List exercise Python
60 Questions to Test Your Knowledge of Python ListsCrush algorithm questions by mastering list fundamentalsGreekDataGuy Show
May 19, 2020·15 min read Ive been doing a lot of algorithm questions lately and discovered I dont understand lists as well as I should This is my compilation of 60 list questions Ive written to evaluate my own knowledge. I hope youll find it as useful as writing it has been for me. Questions 110:1. Check if a list contains an elementThe in operator will return True if a specific element is in a list. li = [1,2,3,'a','b','c']'a' in li#=> True 2. How to iterate over 2+ lists at the same timeYou can zip() lists and then iterate over the zip object. A zip object is an iterator of tuples. Below we iterate over 3 lists simultaneously and interpolate the values into a string. name = ['Snowball', 'Chewy', 'Bubbles', 'Gruff']animal = ['Cat', 'Dog', 'Fish', 'Goat'] age = [1, 2, 2, 6]z = zip(name, animal, age) z #=> print("%s the %s is %s" % (name, animal, age)) #=> Snowball the Cat is 1 #=> Chewy the Dog is 2 #=> Bubbles the Fish is 2 #=> Gruff the Goat is 6 3. When would you use a list vs dictionary?Lists and dictionary generally have slightly different use cases but there is some overlap. The general rule of algorithm questions Ive come to is that if you can use both, use a dictionary because lookups are faster. ListUse a list if you need to store the order of something. Ie: ids of database records in the order theyll be displayed. ids = [23,1,7,9]While both lists and dictionaries are ordered as of python 3.7, a list allows duplicate values while a dictionary doesnt allow duplicate keys. DictionaryUse a dictionary if you want to count occurrences of something. Like the number of pets in a home. pets = {'dogs':2,'cats':1,'fish':5}Each key can only exist once in a dictionary. Note that keys can also be other immutable data structures like tuples. Ie: {('a',1):1, ('b',2):1}. 4. Is a list mutable?Yes. Notice in the code below how the value associated with the same identifier in memory has not changed. x = [1]print(id(x),':',x) #=> 4501046920 : [1]x.append(5) x.extend([6,7]) print(id(x),':',x) #=> 4501046920 : [1, 5, 6, 7] 5. Does a list need to be homogeneous?No. Different types of object can be mixed together in a list. a = [1,'a',1.0,[]]a #=> [1, 'a', 1.0, []] 6. What is the difference between append and extend?.append() adds an object to the end of a list. a = [1,2,3]a.append(4) a #=> [1, 2, 3, 4] This also means appending a list adds that whole list as a single element, rather than appending each of its values. a.append([5,6])a #=> [1, 2, 3, 4, [5, 6]] .extend() adds each value from a 2nd list as its own element. So extending a list with another list combines their values. b = [1,2,3]b.extend([5,6]) b #=> [1, 2, 3, 5, 6] 7. Do python lists store values or pointers?Python lists dont store values themselves. They store pointers to values stored elsewhere in memory. This allows lists to be mutable. Here we initialize values 1 and 2, then create a list including the values 1 and 2. print( id(1) ) #=> 4438537632print( id(2) ) #=> 4438537664a = [1,2,3] print( id(a) ) #=> 4579953480print( id(a[0]) ) #=> 4438537632 print( id(a[1]) ) #=> 4438537664 Notice how the list has its own memory address. But 1 and 2 in the list point to the same place in memory as the 1 and 2 we previously defined. 8. What does del do?del removes an item from a list given its index. Here well remove the value at index 1. a = ['w', 'x', 'y', 'z']a #=> ['w', 'x', 'y', 'z']del a[1]a #=> ['w', 'y', 'z'] Notice how del does not return the removed element. 9. What is the difference between remove and pop?.remove() removes the first instance of a matching object. Below we remove the first b. a = ['a', 'a', 'b', 'b', 'c', 'c']a.remove('b') a #=> ['a', 'a', 'b', 'c', 'c'] .pop() removes an object by its index. The difference between pop and del is that pop returns the popped element. This allows using a list like a stack. a.pop(4) #=> 'c' a #=> ['a', 'a', 'b', 'b', 'c'] By default, pop removes the last element from a list if an index isnt specified. 