')
    data.append(l)

The list "data" is then a list of lists with index data[row][col]

Answer 18

Use:

import copy

def ndlist(*args, init=0):
    dp = init
    for x in reversed(args):
        dp = [copy.deepcopy(dp) for _ in range(x)]
    return dp

l = ndlist(1,2,3,4) # 4 dimensional list initialized with 0's
l[0][1][2][3] = 1

I do think NumPy is the way to go. The above is a generic one if you don't want to use NumPy.

Answer 19

If you want to be able to think it as a 2D array rather than being forced to think in term of a list of lists (much more natural in my opinion), you can do the following:

import numpy
Nx=3; Ny=4
my2Dlist= numpy.zeros((Nx,Ny)).tolist()

The result is a list (not a NumPy array), and you can overwrite the individual positions with numbers, strings, whatever.

Answer 20

That's what dictionary is made for!

matrix = {}

---

You can define keys and values in two ways:

matrix[0,0] = value

or

matrix = { (0,0)  : value }

Result:

   [ value,  value,  value,  value,  value],
   [ value,  value,  value,  value,  value],
   ...

Answer 21

l=[[0]*(L) for _ in range(W)]

Will be faster than:

l = [[0 for x in range(L)] for y in range(W)] 

Answer 22

If you don't have size information before start then create two one-dimensional lists.

list 1: To store rows
list 2: Actual two-dimensional matrix

Store the entire row in the 1st list. Once done, append list 1 into list 2:

from random import randint

coordinates=[]
temp=[]
points=int(raw_input("Enter No Of Coordinates >"))
for i in range(0,points):
    randomx=randint(0,1000)
    randomy=randint(0,1000)
    temp=[]
    temp.append(randomx)
    temp.append(randomy)
    coordinates.append(temp)

print coordinates

Output:

Enter No Of Coordinates >4
[[522, 96], [378, 276], [349, 741], [238, 439]]

Answer 23

by using list :

matrix_in_python  = [['Roy',80,75,85,90,95],['John',75,80,75,85,100],['Dave',80,80,80,90,95]]

by using dict: you can also store this info in the hash table for fast searching like

matrix = { '1':[0,0] , '2':[0,1],'3':[0,2],'4' : [1,0],'5':[1,1],'6':[1,2],'7':[2,0],'8':[2,1],'9':[2,2]};

matrix['1'] will give you result in O(1) time

*nb: you need to deal with a collision in the hash table

Answer 24

# Creates a list containing 5 lists initialized to 0
Matrix = [[0]*5]*5

Be careful about this short expression, see full explanation down in @F.J's answer

Answer 25

Here is the code snippet for creating a matrix in python:

# get the input rows and cols
rows = int(input("rows : "))
cols = int(input("Cols : "))

# initialize the list
l=[[0]*cols for i in range(rows)]

# fill some random values in it
for i in range(0,rows):
    for j in range(0,cols):
        l[i][j] = i+j

# print the list
for i in range(0,rows):
    print()
    for j in range(0,cols):
        print(l[i][j],end=" ")

Please suggest if I have missed something.

Answer 26

Usually, the go-to module is NumPy:

import numpy as np
   
# Generate a random matrix of floats
np.random.rand(cols,rows)

# Generate a random matrix of integers
np.random.randint(1, 10, size=(cols,rows))

Answer 27

Try this:

rows = int(input('Enter rows\n'))
my_list = []
for i in range(rows):
    my_list.append(list(map(int, input().split())))

Answer 28

In case if you need a matrix with predefined numbers you can use the following code:

def matrix(rows, cols, start=0):
    return [[c + start + r * cols for c in range(cols)] for r in range(rows)]


assert matrix(2, 3, 1) == [[1, 2, 3], [4, 5, 6]]

Answer 29

User Define function to input Matrix and print

def inmatrix(m,n):
    #Start function and pass row and column as parameter
    a=[] #create a blank matrix
    for i in range(m): #Row input
        b=[]#blank list
        for j in range(n): # column input
            elm=int(input("Enter number in Pocket ["+str(i)+"]["+str(j)+"] ")) #Show Row And column  number 
            b.append(elm) #add value to b list
        a.append(b)# Add list to matrix
    return  a #return Matrix 

def Matrix(a): #function for print Matrix
    for i in range(len(a)): #row
        for j in range(len(a[0])): #column
            print(a[i][j],end=" ") #print value with space
        print()#print a line After a row print

m=int(input("Enter number of row")) #input row
n=int(input("Enter number of column"))
a=inmatrix(m,n) #call input matrix function 

print("Matrix is ... ")

Matrix(a) #print matrix function

Answer 30

If you want to create a 2d matrix which dimension is defined by two variables and initialise it with a default value for all its elements. You can use this simple syntax

n_rows=3
n_cols=4
aux_matrix= [[1]*n_cols]*n_rows