Accessing Values in a List¶

Use square bracket to access the value in a list. The first item in a list is accessed with index 0. See examples below:

num_list = [4, 12, 'a', 'b', 5.0]

first_item = num_list[0]            # value: 4, first item in the list

last_item = num_list[4]             # value: 5.0, last item in the list

last_item = num_list[-1]            # use -1 to access elements starting from the right

first_two_items = num_list[0:2]     # value: [4, 12] Note: This is called list slicing

items_in_middle = num_list[1:-1]    # value: [12, 'a', 'b']

invalid_index = num_list[5]         # error, list is 0-indexed

Updating a List¶

You can change the value of existing items, add new item or remove existing item(s) from a list.

>>> a = [1, 2, 3, 4]
>>> a[1] = 5            # change the 2nd item from 2 to 5
>>> a
[1, 5, 3, 4]
>>> a.remove(3)         # remove the number 3
>>> a
[1, 5, 4]
>>> a.pop()             # remove last item
4
>>> a
[1, 5]
>>> a.append(10)        # add the number 10 to the end of the list
>>> a
[1, 5, 10]

Other useful list operations¶

Python includes a few useful list functions and methods.

>>> x = [2, 6, 3, 1, 10, 5]
>>> len(x)                      # get number of items in the list
6
>>> x.sort()                    # sort the list in ascending order
>>> x
[1, 2, 3, 5, 6, 10]
>>> x.reverse()                 # reverse the order of the elements in the list
>>> x
[10, 6, 5, 3, 2, 1]
>>> max(x)                      # get the maximum value in the list
10
>>> min(x)                      # get the minimum value in the list
1
>>> x.count(5)                  # get the number of 5 in the list
1

List Comprehensions¶

List comprehension offers a concise way of creating a new list from existing sequences.

>>> x= [1, 2, 3, 4, 5 ]
>>> y = [ i*3 for i in x ] # create a new list with each element multiplied by 3.
>>> y
[3, 6, 9, 12, 15]
>>> even = [ i for i in range(1, 10) if i%2==0 ] # create a list with even numbers
>>> even
[2, 4, 6, 8]

Summary of list operations¶

Each example in the table below assumes a list ‘s’ with initial elements 1 to 5, and a list ‘t’ with elements ‘a’ and ‘b’, when applicable.

>>> s = [1, 2, 3, 4, 5 ]
>>> t = ['a', 'b' ]

Operation	Explanation	Example	Output on s
s[i] = x	item i of s is replaced by x	s[2] = 0	[1, 2, 0, 4, 5]
s[i:j] = t	slice of s from i to j is replaced by the contents of the iterable t	s[1:5] = t	[1, ‘a’, ‘b’]
del s[i:j]	same as s[i:j] = []	del s[2:4]	[1, 2, 5]
s[i:j:k] = t	the elements of s[i:j:k] are replaced by those of t with step k	s[1:4:2] = t	[1, ‘a’, 3, ‘b’, 5]
del s[i:j:k]	removes the elements of s[i:j:k] from the list with step k	del s[0:5:2]	[2, 4]
s.append(x)	same as s[len(s):len(s)] = [x]	s.append(t)	[1, 2, 3, 4, 5, [‘a’, ‘b’]]
s.extend(x)	same as s[len(s):len(s)] = x	s.extend(t)	[1, 2, 3, 4, 5, ‘a’, ‘b’]
s.count(x)	return number of occurrences of x in s	s[0] = 3; s.count(3)	2
s.index(x[,i[,j]])	return smallest k such that s[k] == x and i<=k<j	s.index(3)	2
s.insert(i, x)	same as s[i:i] = [x]	s.insert(2, 6)	[1, 2, 6, 3, 4, 5]
s.pop([i])	same as x = s[i]; del s[i]; return x	s.pop(2)	3
s.remove(x)	same as del s[s.index(x)]	s.remove(3)	[1, 2, 4, 5]
s.reverse()	reverses the items of s in place	s.reverse()	[5, 4, 3, 2, 1]
s.sort([cmp[, key[, reverse ]]])	sort the items of s in place	s[0] = 4; s.sort()	[2, 3, 4, 4, 5]