Introduction to Computing Using Python Dictionaries Userdefined Indexes

Introduction to Computing Using Python Dictionaries User-defined Indexes

Introduction to Computing Using Python User-defined indexes and dictionaries Goal: a container of employee records indexed by employee SS# Problems: • the range of SS#s is huge • SS#s are not really integers >>> employee[987654321] ['Yu', 'Tsun'] >>> employee[864209753] ['Anna', 'Karenina'] >>> employee[100010010] ['Hans', 'Castorp'] Solution: the dictionary class dict key value '864 -20 -9753' ['Anna', 'Karenina'] '987 -65 -4321' ['Yu', 'Tsun'] '100 -01 -0010' ['Hans', 'Castorp'] A dictionary contains (key, value) pairs >>> employee = { '864 -20 -9753': ['Anna', 'Karenina'], '987 -65 -4321': ['Yu', 'Tsun'], '100 -01 -0010': ['Hans', 'Castorp']} >>> employee['987 -65 -4321'] ['Yu', 'Tsun'] >>> employee['864 -20 -9753'] ['Anna', 'Karenina'] A key can be used as an index to access the corresponding value

Introduction to Computing Using Python Properties of dictionaries Dictionaries are not ordered Dictionaries are mutable • new (key, value) pairs can be added • the value corresponding to a key can be modified The empty dictionary is {} Dictionary keys must be immutable >>> employee = { '864 -20 -9753': ['Anna', 'Karenina'], '987 -65 -4321': ['Yu', 'Tsun'], '100 -01 -0010': ['Hans', 'Castorp']} >>> employee {'100 -01 -0010': ['Hans', 'Castorp'], '864 -209753': ['Anna', 'Karenina'], '987 -65 -4321': ['Yu', 'Tsun']} >>> employee['123 -45 -6789'] = 'Holden Cafield' >>> employee {'100 -01 -0010': ['Hans', 'Castorp'], '864 -209753': ['Anna', 'Karenina'], '987 -65 -4321': ['Yu', 'Tsun'], '123 -45 -6789': 'Holden Cafield'} >>> employee['123 -45 -6789'] = 'Holden Caulfield' >>> employee {'100 -01 -0010': ['Hans', 'Castorp'], '864 -20>>> employee = {[1, 2]: 1, [2, 3]: 3} 9753': ['Anna', 'Karenina'], '987 -65 -4321': Traceback (most recent call last): ['Yu', 'Tsun'], '123 -45 -6789': 'Holden File "<pyshell#2>", line 1, in <module> Caulfield’} employee = {[1, 2]: 1, [2, 3]: 3} Type. Error: unhashable type: 'list'

Introduction to Computing Using Python Dictionary operators Class dict supports some of the same operators as class list >>> days = {'Mo': 1, 'Tu': 2, 'W': 3} >>> days['Mo'] 1 >>> days['Th'] = 5 >>> days {'Mo': 1, 'Tu': 2, 'Th': 5, 'W': 3} >>> days['Th'] = 4 >>> days {'Mo': 1, 'Tu': 2, 'Th': 4, 'W': 3} >>> 'Fr' in days False >>> len(days) 4 Class dict does not support all the operators that class list supports • + and * for example

Introduction to Computing Using Python Dictionary methods Operation Explanation d. items() Returns a view of the (key, value) pairs in d d. keys() Returns a view of the keys of d d. pop(key) Removes the (key, value) pair with key from d and returns the value d. update(d 2) Adds the (key, value) pairs of dictionary d 2 to d d. values() Returns a view of the values of d The containers returned by d. items(), d. keys(), and d. values() (called views) can be iterated over >>> days {'Mo': 1, 'Tu': 2, 'Th': 4, 'W': 3} >>> days. pop('Tu') 2 >>> days {'Mo': 1, 'Th': 4, 'W': 3} >>> days 2 = {'Tu': 2, 'Fr': 5} >>> days. update(days 2) >>> days {'Fr': 5, 'W': 3, 'Th': 4, 'Mo': 1, 'Tu': 2} >>> days. items() dict_items([('Fr', 5), ('W', 3), ('Th', 4), ('Mo', 1), ('Tu', 2)]) >>> days. keys() dict_keys(['Fr', 'W', 'Th', 'Mo', 'Tu']) >>> vals = days. values() >>> vals dict_values([5, 3, 4, 1, 2]) >>> for val in vals: print(val, end=' ') 5 3 4 1 2 >>>

Introduction to Computing Using Python Dictionary vs. multi-way if statement Uses of a dictionary: • • container with custom indexes alternative to the multi-way if statement def complete(abbreviation): 'returns day of the week corresponding to abbreviation' if abbreviation == 'Mo': return 'Monday' elif abbreviation == 'Tu': return 'Tuesday' elif. . . else: # abbreviation must be Su return 'Sunday' def complete(abbreviation): 'returns day of the week corresponding to abbreviation' days = {'Mo': 'Monday', 'Tu': 'Tuesday', 'We': 'Wednesday', 'Th': 'Thursday', 'Fr': 'Friday', 'Sa': 'Saturday', 'Su': 'Sunday'} return days[abbreviation]

Introduction to Computing Using Python Dictionary as a container of counters Uses of a dictionary: • • • container with custom indexes alternative to the multi-way if statement container of counters Problem: computing the number of occurrences of items in a list >>> grades = [95, 96, 100, 85, 90, 95, 100] >>> frequency(grades) {96: 1, 90: 1, 100: 3, 85: 1, 95: 3} >>> Solution: Iterate through the list and, for each grade, increment the counter corresponding to the grade. Problems: • impossible to create counters before seeing what’s in the list • how to store grade counters so a counter is accessible using the corresponding grade Solution: a dictionary mapping a grade (the key) to its counter (the value)

