Lecture 15: Dictionaries

Transcription

1 Lecture 15: Dictionaries CS1068+ Introductory Programming in Python Dr Kieran T. Herley 2018/19 Department of Computer Science University College Cork

2 Summary Python s dictionary concept. 1

3 Dictionaries

4 Dictionaries Dictionary values container to capture associations between keys and Examples Number-word associations to english = {1: one, 2: two, 3: three } Location-temperature associations temp = { athenry : 6.0, ballyhaise : 5.0, casement : 6.0} Person-dob associations dob = { aishling : 1/1/1995, bob : 2/2/1996 } 2

5 Dictionary basics Creating a dictionary temp = { athenry : 6.0, ballyhaise : 5.0, casement : 6.0} Terminology Contains key-value pairs called items e.g. ("casement", 6.0) Values (here 6.0) can be anything (even lists, functions etc.) Keys e.g. casement (almost) any comparable type 1 Keys in dictionary must be distinct 1 Mutable ( changeable ) objects e.g. lists, dictionaries cannot be used as keys. 3

6 Dictionary basics Suppose d = {1: "one", 2: "two", 3: "three"} Code Result Explanation len(d) 3 Return the number if itels in dictionary. d[2] two Return the value of item with key specified key. d[17] KeyError Error if key is not in the dictionary (raises ); Use test k in d to check first. k in d True Return True if key k is present in dictionary d, False otherwise. 4

7 Changing dictionary contents Suppose d = {1: "one", 2: "two", 3: "three"} Code d[2] = "zwei" Explanation Update (key present): Re-associate key 2 with value zwei. Dictionary now {1: "one", 2: "zwei", 3: "three"} 5

8 Changing dictionary contents Suppose d = {1: "one", 2: "two", 3: "three"} Code d[2] = "zwei" Explanation Update (key present): Re-associate key 2 with value zwei. Dictionary now {1: "one", 2: "zwei", 3: "three"} d[7] = "seven" Insert (key absent): Add fresh item with key 4 and value four. Dictionary now {1: "one", 2: "zwei", 3: "three", 7: "seven"} 5

9 Dictionaries vs Lists Issue Structure to keep track of CS6507 marks Bad approach : parallel arrays names = [ tom, dick, harry ] results = [55, 70, 35] Better approach : dictionary marks = { tom : 55, dick : 70, harry : 35} 6

10 Dictionary applications

11 Literary interlude: Wuthering Heights in Morse code Task: write a function to translate a passage of text into Morse code Idea: Use a dictionary to capture letters and their Morse equivalents Process text character by character and for each look up and print the appropriate code 7

12 Representing the cpde MORSE = {"a": ".-", "b": "-...", "c": "-.-.", "d": "--.", "e" : ".", "f" : "..-.",... } 8

13 Translating into Morse code def generate morse( plain ): for ch in plain : if ch. isalpha (): print(morse[ch.lower()]) MORSE =... text = I have just returned from a visit to my landlord the solitary neighbour that I shall be troubled with.... generate morse( text) 9

14 Dictionary traversals Can use for loop walk through dictionary contents (keys): d = {1: one, 2: two, 3: three } for k in d: print(k) yields keys (one per line, in some order). 10

15 Dictionary traversals Can use for loop walk through dictionary contents (keys): d = {1: one, 2: two, 3: three } for k in d: print(k) yields keys (one per line, in some order). Python provides no guarantee that keys are handles in increasing order. 10

16 Dictionary traversals cont d Can also get values or items. Code d.keys() d.values() d.items() Explanation Return a new view of the dictionarys keys. Default. Return a new view of the dictionarys values. Return a new view of the dictionarys items ((key, value) pairs). Note: A view is a list-like object. To translate into a list, use e.g. list(d.keys()) 11

17 Letter frequency analysis Task: Analyze a body of English-language text and print the percentage of letters that are As, Bs and so on. Idea: Keep letter count data in dictionary (key = letter, value = number of appearances) Process text character maintaining the count of letters in dictionary Use letter-count data to determine and print percentages 12

18 Outline def analyze( text ): letter count = 0 freq = {} for c in text : # process a character updating freq and letter count # (ignore non letters and case differences ) # Run through contents of freq to print percentages 13

19 Updating freq Loop body: if c. isalpha (): letter count += 1 c upper = c.upper() if c upper in freq : freq [c upper] += 1 else : freq [c upper] = 1 14

20 Notes c.upper() returns copy of c with all lowercase letters translated in upper c.isalpha() returns True is c consists entirely of letters x in d returns True if key x appears in dictionary d if character already represented, just add one to its count if not, create new dictionary entry (key = character, value = 1) 15

21 Aside Python supports a conditional expresion feature e.g. judgement = pass if mark >= 40 else fail The RHS evaluates to either pass or fail dependeing on whether condition is True or False Could re-express freq update as follows: freq [c upper] = freq[c upper] + 1 if c upper in freq else 1 16

22 Printing percentages first approach for c in freq.keys (): percent = freq[c] 100 / letter count print( '%s': %4.2f%% % (c, percent)) Potential snag: dictionary traversals don t guarantee order. 17

23 Printing percentages better approach Workaround: iterate over sorted keys! 18

24 Printing percentages better approach Workaround: iterate over sorted keys! for c in sorted(freq.keys ()): percent = freq[c] 100 / letter count print( '%s': %4.2f%% % (c, percent)) Note: sorted(lst) returns a copy of list lst but in increasing order. Will see this later. 18

25 More dictionary functions

26 Some dictionary operations k in d d.get(k, dft) d.pop(k, dft) d.popitem() d.copy() Return True if d has a key k, else False. Return the value for k if k is in the dictionary, else dft. If dft is not given, it defaults to None, so that this method never raises a KeyError. If key k is in the dictionary, remove it and return its value, else return dft. If dft is not given and k is not in the dictionary, a KeyError is raised. Remove and return an arbitrary (key, value) pair from the dictionary. Return a shallow copy of the dictionary. Link: docs.python.org/3.4/library/stdtypes.html 19

27 Another dictionary application

28 Birthday analysis Task: Analyze date of birth data for group of individuals and identify clusters (of two or more) who share the same birthday Idea: Use dictionary to keep track of dates and the individuals who have birthdays on that date (Keys = dates, values = list of names) 20

29 The data classlist = [ [ Aoife, 1/1/1990 ], [ Barra, 2/2/1990 ],... [ Turlough, 3/6/1991 ] ] Assume data presented as list of pairs of names and dates of birth (as stings) 21

30 A useful helper function def day of year (date): # Return the day month prefix of parameter 'date' return date [: date. rfind ( / )] Treat as black box for now. 22

31 def common birthdays(persons): # Analyze list ' persons' and output the names of all clusters # of individuals that share a common birthday. Assume each # element of 'persons' is a name & date of birth pair. 1. Add each person to list indicated by his/her dob 2. Sweep through days and for each value (list of names) of len at least two print the date and the names in the list 23

32 Populating days dictionary days = {} for p in persons: name = p[0] dob = p[1] birthday = day of year(dob) if birthday not in days: days[ birthday ] = [name] else : days[ birthday ]. append(name) 24

33 Printing birthday clusters for d in days: names = days[d] if len(names) > 1: print( %s: % d, end = ) for n in names: print( %s % n, end = ) print() 25

34 Back Material

35 Notes and Acknowledgements Reading Code Acknowledgements 26