Create an empty list with certain size in Python

Question

How do I create an empty list that can hold 10 elements?

After that, I want to assign values in that list. For example:

xs = list()
for i in range(0, 9):
   xs[i] = i

However, that gives IndexError: list assignment index out of range. Why?

Answer 1

You cannot assign to a list like xs[i] = value, unless the list already is initialized with at least i+1 elements (because the first index is 0). Instead, use xs.append(value) to add elements to the end of the list. (Though you could use the assignment notation if you were using a dictionary instead of a list.)

Creating an empty list:

>>> xs = [None] * 10
>>> xs
[None, None, None, None, None, None, None, None, None, None]

Assigning a value to an existing element of the above list:

>>> xs[1] = 5
>>> xs
[None, 5, None, None, None, None, None, None, None, None]

Keep in mind that something like xs[15] = 5 would still fail, as our list has only 10 elements.

range(x) creates a list from [0, 1, 2, ... x-1]

# 2.X only. Use list(range(10)) in 3.X.
>>> xs = range(10)
>>> xs
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Using a function to create a list:

>>> def display():
...     xs = []
...     for i in range(9): # This is just to tell you how to create a list.
...         xs.append(i)
...     return xs
... 
>>> print display()
[0, 1, 2, 3, 4, 5, 6, 7, 8]

List comprehension (Using the squares because for range you don't need to do all this, you can just return range(0,9) ):

>>> def display():
...     return [x**2 for x in range(9)]
... 
>>> print display()
[0, 1, 4, 9, 16, 25, 36, 49, 64]

Answer 2

You can .append(element) to the list, e.g.:

s1.append(i)

What you are currently trying to do is access an element (s1[i]) that does not exist.

Answer 3

How do I create an empty list that can hold 10 elements?

All lists can hold as many elements as you like, subject only to the limit of available memory. The only "size" of a list that matters is the number of elements currently in it.

However, that gives IndexError: list assignment index out of range. Why?

The first time through the loop, i is equal to 0. Thus, we attempt xs[0] = 0. This does not work because there are currently 0 elements in the list, so 0 is not a valid index.

We cannot use indexing to write list elements that don't already exist - we can only overwrite existing ones. Instead, we should use the .append method:

xs = list();
for i in range(0, 9):
   xs.append(i)

The next problem you will note is that your list will actually have only 9 elements, because the end point is skipped by the range function. (As side notes: [] works just as well as list(), the semicolon is unnecessary, and only one parameter is needed for range if you're starting from 0.) Addressing those issues gives:

xs = []
for i in range(10):
    xs.append(i)

However, this is still missing the mark - range is not some magical keyword that's part of the language the way for (or, say, def) is.

In 2.x, range is a function, which directly returns the list that we already wanted:

xs = range(10) # 2.x specific!
# In 3.x, we don't get a list; we can do a lot of things with the
# result, but we can't e.g. append or replace elements.

In 3.x, range is a cleverly designed class, and range(10) creates an instance. To get the desired list, we can simply feed it to the list constructor:

xs = list(range(10)) # correct in 3.x, redundant in 2.x

Answer 4

A list is always "iterable" and you can always add new elements to it:

1. insert: list.insert(indexPosition, value)
2. append: list.append(value)
3. extend: list.extend(value)

In your case, you had instantiated an empty list of length 0. Therefore, when you try to add any value to the list using the list index (i), it is referring to a location that does not exist. Therefore, you were getting the error "IndexError: list assignment index out of range".

You can try this instead:

s1 = list();
for i in range(0,9):
   s1.append(i)

print (s1)

To create a list of size 10(let's say), you can first create an empty array, like np.empty(10) and then convert it to list using arrayName.tolist(). Alternately, you can chain them as well.

            **`np.empty(10).tolist()`**

Answer 5

Another option is to use numpy for fixed size arrays (of pointers):

> pip install numpy

import numpy as np


a = np.empty(10, dtype=np.object)
a[1] = 2
a[5] = "john"
a[3] = []

If you just want numbers, you can do with numpy:

a = np.arange(10)

Answer 6

Not technically a list but similar to a list in terms of functionality and it's a fixed length

from collections import deque
my_deque_size_10 = deque(maxlen=10)

If it's full, ie got 10 items then adding another item results in item @index 0 being discarded. FIFO..but you can also append in either direction. Used in say

• a rolling average of stats
• piping a list through it aka sliding a window over a list until you get a match against another deque object.

If you need a list then when full just use list(deque object)

Answer 7

The accepted answer has some gotchas. For example:

>>> a = [{}] * 3
>>> a
[{}, {}, {}]
>>> a[0]['hello'] = 5
>>> a
[{'hello': 5}, {'hello': 5}, {'hello': 5}]
>>> 

So each dictionary refers to the same object. Same holds true if you initialize with arrays or objects.

