1.4 lists

This lesson describes the concept of list in python.

A list can be defined as a collection of objects or a container in which we hold different objects.

mylist = []

Above we defined an empty list. How do we know that it is list? We can always check the type of object in python as below.

print(type(mylist))

<class 'list'>

We can also make a list which is not empty as below

imperialists = ['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola']
print(type(imperialists))

<class 'list'>

In above-mentioned list, all of its (6) elements are strings. However, the elements/members a list are not required to be of same type.

imperialists = ["Bush", {"years": 8}, 2000, (2000, 2008)]
print(type(imperialists))

<class 'list'>

In above list, the first member is string, the second member is dictionary, the third member is integer and the fourth member is a tuple. We will study about string, dictionary, integer and tuple in upcoming lessons.

There are two ways to convert a python object into list

• using []

• usnig list function

a = 1,2
print(type(a))
a_as_list_using_slice_op = [a]
print(type(a_as_list_using_slice_op))

<class 'tuple'>
<class 'list'>

a_as_list_using_list_fn = list(a)
print(type(a_as_list_using_list_fn))

<class 'list'>

However there is a major difference in these two. [] converts the whole object as it is into a list. On the other hand, list function is more like element wise operator. This can be verified by printing the converted lists created above.

print(a_as_list_using_slice_op)
print(a_as_list_using_list_fn)

[(1, 2)]
[1, 2]

a_as_list_using_list_fn is a list with two members while a_as_list_using_slice_op is a list with only one member. This can be verified by checking the length of both lists.

print(len(a_as_list_using_slice_op))
print(len(a_as_list_using_list_fn))

1
2

Then length of list which is created using slice operator [] is always 1, while the length of list created using list function depends upon the number of values in the object. The slice operator is creating a list with one member and that one member is a tuple in this case.

Spend some minutes on understanding the difference between these two ways of creating a list. Try with different objects and see the difference.

Question Write code to prove the above statement.

slicing a list

Since the list is a sequence, we can slice it as well. The slicing of a list can be done using the slice operator []. Consider the following list

imperialists = ['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola']
print(imperialists[0])

Bush

print(imperialists[2])

Trump

print(imperialists[-1])

coca cola

The : operator/symbol is used to slice a list. The syntax is as follows list[start:stop:step]. The default value of start is 0, of stop is length of list and of step is 1.

print(imperialists[2:4])

['Trump', 'Zuckerberg']

Question: What will be the output of the following code

a = [1,5, 7,14]
a[2] = 12
print(a)

Operations on a list

Once we have a list, we can perform different operations on it. Some of them are given below.

append

imperialists = ['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola']
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola']

imperialists.append('clinton')
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton']

append changes the original list and it itself returns None.

new_imperialists = imperialists.append('netanyahu')
print(new_imperialists)

None

print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', 'netanyahu']

If we add a similar but new element in list, then the list will have 2 such elements as its member.

imperialists.append('netanyahu')
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', 'netanyahu', 'netanyahu']

Question: What will be the output of the following code?

a = [1,5, 7,14]
a.append(12)
print(a[-1])

pop

last_element = imperialists.pop(-1)
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', 'netanyahu']

print(last_element)

netanyahu

# uncomment following 1 line
# imperialists.pop(8)

imperialists.pop()

'netanyahu'

print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton']

# uncomment following 1 line
# imperialists.pop('Bush')

Question: What will be the output of the following code?

a = [1,5, 7,14]
a.pop(2)
print(len(a))  # ??
a.pop(2)   # ??

extend

If we want to add multiple elements to a list, using append will put a new list in the previous list

imperialists = ['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton']
uk_imperialists = ['churchil', 'Tony Blair', 'BBC']
imperialists.append(uk_imperialists)
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', ['churchil', 'Tony Blair', 'BBC']]

imperialists = ['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton']
imperialists.extend(uk_imperialists)
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', 'churchil', 'Tony Blair', 'BBC']

extend actually takes any sequence as input. It must not be a list. It can be a string or tuple.

new_one = "Murdoch"
imperialists.extend(new_one)
print(imperialists)

