python data structures

Create data structer using formal name

formal_list = list()
formal_dict = dict()
formal_tuple = tuple()

Literal syntax

literal_list = []
literal_dict = {}
literal_tuple = ()

List

x = [5, 3.0, 10, 200.2]
x = ['JupytherNB', 75, None, True, [34, False], 2, 75] 
x[4] = 52
len(x)
y = [9, 0, 4, 2]
print(x + y) # to concatenate two lists, + operator is used
print(y * 3) # to concatenate multiple copies of the same list, * operator is used
z = [y, x]   # to nest lists to create another list

Slicing

use a colon to specify the start point (inclusive) and end point (noninclusive) of the sub-list

x[0:4] # the last element seen in the output is at index 3
x[:4]
x[4:]
x[-2:]
x[0:4:2] # steps of 2
x[4::-2] # start from the element at index 4 and go backward to the beginning with steps of 2

Modifing

x.append(-23)
x.remove(75)
y.sort() # to sort the element of y
x.insert(2, 10) # insert(pos, elmnt) method inserts the specified elmntat the specified position (pos) and shift the rest to the right
print(x.pop(3)) # pop(pos) method removes (and returns) the element at the specified position (pos)
del x[1] # delete an element from a list by its index
x.pop() # by default the position is -1, which means that it removes the last element

Coping

• https://docs.python.org/3/library/copy.html

Shallow copies

>>>list1 = ['A+', 'A', 'B', 'C+']
>>>list2 = list1
>>>list2.append('D')
>>>print(list2)
>>>print(list1)
['A+', 'A', 'B', 'C+', 'D']
['A+', 'A', 'B', 'C+', 'D']

Deep copy

>>>list3 = list1[:] # the use of slicing; that is, using [:] we make a deep copy of the entire list1
>>>list3.append('E')
>>>print(list3)
>>>print(list1)
['A+', 'A', 'B', 'C+', 'D', 'E']
['A+', 'A', 'B', 'C+', 'D']


list4 = list1.copy() # the use of copy() method
list4.append('E')
print(list4)
print(list1)

list5 = list(list1) #the use of list() constructor
list5.append('E')
print(list5)
print(list1)

List functions

• len(list)
• min(list)
• max(list)
• sum(list)
• sorted(list) # return new list sorted in ascending order
• reversed(list) # return reversed iterator
• list.count(element) # number of times element appear on list
• list.extend(iterable)
• list.clear() # remove elements from list

Tuples

tuple1 = ('Machine', 'Learning', 'with', 'Python', '1.0.0')
tuple1[0]
tuple1[::2]
len(tuple1)
tuple1 = ('Jupyter', 'NoteBook') # redefine tuple1
tuple2 = tuple1 + ('Good', 'Morning')

Imutable so can be use as key in dictionaries or elements in set because they are hashable (have unique value to identify them)

coordinates={(0,0):"Origin",(1,1):"Point A",(2,4):"Point B"}
print(coordinates[(0,0)]

Set

colors={("r","g","b"),("y","m","c"),("b","w")}
print(("b","w") in colors) # True

sequence packing

In Python, a sequence of comma separated objects without parentheses is packed into a tuple.

tuple1 = 'Machine', 'Learning', 'with', 'Python', '1.0.0' # sequence packing
x, y, z, v, w = tuple1 # the use of sequence unpacking
### unpacking the list
list6 = ['Machine', 'Learning', 'with', 'Python', '1.0.0']
x, y, z, v, w = list6

if we want to create a one-element tuple, the comma is required

tuple3 = 'Machine',
type(tuple3)

Dictionaries

dict1 = {1:'value for key 1', 'key for value 2':2, (1,0):True, False:[100,50], 2.5:'Hello'}
dict1['key for value 2'] # accessed item using the keys
dict1['key for value 2'] = 30 # change an element
dict1[10] = 'Bye' # add new item
del dict1['key for value 2']
(1,0) in dict1.keys()   # check the membership in keys
(1,0) in dict1 # equivalent to: in dict1.keys()
"Hello" in dict1.values() # check the membership in values
dict2 = dict([('Police', 102), ('Fire', 101), ('Gas', 104)]) # create dict from list of tuples pair
dict3 = dict(Country='USA', phone_numbers=dict2, population_million=18.7) # the use of keywords arguments = object

### Set

set1 = {'a', 'b', 'c', 'd', 'e'} set2 = {'b', 'b', 'c', 'f', 'g'} set1 | set2 # union using an operator set1.union(set2) set1 & set2 # intersection set1.intersection(set2) set1 - set2 # difference: elements of set1 not in set2 set1.difference(set2) set1 ^ set2 # symmetric difference: elements only in one set, not in both set1.symmetric_difference(set2) 'b' in set1 # check membership