10. Remove duplicates from a listIf youre not concerned about maintaining the order of a list, then converting to a set and back to a list will achieve this. li = [3, 2, 2, 1, 1, 1]list(set(li)) #=> [1, 2, 3] Questions 1120:11. Find the index of the 1st matching elementFor example, you want to find the first apple in a list of fruit. Use the .index() method. fruit = ['pear', 'orange', 'apple', 'grapefruit', 'apple', 'pear']fruit.index('apple') #=> 2 fruit.index('pear') #=> 0 12. Remove all elements from a listRather than creating a new empty list, we can clear the elements from an existing list with .clear(). fruit = ['pear', 'orange', 'apple']print( fruit ) #=> ['pear', 'orange', 'apple']print( id(fruit) ) #=> 4581174216fruit.clear()print( fruit ) #=> [] print( id(fruit) ) #=> 4581174216 Or with del. fruit = ['pear', 'orange', 'apple']print( fruit ) #=> ['pear', 'orange', 'apple']print( id(fruit) ) #=> 4581166792del fruit[:]print( fruit ) #=> [] print( id(fruit) ) #=> 4581166792 13. Iterate over both the values in a list and their indicesenumerate() adds a counter to the list passed as an argument. Below we iterate over the list and pass both value and index into string interpolation. grocery_list = ['flour','cheese','carrots']for idx,val in enumerate(grocery_list):print("%s: %s" % (idx, val)) #=> 0: flour #=> 1: cheese #=> 2: carrots 14. How to concatenate two listsThe + operator will concatenate 2 lists. one = ['a', 'b', 'c']two = [1, 2, 3]one + two #=> ['a', 'b', 'c', 1, 2, 3] 15. How to manipulate every element in a list with list comprehensionBelow we return a new list with 1 added to every element. li = [0,25,50,100][i+1 for i in li]#=> [1, 26, 51, 101] 16. Count the occurrence of a specific object in a listThe count() method returns the number of occurrences of a specific object. Below we return the number of times the string, fish exists in a list called pets. pets = ['dog','cat','fish','fish','cat']pets.count('fish') #=> 2 17. How to shallow copy a list?.copy() can be used to shallow copy a list. Below we create a shallow copy of round1, assign it to a new name, round2, and then remove the string sonny chiba. round1 = ['chuck norris', 'bruce lee', 'sonny chiba']round2 = round1.copy()round2.remove('sonny chiba')print(round1) #=> ['chuck norris', 'bruce lee', 'sonny chiba'] print(round2) #=> ['chuck norris', 'bruce lee'] 18. Why create a shallow copy of a list?Building off the previous example, modifying round2 will modify round1 if we dont create a shallow copy. round1 = ['chuck norris', 'bruce lee', 'sonny chiba']round2 = round1round2.remove('sonny chiba')print(round1) #=> ['chuck norris', 'bruce lee'] print(round2) #=> ['chuck norris', 'bruce lee'] Without a shallow copy, round1 and round2 are just names pointing to the same list in memory. Thats why it appears that changing the value of one changes the value of the other. 19. How to deep copy a list?For this we need to import the copy module, then call copy.deepcopy(). Below we create a deep copy of a list, round1 called round2, update a value in round2, then print both. In this case, round1 isnt affected. round1 = [['Arnold', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom'] ]import copy round2 = copy.deepcopy(round1)round2[0][0] = 'Jet Lee'print(round1) #=> [['Arnold', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom']]print(round2) #=> [['Jet Lee', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom']] Above we can see that changing the nested array in round2 did not update round1. 