['Bush', 'Obama', 'Trump', 'Zuckerberg', 'Bezos', 'coca cola', 'clinton', 'churchil', 'Tony Blair', 'BBC', 'M', 'u', 'r', 'd', 'o', 'c', 'h']

imperialists = ['Bush', 'Obama', 'Trump', 'clinton']
capitalists = ('Zuckerberg', 'Bezos', 'coca cola')
imperialists.extend(capitalists)
print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola']

Question: What will be the output of the following code?

a = [1,5, 7,14]
b = [2, 3]
print(len(a.extend(b)))  # ?

using + operator

We can also append lists using the + operator.

media_houses = ['bbc', 'cnn', 'reuters', 'springer']
print(imperialists + media_houses)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'springer']

So when we use + operator between two lists, a new list is created which contains all the members of both lists. The original lists remain unchanged.

print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola']

imperialists = imperialists + media_houses
print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'springer']

Above we are recreating the list imperialists by adding media_houses to it.

morons = ['sam haris', 'richard dawkins', 'baghdadi', 'bin ladan']
imperialists += morons
print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'springer', 'sam haris', 'richard dawkins', 'baghdadi', 'bin ladan']

+= operator means that the list on the left side of += operator is updated by adding the list on the right side of += operator.

Question: What will be the output of the following code?

a = [1,5, 7,14]
b = [2, 3]
print(len(a + b))  # ?

remove

imperialists.remove('springer')
print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'sam haris', 'richard dawkins', 'baghdadi', 'bin ladan']

If we repeat the above operation, it will result in error because springer has already been removed from the list imperialists.

# uncomment following 2 line
# imperialists.remove('springer')
# print(imperialists)

insert

puts an the value before the index

imperialists.insert(-1, 'DW')
print(imperialists)

['Bush', 'Obama', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'sam haris', 'richard dawkins', 'baghdadi', 'DW', 'bin ladan']

Finding position/index of a member of a list

imperialists.index('bbc')

7

# uncomment following 1 line
# imperialists.index('bbc', 8)

# uncomment following 1 line
# imperialists.index('bbc', 3, 6)

if an element is present in a list twice, index of its first position is returned.

last_value = imperialists[-1]
imperialists.insert(2, last_value)
print(imperialists)

['Bush', 'Obama', 'bin ladan', 'Trump', 'clinton', 'Zuckerberg', 'Bezos', 'coca cola', 'bbc', 'cnn', 'reuters', 'sam haris', 'richard dawkins', 'baghdadi', 'DW', 'bin ladan']

imperialists.index(last_value)

2

Question: What will be the output of the following code?

a = [1,5, 7,14]
a.insert(2, 12)
print(a)  # ?

reverse

imperialists = ['bbc', 'cnn', 'reuters', 'springer', 'voa']
imperialists.reverse()

The function does not return anything itself but the original list is reversed.

print(imperialists)

['voa', 'springer', 'reuters', 'cnn', 'bbc']

Question: What will be the output of the following code?

a = [1,5, 7,14]
a.reverse()
print(a)  # ?

sort

We can sort the list using the sort function. If the contents of the list strings, then the list will be sorted in lexicographical order. If the contents are numbers, then the list will be sorted in ascending order.

---

print(imperialists)

['voa', 'springer', 'reuters', 'cnn', 'bbc']

The function does not return anything itself but the original list is sorted.

imperialists.sort()

print(imperialists)

['bbc', 'cnn', 'reuters', 'springer', 'voa']

imperialists.sort(reverse=True)
print(imperialists)

['voa', 'springer', 'reuters', 'cnn', 'bbc']

numbers = [4,3,6,2]
numbers.sort()
print(numbers)

[2, 3, 4, 6]

# uncomment following 2 line
# imperialists = ['bbc', 1, 'cnn', 3, 'voa', 2]
# imperialists.sort()

Question What will be the output of the following code

x = [1,2]
y = [3,4, 5]
print(len(x + y))

*

a = ['Najaf']
print(a * 3)

['Najaf', 'Najaf', 'Najaf']

a = ['Najaf', '->', 'Karbala']
print(a * 3)

['Najaf', '->', 'Karbala', 'Najaf', '->', 'Karbala', 'Najaf', '->', 'Karbala']

Notes

I have been using the word function for append, sort etc. However, in Object-Oriented Programming, it can be seen that they are actually called methods.

There are a lot more powerful list manipulations which can be done by combining conditional and looping statements. We will come back to them once looping and conditioning statements are covered.