Lists, tuples and dictionaries

Question 1

How do you create a new list or tuple. Give examples.

Two methods for each structure

Answer 1

my_list = [1,2,3]
my_list2 = list([1,2,3])

my_tuple = (1,2,3)
my_tuple2 = tuple([1,2,3])

Question 2

strings are represented as lists or tuples?

Answer 2

tuples (strings are unmutable objects)

#---------------example ---------------
   s = 'felipe'
   s[0] = 'F'      #This is not possible

Question 3

ACCESING LIST AND TUPLE ELEMENTS

How do you acces the first element in a list or tuple? What about the last one?

Answer 3

my_list = list(range(10))
my_list[0] #first element
my_list[-1] #last element

Question 4

How can you tell if some_value IS part of a list or tuple?

How can you tell if some_velue IS NOT part of a list or tuple?

Answer 4

#IS part of
som_value in my_list      #returns boolean

#IS NOT part of
some_value not in my_list      #returns boolean

Question 5

this_dict = {‘k1’ : v1, ‘k2’ : v2}

(a) How do you retrieve the value corresponding to the item with key = ‘k2’?
(b) How can you check if exists an item with key = ‘some_key’?
(c) How can you check if axists an item with value = some_value?

Answer 5

(a) this_dict[‘k2’]
(b) ‘some_key’ in this_dict.keys() # returns boolean
(c) ‘some_value’ in this_dict.values() # returns boolean

Question 6

How do you make a for loop over keys, values and items of a dictionary. Give an example.

( Consider d = {‘k1’ : v1, ‘k2’ : v2, ‘k3’ : v3} )

Answer 6

iterate over items

To iterate over keys and values use d.keys(), and d.values() methods.

example: for k in d.keys():

for k, v in d:
statements

Question 7

SLICING

Consider the list a = list( range(100) )

(a) get the first 20 elements
(b) get the last 20 elements
(c) get the elements from index 50 to the last one
(d) get the elements in even indexes

Answer 7

(a) ———————
a[:20]
a[0:20]

(b) ———————
a[-20:0]
a[-20:-1]

# (c) ---------------------         
a[50:]                               

# (d) --------------------- 
 a[::2]

Question 8

RANGE

Indicate the statement to produce a range with the next characteristics:

(b) start = 0, n = 100
(a) 5, … , 20
(c) multiples of three (start = 3; end 99)

Answer 8

# (a) --------------------- 
range(101)

# (b) --------------------- 
range(5, 21)

# (c) --------------------- 
range(0:100:3)

Question 9

ADDING NEW ELEMEMTS

How do you add new elements to a list, tuple, and dictionary.

(Can yu add more than one element at once?)

Answer 9

lists:

my_list.append(new_element)
my_list.extend([other_list]) #add a list of elements
my_list.insert(i, element)

#tuples:
tricky question. tuples are unmutable objects, which means you can not add new elements.

#dictionaries
some_dict[new_key] = new_value

di_2 = {new_key, new_value}
some_dict.update(di_2);

Question 10

.pop( )

How is the pop method used in lists, tuples and dictionaries?

Answer 10

#lists:
some_list.pop( )       # returns the last element of the list
some_list.pop(idx)   # returns the element in index = idx

#tuples:
tricky question. tuples are unmutable objects, therefore they do not have a poo method

#dictionaries
some_dict.popitem()           # returns the last inserted item
some_dicti.pop('some_key')   # returns item with key = 'some_key'

REMEMBER: The pop method deletes the corresponding element/item from the list or dictionarie.

Question 11

REMOVING ELEMENTS

Cosider the following list: some_list = [1, 2, 2, 2, 3, 2, 4, 5]

How can you remove all the ocurrences of the number 2?