20. What is the difference between a deep copy and a shallow copy?Building off the previous example, creating a shallow copy and then modifying it would have affected the original list.. round1 = [['Arnold', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom'] ]import copy round2 = round1.copy()round2[0][0] = 'Jet Lee'print(round1) #=> [['Jet Lee', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom']]print(round2) #=> [['Jet Lee', 'Sylvester', 'Jean Claude'], ['Buttercup', 'Bubbles', 'Blossom']] Why does this happen? Creating a shallow copy does create a new object in memory, but its filled with the same references to existing objects that the previous list has. Creating a deep copy creates copies of the original objects and points to these new versions. So the new list is completely unaffected by changes to the old list and vice versa. Questions 21-30:21. What is the difference between a list and a tuple.Tuples cannot be updated after creation. Adding/removing/updating an existing tuple requires creating a new tuple. Lists can be modified after creation. Tuples often represent an object like a record loaded from a database where elements are of different datatypes. Lists are generally used to store an ordered sequence of a specific type of object (but not always). Both are sequences and allow duplicate values. 22. Return the length of a listlen() can returns the length of a list. li = ['a', 'b', 'c', 'd', 'e']len(li) #=> 5 But note it counts top level objects, so a nested list of several integers will only be counted as a single object. Below, li has a length of 2, not 5. li = [[1,2],[3,4,5]]len(li) #=> 2 23. What is the difference between a list and a set?While a list is ordered, a set is not. Thats why using set to find unique values in a list, like list( set([3, 3, 2, 1]) ) loses the order. While lists are often used to track order, sets are often used to track existence. Lists allow duplicates, but all values in a set are unique by definition. 24. How to check if an element is not in a list?For this we use the in operator, but prefix it with not. li = [1,2,3,4]5 not in li #=> True 4 not in li #=> False 25. Multiply every element in a list by 5 with the map function.map() allows iterating over a sequence and updating each value with another function. map() returns a map object but Ive wrapped it with a list comprehension so we can see the updated values. def multiply_5(val):return val * 5a = [10,20,30,40,50][val for val in map(multiply_5, a)] #=> [50, 100, 150, 200, 250] 26. Combine 2 lists into a list of tuples with the zip functionzip() combines multiple sequences into an iterator of tuples, where values at the same sequence index are combined in the same tuple. alphabet = ['a', 'b', 'c']integers = [1, 2, 3]list(zip(alphabet, integers)) 27. Insert a value at a specific index in an existing listThe insert() method takes an object to insert and the index to insert it at. li = ['a','b','c','d','e']li.insert(2, 'HERE')li #=> ['a', 'b', 'HERE', 'c', 'd', 'e'] Note that the element previously at the specified index is shifted to the right, not overwritten. 28. Subtract values in a list from the first element with the reduce functionreduce() needs to be imported from functools. Given a function, reduce iterates over a sequence and calls the function on every element. The output from the previous element is passed as an argument when calling the function on the next element. from functools import reducedef subtract(a,b):return a - bnumbers = [100,10,5,1,2,7,5]reduce(subtract, numbers) #=> 70 Above we subtracted 10, 5, 1, 2, 7 and 5 from 100. 29. Remove negative values from a list with the filter functionGiven a function, filter() will remove any elements from a sequence on which the function doesnt return True. Below we remove elements less than zero. def remove_negatives(x):return True if x >= 0 else False a = [-10, 27, 1000, -1, 0, -30][x for x in filter(remove_negatives, a)] #=> [27, 1000, 0] 30. Convert a list into a dictionary where list elements are keysFor this we can use a dictionary comprehension. li = ['The', 'quick', 'brown', 'fox', 'was', 'quick']d = {k:1 for k in li} d #=> {'The': 1, 'quick': 1, 'brown': 1, 'fox': 1, 'was': 1} Questions 31-40:31. Modify an existing list with a lambda functionLets take the previous map function we wrote and turn it into a one-liner with a lambda. a = [10,20,30,40,50]list(map(lambda val:val*5, a))#=> [50, 100, 150, 200, 250] I could have left it as a map object until I needed to iterate over it but I converted to a list to show the elements inside. 