You could do this instead:

>>> b = [{} for i in range(0, 3)]
>>> b
[{}, {}, {}]
>>> b[0]['hello'] = 6
>>> b
[{'hello': 6}, {}, {}]
>>> 

Answer 8

Try this instead:

lst = [None] * 10

The above will create a list of size 10, where each position is initialized to None. After that, you can add elements to it:

lst = [None] * 10
for i in range(10):
    lst[i] = i

Admittedly, that's not the Pythonic way to do things. Better do this:

lst = []
for i in range(10):
    lst.append(i)

Or even simpler, in Python 2.x you can do this to initialize a list with values from 0 to 9:

lst = range(10)

And in Python 3.x:

lst = list(range(10))

Answer 9

One simple way to create a 2D matrix of size n using nested list comprehensions:

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

Answer 10

I'm a bit surprised that the easiest way to create an initialised list is not in any of these answers. Just use a generator in the list function:

list(range(9))

Answer 11

Make it more reusable as a function.

def createEmptyList(length,fill=None):
    '''
    return a (empty) list of a given length
    Example:
        print createEmptyList(3,-1)
        >> [-1, -1, -1]
        print createEmptyList(4)
        >> [None, None, None, None]
    '''
    return [fill] * length

Answer 12

This code generates an array that contains 10 random numbers.

import random
numrand=[]
for i in range(0,10):
   a = random.randint(1,50)
   numrand.append(a)
   print(a,i)
print(numrand)

Answer 13

I came across this SO question while searching for a similar problem. I had to build a 2D array and then replace some elements of each list (in 2D array) with elements from a dict. I then came across this SO question which helped me, maybe this will help other beginners to get around. The key trick was to initialize the 2D array as an numpy array and then using array[i,j] instead of array[i][j].

For reference this is the piece of code where I had to use this :

nd_array = []
for i in range(30):
    nd_array.append(np.zeros(shape = (32,1)))
new_array = []
for i in range(len(lines)):
    new_array.append(nd_array)
new_array = np.asarray(new_array)
for i in range(len(lines)):
    splits = lines[i].split(' ')
    for j in range(len(splits)):
        #print(new_array[i][j])
        new_array[i,j] = final_embeddings[dictionary[str(splits[j])]-1].reshape(32,1)

Now I know we can use list comprehension but for simplicity sake I am using a nested for loop. Hope this helps others who come across this post.

Answer 14

Here's my code for 2D list in python which would read no. of rows from the input :

empty = []
row = int(input())

for i in range(row):
    temp = list(map(int, input().split()))
    empty.append(temp)

for i in empty:
    for j in i:
        print(j, end=' ')
    print('')

Answer 15

s1 = []
for i in range(11):
   s1.append(i)

print s1

To create a list, just use these brackets: "[]"

To add something to a list, use list.append()

Answer 16

I'm surprised nobody suggest this simple approach to creating a list of empty lists. This is an old thread, but just adding this for completeness. This will create a list of 10 empty lists

x = [[] for i in range(10)]

Answer 17

varunl's currently accepted answer

 >>> l = [None] * 10
 >>> l
 [None, None, None, None, None, None, None, None, None, None]

Works well for non-reference types like numbers. Unfortunately if you want to create a list-of-lists you will run into referencing errors. Example in Python 2.7.6:

>>> a = [[]]*10
>>> a
[[], [], [], [], [], [], [], [], [], []]
>>> a[0].append(0)
>>> a
[[0], [0], [0], [0], [0], [0], [0], [0], [0], [0]]
>>> 

As you can see, each element is pointing to the same list object. To get around this, you can create a method that will initialize each position to a different object reference.

def init_list_of_objects(size):
    list_of_objects = list()
    for i in range(0,size):
        list_of_objects.append( list() ) #different object reference each time
    return list_of_objects


>>> a = init_list_of_objects(10)
>>> a
[[], [], [], [], [], [], [], [], [], []]
>>> a[0].append(0)
>>> a
[[0], [], [], [], [], [], [], [], [], []]
>>> 

There is likely a default, built-in python way of doing this (instead of writing a function), but I'm not sure what it is. Would be happy to be corrected!

Edit: It's [ [] for _ in range(10)]

Example :

>>> [ [random.random() for _ in range(2) ] for _ in range(5)]
>>> [[0.7528051908943816, 0.4325669600055032], [0.510983236521753, 0.7789949902294716], [0.09475179523690558, 0.30216475640534635], [0.3996890132468158, 0.6374322093017013], [0.3374204010027543, 0.4514925173253973]]

Answer 18

There are two "quick" methods:

x = length_of_your_list
a = [None]*x
# or
a = [None for _ in xrange(x)]

It appears that [None]*x is faster:

>>> from timeit import timeit
>>> timeit("[None]*100",number=10000)
0.023542165756225586
>>> timeit("[None for _ in xrange(100)]",number=10000)
0.07616496086120605

But if you are ok with a range (e.g. [0,1,2,3,...,x-1]), then range(x) might be fastest:

>>> timeit("range(100)",number=10000)
0.012513160705566406