How to initialize a two-dimensional array in Python?

Question

I'm beginning python and I'm trying to use a two-dimensional list, that I initially fill up with the same variable in every place. I came up with this:

def initialize_twodlist(foo):
    twod_list = []
    new = []
    for i in range (0, 10):
        for j in range (0, 10):
            new.append(foo)
        twod_list.append(new)
        new = []

It gives the desired result, but feels like a workaround. Is there an easier/shorter/more elegant way to do this?

Answer 1

To initialize a two-dimensional list in Python, use

t = [ [0]*3 for i in range(3)]

But don't use [[v]*n]*n, it is a trap!

>>> a = [[0]*3]*3
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]

Answer 2

A pattern that often came up in Python was

bar = []
for item in some_iterable:
    bar.append(SOME EXPRESSION)

which helped motivate the introduction of list comprehensions, which convert that snippet to

bar = [SOME_EXPRESSION for item in some_iterable]

which is shorter and sometimes clearer. Usually, you get in the habit of recognizing these and often replacing loops with comprehensions.

Your code follows this pattern twice

twod_list = []                                       \                      
for i in range (0, 10):                               \
    new = []                  \ can be replaced        } this too
    for j in range (0, 10):    } with a list          /
        new.append(foo)       / comprehension        /
    twod_list.append(new)                           /

Answer 3

You can use a list comprehension:

x = [[foo for i in range(10)] for j in range(10)]
# x is now a 10x10 array of 'foo' (which can depend on i and j if you want)

Answer 4

This way is faster than the nested list comprehensions

[x[:] for x in [[foo] * 10] * 10]    # for immutable foo!

Here are some python3 timings, for small and large lists

$python3 -m timeit '[x[:] for x in [[1] * 10] * 10]'
1000000 loops, best of 3: 1.55 usec per loop

$ python3 -m timeit '[[1 for i in range(10)] for j in range(10)]'
100000 loops, best of 3: 6.44 usec per loop

$ python3 -m timeit '[x[:] for x in [[1] * 1000] * 1000]'
100 loops, best of 3: 5.5 msec per loop

$ python3 -m timeit '[[1 for i in range(1000)] for j in range(1000)]'
10 loops, best of 3: 27 msec per loop

Explanation:

[[foo]*10]*10 creates a list of the same object repeated 10 times. You can't just use this, because modifying one element will modify that same element in each row!

x[:] is equivalent to list(X) but is a bit more efficient since it avoids the name lookup. Either way, it creates a shallow copy of each row, so now all the elements are independent.

All the elements are the same foo object though, so if foo is mutable, you can't use this scheme., you'd have to use

import copy
[[copy.deepcopy(foo) for x in range(10)] for y in range(10)]

or assuming a class (or function) Foo that returns foos

[[Foo() for x in range(10)] for y in range(10)]

Answer 5

To initialize a two-dimensional array in Python:

a = [[0 for x in range(columns)] for y in range(rows)]

Answer 6

[[foo for x in xrange(10)] for y in xrange(10)]

Answer 7

Usually when you want multidimensional arrays you don't want a list of lists, but rather a numpy array or possibly a dict.

For example, with numpy you would do something like

import numpy
a = numpy.empty((10, 10))
a.fill(foo)

Answer 8

For those who are confused why [['']*m]*n is not good to use.

Python uses a system known as “Call by Object Reference” or “Call by assignment”.(More info)

Best way is [['' for i in range(columns)] for j in range(rows)]
This will solve all the problems.

For more Clarification
Example:

>>> x = [['']*3]*3
[['', '', ''], ['', '', ''], ['', '', '']]
>>> x[0][0] = 1
>>> print(x)
[[1, '', ''], [1, '', ''], [1, '', '']]

>>> y = [['' for i in range(3)] for j in range(3)]
[['', '', ''], ['', '', ''], ['', '', '']]
>>> y[0][0]=1
>>> print(y)
[[1, '', ''], ['', '', ''], ['', '', '']]

Answer 9

You can do just this:

[[element] * numcols] * numrows

For example:

>>> [['a'] *3] * 2
[['a', 'a', 'a'], ['a', 'a', 'a']]

But this has a undesired side effect:

>>> b = [['a']*3]*3
>>> b
[['a', 'a', 'a'], ['a', 'a', 'a'], ['a', 'a', 'a']]
>>> b[1][1]
'a'
>>> b[1][1] = 'b'
>>> b
[['a', 'b', 'a'], ['a', 'b', 'a'], ['a', 'b', 'a']]

Answer 10

twod_list = [[foo for _ in range(m)] for _ in range(n)]

for n is number of rows, and m is the number of column, and foo is the value.