Answer 11

some_list = [v for v in some_list if v != 2]

Question 12

LIST COMPRENHENSION

(a) Make a list containing the numbers between 0 and 20 which are not multiples of 3.
(b) Make a list with the first 10 cubes (i.e. 1, 8, 27, …)

Answer 12

(a) some_list = [n for n in range(21) if n%3 != 0]

(b) other_list = [ n**3 for n in range(1:11:1) ]

Question 13

FINDING ALL THE OCURRENCES

Cosider the following list of numbers:

collection = [1, 2, 2, 2, 3, 2, 4, 5]

Use the method enumerate() to find the index of all the ocurrences of number 2.

(Give a long and a short solution.)

Answer 13

collection = [1,2,2,2,3,2,4]

#long version
def find_ocurrences(some_list, element):
    index_list = []
    for i, value in enumerate(collection):
        if value == 2:
            index_list.append(i)

return index_list

print(find_ocurrences(collection, 2))

#short version
ocurrences = [i for i, v in enumerate if v == 2]

Question 14

Sorted() and Reversed()

Cosider the following list of numbers: collection = [3, 4, 2, 1, 8, 4, 5, 11, 3]

Make a list containig the same elements as collection[], but arrenged from max to min. The original list must remain unchanged.

Answer 14

collection = [3, 4, 2, 1, 8, 4, 5, 11, 3]

#option 1
new_list = sorted(collection, reverse = True)

#option 2
new_list = list(reversed(sorted(collection)))

Question 15

ZIP

Consider the next data:

             JAN      FEB     MAR SALES      100       80       120 COST       70         40       70

sales = [100, 80, 120]
cost = [70, 40, 70]

Use zip to calculate the margin at every month. Try using list conprenhension.

Answer 15

method 1

sales = (100, 80, 120)
cost = (70, 40, 70)

margin1 = []
for s, c in zip(sales, cost):
margin1.append(s-c)

#method 2
margin2 = [s-c for s,c in zip(sales, cost)]

Question 16

Dictionaries are also known as ——- or ——-

Answer 16

hash maps or associative arrays

Question 17

There are some students standing in a row, each of them has some number written on their back. The students are about to divide into two teams by counting off by twos: those standing at the even positions (0-based) will go to team A, and those standing at the odd position will join the team B.

Calculate the difference between the sums of numbers written on the backs of the students that will join team A, and those written on the backs of the students that will join team B.

Answer 17

def difference(list):
    A_sum = sum(list[::2])
    B_sum = sum(list[1::2])
    return  A_sum - B_sum

Question 18

toDo = [‘t1’, ‘t2’, ‘t3’, ‘t4’, ‘t5’, ‘t6’, ‘t7’, ‘t8’, ‘t9’, ‘t10’]

Given the list of task ids in your toDo list, remove each kth task from it and return the list of remaining tasks.

(a) Try using del
(b) Try using enumerate

Answer 18

#(a)
del toDo[k-1::k]

#(b)
toDo =  [t for n, t in enumerate(toDo, start = 1) if n%k != 0]

Question 19

Multiple level sorting

Consider a list of tuples with students info; each tuple contains the name, age, and grade (‘A’, ‘B’, ‘C’) of the student (i.e. student = (‘John’ 12 ‘B’).

Sort the students by name and by age.

Consider the list

student_tuples = [ (‘john’, ‘B’, 15), (‘jane’, ‘B’, 12), (‘dave’, ‘B’, 10), (‘john’, ‘A’, 16) ]

Answer 19

import operator

student_tuples = [ (‘john’, ‘A’, 15), (‘jane’, ‘B’, 12), (‘dave’, ‘B’, 10) ]
sorted(student_tuples, key= operator.itemgetter(1,2))

Question 20

DICT CONSTRUCTION

Use zip to construct a dictionary like the next one:

alphabet = {‘a’ : 1, ‘b’ : 2, …, ‘z’ : 26}

Answer 20

#option 1:
alphabet = dict(zip('abcdefghijklmnopqrstuvwxyz', range(1,27))

#option 2:
alphabet = dict(zip(list(string.ascii_lowercase), range(1,27))

Question 21

SET COMPRENHENSION

A pair of numbers is considered to be cool if their product is divisible by their sum. More formally, a pair (i, j) is cool if and only if (i * j) % (i + j) = 0.

