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?
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
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]]
but
t = [ [0]*3 for i in range(3)]
works great.
Method 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) /
Method 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)
Method 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)]
Method 5
To initialize a two-dimensional array in Python:
a = [[0 for x in range(columns)] for y in range(rows)]
Method 6
[[foo for x in xrange(10)] for y in xrange(10)]
Method 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)
Method 8
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']]
Method 9
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.
Method 10
For those who are confused why [['']*m]*n is not good to use.
Reason:- Python uses calls by reference, so changing one value in above case cause changing of other index values also.
Best way is [['' for i in range(m)] for j in range(n)]
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, '', ''], ['', '', ''], ['', '', '']]
Method 11
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
...
Method 12
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.
Method 13
t = [ [0]*10 for i in [0]*10]
for each element a new [0]*10 will be created ..
Method 14
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.
Method 15
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/
Method 16
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)]
Method 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)
Method 18
Matrix={}
for i in range(0,3):
for j in range(0,3):
Matrix[i,j] = raw_input("Enter the matrix:")
Method 19
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]])
Method 20
row=5 col=5 [[x]*col for x in [b for b in range(row)]]
The above will give you a 5×5 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
Method 21
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)<div class="su-row"></div>)[col]
if(__name__ == "__main__"):
twodArray = TwoDArray(4,5,5)#sample input
print(twodArray[2,3])
Method 22
I use it this way to create mxn matrix where m = no(rows) and n = no(columns).
arr = [[None]*(n) for _ in range(m)]
Method 23
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
Method 24
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
Method 25
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())]
Method 26
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)
Method 27
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)
Method 28
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
Method 29
lst=[[0]*n]*m np.array(lst)
initialize all matrix m=rows and n=columns
Method 30
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))
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0