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
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
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.
range
to create the internal lists directly: [range(5) for x in range(5)]
[0] * w
part is nice, but [[0] * w] * h]
will produce unexpected behavior. Try mat = [[0] * 3] * 3; mat[0][1] = 10; print(mat == [[0, 10, 0], [0, 10, 0], [0, 10, 0]])
and mat = [[0] * 3 for i in range(3)]; mat[0][1] = 10; print(mat == [[0, 10, 0], [0, 0, 0], [0, 0, 0]])
.
test = [[0]*3]*5; test[1][1]=7; print(test)
Oct 15, 2018 at 9:18
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.
np.matrix
to represent it. The proper way to represent a matrix in numpy is with an array
.
Nov 5, 2017 at 20:39
array
s instead of matrices. While it isn't always encouraged, there are legitimate reasons for using matrix
-- context matters.
matrix
? Since @
operator was introduced, there seems to be one less reason since this post was written.
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]]
[0]*5
?
Mar 22, 2015 at 23:00
[[0]*5 for _ in range(5)]
with anonymous loop counter you're not using
Nov 11, 2018 at 20:38
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)]
matrix = [[]]
then needs .append
to actually create an index. Because otherwise matrix[0][0] = 1
does not yet work.
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.
import collections; Matrix = collections.defaultdict(float)
, to substitute zeros for uninitialized elements.
Oct 22, 2015 at 16:17
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)]
extend
would also be helpful in the first case: If you start with m = [[]]
, then you could add to the inner list (extend a row) with m[0].extend([1,2])
, and add to the outer list (append a new row) with m.append([3,4])
, those operations would leave you with [[1, 2], [3, 4]]
.
Oct 9, 2013 at 16:59
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.)
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
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
[[], [], []]
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.
Use:
matrix = [[0]*5 for i in range(5)]
The *5 for the first dimension works because at this level the data is immutable.
matrix = [[0]*cols for _ in range(rows)]
Feb 20, 2019 at 1:20
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)
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)
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]]
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
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)
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]
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],
...
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.
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.
numpy.matrix
equivalent to numpy.zeros
without zeros without being list?
Mar 19, 2018 at 22:14
l=[[0]*(L) for _ in range(W)]
Will be faster than:
l = [[0 for x in range(L)] for y in range(W)]
[[0]*(L) for i in range(W)]
should be [[0]*(L) for _ in range(W)]
since i
isn't used anywhere
Jan 8, 2019 at 3:39
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]]
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
# 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
Matrix[0], Matrix[1], ..., Matrix[4]
all point to the same array, so after Matrix[0][0] = 3
, you would expect Matrix[0][0] == Matrix[1][0] == ... == Matrix[4][0] == 3
.
Apr 3, 2014 at 19:38
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.
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))
Try this:
rows = int(input('Enter rows\n'))
my_list = []
for i in range(rows):
my_list.append(list(map(int, input().split())))
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]]
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
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