Answer 11

Code:

num_rows, num_cols = 4, 2
initial_val = 0
matrix = [[initial_val] * num_cols for _ in range(num_rows)]
print(matrix) 
# [[0, 0], [0, 0], [0, 0], [0, 0]]

initial_val must be immutable.

Answer 12

If it's a sparsely-populated array, you might be better off using a dictionary keyed with a tuple:

dict = {}
key = (a,b)
dict[key] = value
...

Answer 13

t = [ [0]*10 for i in [0]*10]

for each element a new [0]*10 will be created ..

Answer 14

use the simplest think to create this.

wtod_list = []

and add the size:

wtod_list = [[0 for x in xrange(10)] for x in xrange(10)]

or if we want to declare the size firstly. we only use:

   wtod_list = [[0 for x in xrange(10)] for x in xrange(10)]

Answer 15

Incorrect Approach: [[None*m]*n]

>>> m, n = map(int, raw_input().split())
5 5
>>> x[0][0] = 34
>>> x
[[34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None]]
>>> id(x[0][0])
140416461589776
>>> id(x[3][0])
140416461589776

With this approach, python does not allow creating different address space for the outer columns and will lead to various misbehaviour than your expectation.

Correct Approach but with exception:

y = [[0 for i in range(m)] for j in range(n)]
>>> id(y[0][0]) == id(y[1][0])
False

It is good approach but there is exception if you set default value to None

>>> r = [[None for i in range(5)] for j in range(5)]
>>> r
[[None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None]]
>>> id(r[0][0]) == id(r[2][0])
True

So set your default value properly using this approach.

Absolute correct:

Follow the mike's reply of double loop.

Answer 16

To initialize a 2-dimensional array use: arr = [[]*m for i in range(n)]

actually, arr = [[]*m]*n will create a 2D array in which all n arrays will point to same array, so any change in value in any element will be reflected in all n lists

for more further explanation visit : https://www.geeksforgeeks.org/python-using-2d-arrays-lists-the-right-way/

Answer 17

Initializing a 2D matrix of size m X n with 0

m,n = map(int,input().split())
l = [[0 for i in range(m)] for j in range(n)]
print(l)

Answer 18

I use it this way to create MxN matrix where m = number(rows) and n = number(columns).

arr = [[None]*(n) for _ in range(m)]

Answer 19

Matrix={}
for i in range(0,3):
  for j in range(0,3):
    Matrix[i,j] = raw_input("Enter the matrix:")

Answer 20

If you use numpy, you can easily create 2d arrays:

import numpy as np

row = 3
col = 5
num = 10
x = np.full((row, col), num)

x

array([[10, 10, 10, 10, 10],
       [10, 10, 10, 10, 10],
       [10, 10, 10, 10, 10]])

Answer 21

row=5
col=5
[[x]*col for x in [b for b in range(row)]]

The above will give you a 5x5 2D array

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

It is using nested list comprehension. Breakdown as below:

[[x]*col for x in [b for b in range(row)]]

[x]*col --> final expression that is evaluated
for x in --> x will be the value provided by the iterator
[b for b in range(row)]] --> Iterator.

[b for b in range(row)]] this will evaluate to [0,1,2,3,4] since row=5
so now it simplifies to

[[x]*col for x in [0,1,2,3,4]]

This will evaluate to [[0]*5 for x in [0,1,2,3,4]] --> with x=0 1st iteration
[[1]*5 for x in [0,1,2,3,4]] --> with x=1 2nd iteration
[[2]*5 for x in [0,1,2,3,4]] --> with x=2 3rd iteration
[[3]*5 for x in [0,1,2,3,4]] --> with x=3 4th iteration
[[4]*5 for x in [0,1,2,3,4]] --> with x=4 5th iteration

Answer 22

As @Arnab and @Mike pointed out, an array is not a list. Few differences are 1) arrays are fixed size during initialization 2) arrays normally support lesser operations than a list.