Given two lists a and b, find cool pairs with the first number in the pair from a, and the second one from b.

a = [4, 5, 6, 7, 8]
b = [8, 9, 10, 11, 12]

Answer 21

a = [4, 5, 6, 7, 8]
b = [8, 9, 10, 11, 12]

cool_pairs = {(i,j) for i in a for j in b if (i*j)%(i+j) == 0}

Question 22

LIST COMPRENHENSION

Construct a multiplication table. Given an integer n, return the multiplication table of size n × n.

Example, n = 3:

multiplicationTable(n) = [[1, 2, 3],
[2, 4, 6],
[3, 6, 9,] ]

Answer 22

m_table = [ [i * j for i in range(1, n+1)] for j in range(1, n+1)]

Question 23

DEFINING A NEW DICTIONARY

Give examples for the next three dictionary creation methods:

1- Typing items in batch
2- One item at a time
3- Using zip

Answer 23

# Typing items in batch
d1 = {'k1' : 1, 'k2' : 2, 'k3' : 3}

# one item at a time
d2 = {}
d2['k1'] = 1
d2['k2'] = 2
d2['k3'] = 3

using zip
keys = (‘k1’, ‘k2’, ‘k3’)
values = (1, 2, 3)

d3 = dict(zip(keys, values))

d1, d2, d3

Question 24

DICTIONARY COMPRENHENSION

Use dict comprenhension to create the next dictionaries:

d1 = {1:1, 2:2, 3:3}
d2 = {1:0, 2:0, 3:0}

Answer 24

d1 = {k:v for k,v in zip(range(1:4), range(1,4))}
d2 = {k:0 for k in range(1:4)}

Question 25

DELETING ELEMENTS FROM LIST

Considers the next piece of code:

some_list = [1, 2, 2, 2, 3, 4, 2, 5]
some_list.remove(2)

What is the content of some list after executing this code?

(a) some_list = [1, 3, 4, 5]
(b) some_list = [1, 2, 2, 3, 4, 2, 5]

Answer 25

The remove method only deletes the first ocurrence of the corresponding element, therefore the answer is (b).

some_list = [1, 2, 2, 3, 4, 2, 5]

(Can you tell how to remove all of the ocurrences of some_element in the list?)

Question 26

UPDATING LIST

some_list = [1, 2, 3, 4, 5]

How can you delete the elements 2, 3, 4 of the list.

Answer 26

del some_list[1:4]

Question 27

COPYING A LIST

some_list = [1, 2, 3, 4, 5]

How can you copy the content of some_list[ ] to a new_list[ ]?

Answer 27

# option 1
new_list = some_list.copy()

# option 2
new_list = some_list[:]

Question 28

COUNTING OCURRENCES OF ELEMENTS

numbers = [0, 1, 1, 2, 0, 1, 3, 4, 5, 2, 3, 6, 9, 8, 7, 1, 3]

Create a dictionary ocurrences{ } que indique el número de veces que aparece cada elemento en números[ ].

Answer 28

numbers = [0, 1, 1, 2, 0, 1, 3, 4, 5, 2, 3, 6, 9, 8, 7, 1, 3]

num_set = set(numbers)
ocurrences = {n:numbers.count(n) for n in num_set}

Question 29

MEASURING SIZE

How can you tell how many elements or items are there in a list, tuple or dictionary?

Answer 29

len(some_structure)

Question 30

UNPACKING

parameters = (10, 100 , “cm”)

How can you assing the values stored in parameters to the variables _a, _b and _units?

Answer 30

parameters = (10, 100 , “cm”)

a, b, units = parameters