32. Remove elements in a list after a specific indexUsing the slice syntax, we can return a new list with only the elements up to a specific index. li = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10]li[:10]#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 33. Remove elements in a list before a specific indexThe slice syntax can also return a new list with the values after a specified index. li = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10]li[15:]#=> [16, 17, 18, 19, 10] 34. Remove elements in a list between 2 indicesOr between two indices. li = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10]li[12:17]#=> [13, 14, 15, 16, 17] 35. Return every 2nd element in a list between 2 indicesOr before/after/between indices at a specific interval. Here we return every 2nd value between the indices 10 and 16 using the slice syntax. li = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,10]li[10:16:2]#=> [11, 13, 15] 36. Sort a list of integers in ascending orderThe sort() method mutates a list into ascending order. li = [10,1,9,2,8,3,7,4,6,5]li.sort()li #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 37. Sort a list of integers in descending orderIts also possible to sort in descending order with sort() by adding the argument reverse=True. li = [10,1,9,2,8,3,7,4,6,5]li.sort(reverse=True)li #=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] 38. Filter even values out of a list with list comprehensionYou can add conditional logic inside a list comprehension to filter out values following a given pattern. Here we filter out values divisible by 2. li = [1,2,3,4,5,6,7,8,9,10][i for i in li if i % 2 != 0]#=> [1, 3, 5, 7, 9] 39. Count occurrences of each value in a listOne option is to iterate over a list and add counts to a dictionary. But the easiest option is to import the Counter class from collections and pass the list to it. from collections import Counterli = ['blue', 'pink', 'green', 'green', 'yellow', 'pink', 'orange']Counter(li)#=> Counter({'blue': 1, 'pink': 2, 'green': 2, 'yellow': 1, 'orange': 1}) 40. Get the first element from each nested list in a listA list comprehension is well suited for iterating over a list of other objects and grabbing an element from each nested object. li = [[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15]][i[0] for i in li]#=> [1, 4, 7, 10, 13] Questions 4150:41. What is the time complexity of insert, find and delete for a list?Insert is O(n). If an element is inserted at the beginning, all other elements must be shifted right. Find by index is O(1). But find by value is O(n) because elements need to be iterated over until the value is found. Delete is O(n). If an element is deleted at the beginning, all other elements must to be shifted left. 42. Combine elements in a list into a single string.This can be done with the join() function. li = ['The','quick','brown', 'fox', 'jumped', 'over', 'the', 'lazy', 'dog']' '.join(li) #=> 'The quick brown fox jumped over the lazy dog' 43. Whats the affect of multiplying a list by an integer?Multiplying a list by an integer is called multiple concatenation and has the same affect as concatenating a list to itself n-times. Below we multiple a list by 5. ['a','b'] * 5#=> ['a', 'b', 'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'] This is the same as. ['a','b'] + ['a','b'] + ['a','b'] + ['a','b'] + ['a','b']#=> ['a', 'b', 'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'] 44. Use the any function to return True if any value in a list is divisible by 2We can combine any() with a list comprehension to return True if any values in the returned list evaluate to True. Below the 1st list comprehension returns True because the list has a 2 in it, which is divisible by 2. li1 = [1,2,3]li2 = [1,3]any(i % 2 == 0 for i in li1) #=> True any(i % 2 == 0 for i in li2) #=> False 45. Use the all function to return True if all values in a list are negativeSimilar to the any() function, all() can also be used with a list comprehension to return True only if all values in the returned list are True. li1 = [2,3,4]li2 = [2,4]all(i % 2 == 0 for i in li1) #=> False all(i % 2 == 0 for i in li2) #=> True 46. Can you sort a list with None in it?You