Get familiar with lists

What is a list
Like a String, list also is sequence data type. It is an ordered set of values enclosed in square brackets []. Values in the list can be modified, i.e. it is mutable.

For accessing an element of the list, indexing is used.

Syntax:

listName[index] #(variable name is name of the list).

Index here, has to be an integer value- which can be positive or negative. Positive value of index means counting forward from beginning of the list and negative value means counting backward from end of the list. Remember the result of indexing a list is the value of type accessed from the list.

Let’s look at some example of simple list:

>>> L1 = [1, 2, 3, 4] # list of 4 integer elements.

>>> L2 = [“Delhi”, “Chennai”, “Mumbai”] #list of 3 string elements.

>>> L3 = [ ] # empty list i.e. list with no element

>>> L4 = [“abc”, 10, 20] # list with different types of elements

>>> L5 = [1, 2, [6, 7, 8], 3] # A list containing another list known as nested list

To change the value of element of list, we access the element & assign the new value.

Example:

>>> print L1 # let‟s get the values of list before change
>>> L1 [2] = 5
>>> print L1 # modified list
[1, 2, 5, 4]

NOTE: An Index Error appears, if you try and access element that does not exist in the list.

Creating a list

List can be created in many ways:

i) By enclosing elements in [ ], as we have done in above examples.

ii) Using other Lists

Example:

L5=L1 [:]

Here L5 is created as a copy of L1.

>>> print L5
L6 = L1 [0:2]
>>> print L6
# will create L6 having first two elements of L1.

iii) List comprehension:

Example:

>>> S= [x**2 for x in range (10)]
>>> print S
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

In mathematical terms, S can be defined as S = {x 2 for: x in (0.....9)}. So, we can say that list comprehension is short-hand for creating list.

iv) Using built-in object

L = list () will create an empty list.

L = list [(1, 2, 3, 4)]

A single new list is created every time, you execute [ ]. We have created many different lists each using [ ]. But if a list is assigned to another variable, a new list is not created.

Example:

A = []
B = A # Will also create one list mapped to both
C = []

You can try to check the address of A and B.

if id(A) == id(B):
  print 'yes'

# printing the address
print hex(id(A))
print hex(id(B))

Accessing an element of list

For accessing an element, we use index and we have already seen example doing so. To access an element of list containing another list, we use pair of index. Lets access elements of L5 list. Also a sub-list of list can be accessed using list slice.

List Slices

Slice operator works on list also. We know that a slice of a list is its sub-list. For creating a list slice, we use [n:m] operator.

>>> print L5 [0]
1
>>> print L5 [2]
[6, 7, 8]
>>> print L5 [2] [0]
6
>>> print L5 [2] [2]
8
>>> L1 [1:2]
[2]

list[n:m] will return the part of the list from n th element to m th element, including the first element but excluding the last element. So the resultant list will have m-n elements in it. Slices are treated as boundaries, and the result will contain all the elements between boundaries.

Its Syntax is:

seq = L [start: stop: step]

Where start, stop & step - all three are optional. If you omit first index, slice starts from 0 and omitting of stop will take it to end. Default value of step is 1.

Example

For list L2 containing [“Delhi”, “Chennai”, “Mumbai”]

>>> L2 [0:2]
[“Delhi”, “Chennai”]

>>> list = [10, 20, 30, 40, 50, 60]
>>> list [::2] # produce a list with every alternate element
[10, 30, 50]

>>> list [4:] # will produce a list containing all the elements from 5 th position till end
[50, 60]

>>> list [:3]
[10, 20, 30]
>>> list [:]
[10, 20, 30, 40, 50, 60]

>>> list [-1] # „-1‟ refers to last elements of list
60

Note: Since lists are mutable, it is often recommended to make a copy of it before performing operation that change a list.

Traversing a List

Let us visit each element (traverse the list) of the list to display them on screen. This can be done in many ways:

while loop

i = 0
while i < len [L1]:
    print L1 [i],
i + = 1

will produce following output

1 2 5 4

for loop

for i in L1:
    print i,

for i in range ( len (L1)):
    print L1 [i],

range() function is used to generate, indices from 0 to len -1; with each iteration i gets the index of next element and values of list are printed.

Note: for loop in empty list is never executed

Using enumerate

The enumerate() function adds a counter to an iterable. So for each element in cursor, a tuple is produced with (counter, element); the for loop binds that to row_number and row, respectively.

Example:

>>> elements = ('foo', 'bar', 'baz')
>>> for elem in elements:
...     print elem
... 
foo
bar
baz
>>> for count, elem in enumerate(elements):
...     print count, elem
... 
0 foo
1 bar
2 baz

By default, enumerate() starts counting at 0 but if you give it a second integer argument, it’ll start from that number instead:

Example:

>>> for count, elem in enumerate(elements, 42):
...     print count, elem
... 
42 foo
43 bar
44 baz

Refered from here

Appending in the list

Appending a list is adding more element(s) at the end of the list. To add new elements at the end of the list, Python provides a method append ( ).