Introduction to Computing Using Python Dictionary as a container of counters Problem: computing the number of occurrences of items in a list >>> grades = [95, 96, 100, 85, 90, 95, 100] � counters � � � 95 96 100 85 90 1 2 3 1 1 def frequency(item. List): 'returns frequency of items in item. List' counters = {} for item in item. List: if item in counters: # increment item counters[item] += 1 else: # create item counters[item] = 1 return counters

Introduction to Computing Using Python Exercise Implement function wordcount() that takes as input a text—as a string— and prints the frequency of each word in the text; assume there is no punctuation in the text. def word. Count(text): 'prints frequency of each word in text' >>> text = 'all animals are equal but some animals are more equal than other' >>> word. List word. Count(text) = text. split() # split text into list of words all def num. Chars(filename): appears 1 time. animals appears 2 times. counters ={} the dictionary in of file counters 'returns number of# characters filename' somefor word appears 1 time. in word. List: equal infile appears 2 counters: times. if word= in # counter for word exists open(filename, 'r') but appears time. counters[word] += 1 content = 1 infile. read() other infile. close() appears 1 time. else: # counter for word doesn't exist are appears 2 times. counters[word] = 1 than appears 1 time. return len(content) morefor word appears 1 time. in counters: # print word counts >>> if counters[word] == 1: print('{: 8} appears {} time. '. format(word, counters[word])) else: print('{: 8} appears {} times. '. format(word, counters[word]))

Introduction to Computing Using Python Exercise Implement function lookup() that implements a phone book lookup application. Your function takes, as input, a dictionary representing a phone book, mappingtuples (containing the first and last name) to strings >>> phonebook = { (containing phone numbers) ('Anna', 'Karenina'): '(123)456 -78 -90', ('Yu', 'Tsun'): '(901)234 -56 -78', ('Hans', 'Castorp'): '(321)908 -76 -54'} def lookup(phonebook): >>> lookup(phonebook) '''implements interactive phone book service using the input Enter the first name: Anna phonebook dictionary''' Enter the last name: Karenina while True: (123)456 -78 -90 first = input('Enter the first name: ') Enter the first name: last = input('Enter the last name: ') person = (first, last) # construct the key if person in phonebook: # if key is in dictionary print(phonebook[person]) # print value else: # if key not in dictionary print('The name you entered is not known. ')

Dict Comprehensions in Python By Dr. Ziad Al-Sharif

Remember Comprehensions • Syntax expr for item in iterable if condition expr 1 if condition else expr 2 for item in iterable >>> grades = [95, 96, 100, 85, 90, 95, 100] >>> L = [ x+1 for x in grades if x<100] >>> L [96, 97, 86, 91, 96] >>> L = [ x+1 if x<100 else x for x in grades] >>> L [96, 97, 100, 86, 91, 96, 100] >>>

Remember Comprehensions • Syntax expr 1 if cond 1 else expr 2 for item in iterable if cond 2 >>> grades = [95, 96, 100, 85, 90, 95, 100] >>> L = [ x+2 if x<90 else x+1 for x in grades if x<100] >>> L [96, 97, 87, 96, 91, 96] >>>

Dict Comprehensions • Semantics >>> >>> {0: >>> d = dict() for i in range(4): d[i] = chr(65+i) d 'A', 1: 'B', 2: 'C', 3: 'D'} is semantically equivalent to: >>> dict([(i, chr(65+i)) for i in range(4)]) {0: 'A', 1: 'B', 2: 'C', 3: 'D'} >>> is semantically equivalent to: >>> {i : chr(65+i) for i in range(4)}

Examples >>> d 1 = {i: chr(65+i) for i in range(4)} >>> d 1 {0 : 'A', 1 : 'B', 2 : 'C', 3 : 'D'} >>> d 2 = {k+1: v for k, v in d 1. items()} >>> d 2 {1: 'A', 2: 'B', 3: 'C', 4: 'D'} >>> d 3 = {k: v. lower() for k, v in d 2. items()} {1: 'a', 2: 'b', 3: 'c', 4: 'd'} >>>

Examples >>> def invert(d): . . . return {v : k for k, v in d. items()}. . . >>> d = {0 : 'A', 1 : 'B', 2 : 'C', 3 : 'D'} >>> invert(d) {'A' : 0, 'B' : 1, 'C' : 2, 'D' : 3} >>> d = {(k, v): k+v for k in range(4) for v in range(4)} >>> d {(0, 0): (1, 0): (2, 0): (3, 0): } >>> 0, 1, 2, 3, (0, (1, (2, (3, 1): 1): 1, 2, 3, 4, (0, (1, (2, (3, 2): 2): 2, 3, 4, 5, (0, (1, (2, (3, 3): 3): 3, 4, 5, 6

Dict Comprehensions • PEP 202 introduces a syntactical extension to Python called the "list comprehension". • https: //www. python. org/dev/peps/pep-0202/ • PEP 274 proposes a similar syntactical extension called the "dictionary comprehension" or "dict comprehension" for short. • https: //www. python. org/dev/peps/pep-0274/ • You can use dict comprehensions in ways very similar to list comprehensions, • except that they produce Python dictionary objects instead of list objects.