Maybe an overkill in most cases, but here is a basic 2d array implementation that leverages hardware array implementation using python ctypes(c libraries)

import ctypes
class Array:
    def __init__(self,size,foo): #foo is the initial value
        self._size = size
        ArrayType = ctypes.py_object * size
        self._array = ArrayType()
        for i in range(size):
            self._array[i] = foo
    def __getitem__(self,index):
        return self._array[index]
    def __setitem__(self,index,value):
        self._array[index] = value
    def __len__(self):
        return self._size

class TwoDArray:
    def __init__(self,columns,rows,foo):
        self._2dArray = Array(rows,foo)
        for i in range(rows):
            self._2dArray[i] = Array(columns,foo)

    def numRows(self):
        return len(self._2dArray)
    def numCols(self):
        return len((self._2dArray)[0])
    def __getitem__(self,indexTuple):
        row = indexTuple[0]
        col = indexTuple[1]
        assert row >= 0 and row < self.numRows() \
               and col >=0 and col < self.numCols(),\
               "Array script out of range"
        return ((self._2dArray)[row])[col]

if(__name__ == "__main__"):
    twodArray = TwoDArray(4,5,5)#sample input
    print(twodArray[2,3])

Answer 23

An empty 2D matrix may be initialized in the following manner:

temp=[[],[]]

Answer 24

This is the best I've found for teaching new programmers, and without using additional libraries. I'd like something better though.

def initialize_twodlist(value):
    list=[]
    for row in range(10):
        list.append([value]*10)
    return list

Answer 25

Here is an easier way :

import numpy as np
twoD = np.array([[]*m]*n)

For initializing all cells with any 'x' value use :

twoD = np.array([[x]*m]*n

Answer 26

Often I use this approach for initializing a 2-dimensional array

n=[[int(x) for x in input().split()] for i in range(int(input())]

Answer 27

The general pattern to add dimensions could be drawn from this series:

x = 0
mat1 = []
for i in range(3):
    mat1.append(x)
    x+=1
print(mat1)


x=0
mat2 = []
for i in range(3):
    tmp = []
    for j in range(4):
        tmp.append(x)
        x+=1
    mat2.append(tmp)

print(mat2)


x=0
mat3 = []
for i in range(3):
    tmp = []
    for j in range(4):
        tmp2 = []
        for k in range(5):
            tmp2.append(x)
            x+=1
        tmp.append(tmp2)
    mat3.append(tmp)

print(mat3)

Answer 28

The important thing I understood is: While initializing an array(in any dimension) We should give a default value to all the positions of array. Then only initialization completes. After that, we can change or receive new values to any position of the array. The below code worked for me perfectly

N=7
F=2

#INITIALIZATION of 7 x 2 array with deafult value as 0
ar=[[0]*F for x in range(N)]

#RECEIVING NEW VALUES TO THE INITIALIZED ARRAY
for i in range(N):
    for j in range(F):
        ar[i][j]=int(input())
print(ar)

Answer 29

lst=[[0]*n]*m
np.array(lst)

initialize all matrix m=rows and n=columns

Answer 30

Another way is to use a dictionary to hold a two-dimensional array.

twoD = {}
twoD[0,0] = 0
print(twoD[0,0]) # ===> prints 0

This just can hold any 1D, 2D values and to initialize this to 0 or any other int value, use collections.

import collections
twoD = collections.defaultdict(int)
print(twoD[0,0]) # ==> prints 0
twoD[1,1] = 1
print(twoD[1,1]) # ==> prints 1

Answer 31

from random import randint
l = []

for i in range(10):
    k=[]
    for j in range(10):
        a= randint(1,100)
        k.append(a)

    l.append(k)




print(l)
print(max(l[2]))

b = []
for i in range(10):
    a = l[i][5]
    b.append(a)

print(min(b))