Syntax is:

# List. append (item)

L1. append (70)

This will add 70 to the list at the end, so now 70 will be the 5 th element of the list, as it already have 4 elements.

Using append(), only one element at a time can be added. For adding more than one element, extend() method can be used, this can also be used to add elements of another list to the existing one.

Example:

>>> A = [100, 90, 80, 50]
>>> L1. extend (A)
>>> print L1
[1, 2, 5, 4, 70, 100, 90, 80, 50]

will add all the elements of list A at the end of the list L1.

>>>print A
[100, 90, 80, 50]

Remember: A remains unchanged.

Updating array elements

Updating an element of list is, accomplished by accessing the element & modifying its value in place. It is possible to modify a single element or a part of list. For first type, we use index to access single element and for second type, list slice is used. We have seen examples of updations of an element of list.

Lets update a slice.

Example1:

>>> L1 [1:2] = [10, 20]
>>> print L1
# will produce
[1, 10, 20, 4, 70, 100, 90, 80, 50]

Example2:

>>> A=[10, 20, 30, 40]
>>> A [1:4] = [100]
>>> print A
# will produce
[10, 100]

As lists are sequences, they support many operations of strings. For example, operator + & * results in concatenation & repetition of lists. Use of these operators generate a new list.

Example:

>>> a= L1+L2

# will produce a 3 rd list a containing elements from L1 & then L2. a will contain
[1, 10, 20, 4, 70, 100, 90, 80, 50, “Delhi”, “Chennai”, “Mumbai”]

>>> [1, 2, 3] + [4, 5, 6]
[1, 2, 3, 4, 5, 6]

>>> ['Hi!']* 3
['Hi!', 'Hi!', 'Hi!']

It is important to know that + operator in lists expects the same type of sequence on both the sides otherwise you get a type error.

If you want to concatenate a list and string, either you have to convert the list to string or string to list.

Example:

>>> str([11, 12]) + “34”
'[11, 12] 34'
>>> “[11,12]” + “34”
'[11, 12] 34'


>>> [11, 12] + list (“34”)
[11, 12, '3', '4']
>>> [11, 12] + [“3”, “4”]

Deleting Elements

It is possible to delete/remove element(s) from the list. There are many ways of doing:

(i) If index is known, we can use pop() or del

(ii) If the element is known, not the index, remove () can be used.

(iii) To remove more than one element, del() with list slice can be used.

(iv) Using assignment operator.

Let us study all the above methods in details:

Pop()

It removes the element from the specified index, and also return the element which was removed.

Example:

>>> L1 = [1, 2, 5, 4, 70, 10, 90, 80, 50]
>>> a= L1.pop (1) # here the element deleted will be returned to ‘a’
>>> print L1
[1, 5, 4, 70, 10, 90, 80, 50]
>>> print a
2

# If no index value is provided in pop ( ), then last element is deleted.
>>>L1.pop ( )
50

del removes the specified element from the list, but does not return the deleted value.

>>> del L1 [4]
>>> print L1
[1, 5, 4, 70, 90, 80]

remove ()

In case, we know the element to be deleted not the index, of the element, then remove() can be used.

>>> L1. remove (90) # will remove the value 90 from the list

>>> print L1
[1, 5, 4, 70, 80]

del () with slicing

Example:

>>> del L1 [2:4]
>>>print L1
[1, 5, 80]

will remove 2 nd and 3 rd element from the list. As we know that slice selects all the elements up to 2nd index but not the 2nd index element. So 4th element will remain in the list.

>>> L5 [1:2] = []

# Will delete the slice
>>>print L5
[1, [6, 7, 8], 3]

NOTE:

All the methods, modify the list, after deletions.

If an out of range index is provided with del () and pop (), the code will result in to run-time error.

del can be used with negative index value also.

Other functions & methods

insert ()

This method allows us to insert an element, at the given position specified by its index, and the remaining elements are shifted to accommodate the new element.

insert() requires two arguments- index value and item value.

syntax is:

list. insert (index, item)

Index specifies the position (starting from 0) where the element is to be inserted. Item is the element to be inserted in the list. Length of list changes after insert operation.

Example

>>> L1.insert (3,100)
>>>print L1
# will produce
[1, 5, 80, 100]

Note: If the index specified is greater then len (list) the object is inserted in the last and if index is less than zero, the object is inserted at the beginning.