cannot sort a list with None in it because comparison operators (used by sort()) cant compare an integer with None. li = [10,1,9,2,8,3,7,4,6,None]li.sort()li #=> TypeError: '<' not supported between instances of 'NoneType' and 'int' 47. What kind of copy would the list constructor create from an existing list?The list constructor creates a shallow copy of a passed in list. That said, this is less pythonic than using .copy(). li1 = ['a','b']li2 = list(li1) li2.append('c')print(li1) #=> ['a', 'b'] print(li2) #=> ['a', 'b', 'c'] 48. Reverse the order of a listA list can be mutated into reverse order with the reverse() method. li = [1,2,3,4,5,6,7,8,9,10]li.reverse() li #=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] Note that this mutates the object instead of returning a new object. 49. What is the difference between reverse and reversed?reverse() reverses the list in place. reversed() returns an iterable of the list in reverse order. li = [1,2,3,4,5,6,7,8,9,10]list(reversed(li)) #=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] 50. What is the difference between sort and sorted?sort() modifies the list in place. sorted() returns a new list in reverse order. li = [10,1,9,2,8,3,7,4,6,5]li.sort() li #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]li = [10,1,9,2,8,3,7,4,6,5] sorted(li) #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] Questions 5160:51. Return the minimum value in a listThe min() function returns the minimum value in a list. li = [10,1,9,2,8,3,7,4,6,5]min(li) #=> 1 52. Return the maximum value in a listThe max() function returns the maximum value in a list. li = [10,1,9,2,8,3,7,4,6,5]max(li) #=> 10 53. Return the sum of values in a listThe sum() function returns the sum of all values in a list. li = [10,1,9,2,8,3,7,4,6,5]sum(li) #=> 55 54. Use a list as a stackYou can use append() and pop() to treat a list like a stack. Stacks function per LIFO (last in first out). stack = []stack.append('Jess')stack.append('Todd') stack.append('Yuan')print(stack) #=> ['Jess', 'Todd', 'Yuan']print(stack.pop()) #=> Yuanprint(stack) #=> ['Jess', 'Todd'] One benefit of a stack is that elements can be added and removed in O(1) time because the list does not need to be iterated over. 55. Find the intersection of 2 listsWe can do this by utilizing set() with an ampersand. li1 = [1,2,3]li2 = [2,3,4]set(li1) & set(li2) #=> {2, 3} 56. Find the difference between a set and another setWe cant subtract lists, but we can subtract sets. li1 = [1,2,3]li2 = [2,3,4]set(li1) - set(li2) #=> {1}set(li2) - set(li1) #=> {4} 57. Flatten a list of lists with a list comprehensionsUnlike Ruby, Python3 doesnt have an explicit flatten function. But we can use list comprehension to flatten a list of lists. li = [[1,2,3],[4,5,6]][i for x in li for i in x]#=> [1, 2, 3, 4, 5, 6] 58. Generate a list of every integer between 2 valuesWe can create a range between 2 values and then convert that to a list. list(range(5,10))#=> [5, 6, 7, 8, 9] 59. Combine 2 lists into a dictionaryUsing zip() and the list() constructor we can combine 2 lists into a dictionary where one list becomes the keys and the other list becomes the values. name = ['Snowball', 'Chewy', 'Bubbles', 'Gruff']animal = ['Cat', 'Dog', 'Fish', 'Goat']dict(zip(name, animal)) #=> {'Snowball': 'Cat', 'Chewy': 'Dog', 'Bubbles': 'Fish', 'Gruff': 'Goat'} 60. Reverse the order of a list using the slice syntaxWhile we can reverse a list with reverse() and reversed(), it can also be done with the slice syntax. This returns a new list by iterating back over the list from end to beginning. li = ['a','b',3,4]li[::-1] #=> [4, 3, 'b', 'a'] ConclusionAfter working through this I feel more prepared to tackle algorithm questions without second guessing specific list method. Removing constant googling can free up energy for higher order thinking, like getting a function to do what its supposed to, and reducing function complexity. Again, I hope you found this useful as well. You may be interested in my previous post on strings questions, or the weekly series on algorithm questions Ive started to publish. |
