Python’s list type (part 4) — Operations on lists

5 min readApr 15, 2022

In part 4, we will learn

Regular operations on lists
Inplace operations on lists
Unpack a list

Operations that don’t change a list

These operations operate on lists and return some values, but don’t modify the contents of the lists.

First, initialze two list

x = [2, 7, -3, 11.5, True, 2]
y = ["A", 1, 58.5, 94, [12, 20]]print(x)
print(y)[2, 7, -3, 11.5, True, 2]
['A', 1, 58.5, 94, [12, 20]]

Count the number of elements

len(x)6

Check membership

# Check if 2 is in x
2 in xTrue# Check if 99 is in x
99 in xFalse

Count the number of occurrences of a value

# Count how many times True occurs
x.count(True)1# Count how many times 2 occurs
x.count(2)2# Count how many times 99 occurs
x.count(99)0

Concatenate 2 lists

# Elements of x then elements of y
x + y[2, 7, -3, 11.5, True, 2, 'A', 1, 58.5, 94, [12, 20]]# Elements of y then elements of x
y + x['A', 1, 58.5, 94, [12, 20], 2, 7, -3, 11.5, True, 2]# Note that this operation doesn't change x and y
print(x)
print(y)[2, 7, -3, 11.5, True, 2]
['A', 1, 58.5, 94, [12, 20]]# To save the result from a concatenation
# You need to assign it to some variable
z = x + y
z[2, 7, -3, 11.5, True, 2, 'A', 1, 58.5, 94, [12, 20]]

Replicate a list

# Replicate x 3 times
x * 3[2, 7, -3, 11.5, True, 2, 2, 7, -3, 11.5, True, 2, 2, 7, -3, 11.5, True, 2]# Similarly, x is unchanged
x[2, 7, -3, 11.5, True, 2]

Sort a list (if possible)

# We can sort x because 
# True can be treated as 1
sorted(x)[-3, True, 2, 2, 7, 11.5]# However, we cannot sort y
# Because its element are not compatible
# (cannot compare a string to a number or a list)
# Try it yourself to see the error# Sort values in descending order
sorted(x, reverse=True)[11.5, 7, 2, 2, True, -3]# Of course, x is unchanged
x[2, 7, -3, 11.5, True, 2]# To save the sorted values
# You need an assignment
z = sorted(x)
z[-3, True, 2, 2, 7, 11.5]

Get the index of the first occurence of a value

# Index of the first value 2
# returns 0 (first position)
x.index(2)0# Index of the first value True
# returns 4 (fifth position)
x.index(True)4# If the value is not in the list
# You will get an error
# Try x.index(99)

Operations that change a list

These operations make some changes to a list, and often return None. They are often called "inplace" operation (happen inplace and save changes to the list).

Note that inplace operations are only available for mutable data types (Ex: lists, dictionaries, sets). For immutable types (Ex: numbers, tuples, ranges, frozensets) the objects cannot change after created, thus will not have inplace methods.

First, initialize a list

x = [2, 7, -3, 11.5, 2, True]
print(x)
print(id(x))[2, 7, -3, 11.5, 2, True]
1893914337600

Reverse a list

# No assignment is needed
x.reverse()
x[True, 2, 11.5, -3, 7, 2]# If you mistakenly assign x = x.reverse()
# You will get None back, because x.reverse() reverses x
# saves the change, and returns None
# then you re assign this None back to x
# then it overwrites the reversed values
# Now try to use y = x.reverse() to avoid messing up x
y = x.reverse()
print(y)None# Since you call x.reverse() twice
# you get back the original order
x[2, 7, -3, 11.5, 2, True]

Sort a list

# Ascending sort
x.sort()
x[-3, True, 2, 2, 7, 11.5]# Descending sort
x.sort(reverse=True)
x[11.5, 7, 2, 2, True, -3]

Append an element to the end of the list

# Append 99
x.append(99)
x[11.5, 7, 2, 2, True, -3, 99]# Append a list ['A', 'B']
x.append(['A', 'B'])
x[11.5, 7, 2, 2, True, -3, 99, ['A', 'B']]

Extend a list by adding multiple elements to the end of the list

# Adding 55, 66 as two last elements 
x.extend([55, 66])
x[11.5, 7, 2, 2, True, -3, 99, ['A', 'B'], 55, 66]

Insert an element into a given position

# Insert 100 into the 2nd position
x.insert(1, 100)
x[11.5, 100, 7, 2, 2, True, -3, 99, ['A', 'B'], 55, 66]

Remove the first occurence of a value

# Initialize a another list
a = [1, 2, 4, 3, 4, 4]
a.remove(4)
a[1, 2, 3, 4, 4]

Delete an element by index

# Delete the second element
del x[1]
x[11.5, 7, 2, 2, True, -3, 99, ['A', 'B'], 55, 66]

Pop out the last value, and delete it from the list

# Last element of x is assigned to p
# and deleted from x
p = x.pop()# Let's check x and p
print(x)
print(p)[11.5, 7, 2, 2, True, -3, 99, ['A', 'B'], 55]
66# We can choose to pop at a given index
# For example, pop the first element
p = x.pop(0)
print(x)
print(p)[7, 2, 2, True, -3, 99, ['A', 'B'], 55]
11.5

Clear all element

# Empty the contents inside the object
# but still keeping the object
x.clear()
x[]# Confirm the object is the same
id(x)1893914337600

Suppose we use a list to store date data. For example

my_date = [2022, 4, 1]
my_date[2022, 4, 1]

Now, suppose we want to extract each of the elements of my_date into three variables y, m, d. One way to do it is through three assignment as follows.

y = my_date[0]
m = my_date[1]
d = my_date[2]

But there’s a faster way to do it.

# Unpack three elements into y, m, d
y, m, d = my_date# Confirm
print(y)
print(m)
print(d)2022
4
1

Sometimes, if we want to extract some elements and ignore the others, we can use _ for the unwanted variables so that we don't have to waste time thinking names for them.

# Get date, month only
_, m, d = my_date
print(m)
print(d)4
1# Get year only
y, _, _ = my_date
print(y)2022# Get year only, but faster
# Here, *_ means we don't care about the rest
# Just ignore all of them, regarless how many
y, *_ = my_date
print(y)2022

Unpack a list