>>> print len(L1)
4
>>> L1.insert (6, 29)
>>> L1.insert (-2, 46)
>>>print L1
# will produce
[46, 1, 5, 80, 100, 29]

reverse ()

This method can be used to reverse the elements of the list in place

syntax is:

list.reverse()

Method does not return anything as the reversed list is stored in the same variable.

Example

>>> L1.reverse()
>>> print L1
# will produce
[29, 100, 80, 5, 1, 46]

Following will also result into reversed list.

>>>L1 [: : -1]

As this slices the whole sequence with the step of -1 i.e. in reverse order.

sort()

For arranging elements in an order Python provides a method sort () and a function sorted (). sort () modifies the list in place and sorted () returns a new sorted list.

Syntax are:

sort ([cmp [, key [, reverse]]])

and

sorted (list [, cmp [, key [, reverse]]])

Parameters mentioned in [] are optional in both the cases. These parameters allow us to customize the function/method.

cmp, argument allow us to override the default way of comparing elements of list. By default, sort determines the order of elements by comparing the elements in the list against each other. To overside this, we can use a user defined function which should take two values and return -1 for less than, 0 for equal to and 1 for greater than.

Key argument is preferred over cmp as it produces list faster.

Example 1:

The parameter key is for specifying a function that transforms each element of list before comparison. We can use predefined functions or a user defined function here. If its user defined then, the function should take a single argument and return a key which can be used for sorting purpose.

Reverse parameter can have a boolean value which is used to specify the order of arranging the elements of list. Value True for reverse will arrange the elements of list in descending order and value False for reverse will arrange the elements in ascending order. Default value of this parameter is False.

sorted () function also behaves in similar manner except for it produce a new sorted list, so original is not changed. This function can also be used to sort any iterable collection. As sort() method does not create a new list so it can be little faster.

>>> L1.sort()
>>> print L1
# will produce
[1, 5, 29, 46, 80, 100]
>>> L2.sort ( )
>>> print L2
# will produce
['Chennai', 'Delhi', 'Mumbai']
>>> L2.sort (key=len)
# will produce
['Delhi', 'Mumbai', 'Chennai']

Here we have specified len() built in function, as key for sorting. So the list will get sorted by the length of the strings, i.e., from shorted to longest. sort will call len() function for each element of list and then these lengths will be used for arranging elements.

Example 2:

>>> L4.sort()
>>> print L4
# will produce
[10, 20, 30, 'abc']
>>>L4.sort (reverse = True)
['abc', 30, 20, 10]
>>> def compare (str):
	return len (str)
>>> L2.sort (key=compare)
>>> L2
['Delhi', 'Mumbai', 'Chennai]

Passing list to a function (List as arguments)

Arguments are passed by assignment. The rationale behind this is two fold:

The parameter passed in is actually a reference to an object (but the reference is passed by value).
Some data types are mutable, but others aren’t.

When a list is passed to the function, the function gets a reference to the list. So if the function makes any changes in the list, they will be reflected back in the list.

In a more generalized form,

If you pass a mutable object into a method, the method gets a reference to that same object and you can mutate it to your heart’s delight, but if you rebind the reference in the method, the outer scope will know nothing about it, and after you’re done, the outer reference will still point at the original object.
If you pass an immutable object to a method, you still can’t rebind the outer reference, and you can’t even mutate the object.

Example:

def try_to_change_list_contents(the_list):
    print('got', the_list)
    the_list.append('four')
    print('changed to', the_list)

outer_list = ['one', 'two', 'three']

print('before, outer_list =', outer_list)
try_to_change_list_contents(outer_list)
print('after, outer_list =', outer_list)

Here parameter outer_list and argument the_list are alias for same object. So any changes made in outer_list will be reflected to the_list as lists as mutable.

Output:

before, outer_list = ['one', 'two', 'three']
got ['one', 'two', 'three']
changed to ['one', 'two', 'three', 'four']
after, outer_list = ['one', 'two', 'three', 'four']

Note: Here, it becomes important to distinguish between the operations which modifies a list and operation which creates a new list. Operations which create a new list will not affect the original (argument) list.

Let’s look at some examples to see when we have different lists and when an alias is created.

>>> a = [2, 4, 6]
>>> b = a

will map b to a. To check whether two variables refer to same object (i.e. having same value), we can use is operator. So in our example:

>>> a is b
# will return "True"
>>> a = [2, 4, 6]
>>> b = [2, 4, 6]
>>> a is b
False

In first example, Python created one list, reference by a & b. So there are two references to the same object b. We can say that object [2, 4, 6] is aliased as it has more than one name, and since lists are mutable. So changes made using a will affect b.

>>> a [1] = 10
>>> print b
# will print
[2, 10, 6]

Thank you 👏

Amit Upadhyay