Question

Hey guys I need help with this assignment. However it contains 7 sub-problems to solve but I figured only 4 of them can be solved in one post so I posted the other on another question so please check them out as well :)

Here is the questions in this assignment:

Note: Two helper functions Some of the testing codes for the functions in this assignment makes use of the print_dict in_key_order (a dict) function which prints dictionary keyvalue pairs in sorted key order. The testing code for function 3 in this assignment makes use of the remove less than 2(a dict) function which removes any keyvalue pairs which have a corresponding value of 1 from the dictionary. # Two helper functions def print_dict in_key order (a dict): all-keys = list(adict.keys()) all_keys.sort) for key in all keys: print (key,a dict [key) def remove less than 2 (a dict): all keys 1ist(a_dict.keys() for key in all keys: if a-dict[key] It del a dict[keyl «Assignment 4 Questions draw rows () Define the draw rows () function which is passed a Python dictionary as a parameter The keys of the dictionary are integers and the corresponding values are strings, e.g., (4: a', 2: f 0 x4: z'. For each key, followed by:followed by a string which is a repetition of the corresponding value. The number of times the corresponding value printed is given by the key value. The keys are printed in numeric order. Note that nothing is printed if the key is a number less than l key-value pair in the dictionary the function prints the For example, the following code: print(. print print(2." draw rows (4: a20x,423 2' prints: 2. 2: ff def draw rows (histogram dict): pass # 2222222222222222222222222222222222222222222222222 # get text-valuation() Define the get_text valuation() function which is passed two parameters, a dictionary and a string of text. The keys of the parameter dictionary are single letters and the corresponding values are integers (the value of the key letter), e.g., ('b': 5, 'a': 6, c': 3). The function returns the total valuation (an integer) of the string of text where: . if the letter from the text is a keyletter of the dictionary then its value is the integer corresponding to the letter in the dictionary " any alphabetic characters from the text which are not in the dictionary are worth 1, and, all non alphabetic characters are worth 0 (use the isalpha() method to check if a character is alphabetic or not). Notes: you will need to change the text to lowercase before you work out the total value of the text . you can assume that all the keys in the dictionary are lowercase characters. For example, the following code: 2, "s": 2, "h":4, "t":3, "m": 7, "g", "v":8) {"r": letter-value, dict letters"BLAH" print("1.", letters, "-", get_text_valuation(letter_value_dict, letters)) letters 'thought provoking print("2.", letters, "-*, get_text_valuation(letter_value_dict, letters)) letters"too much month at the end of the money" print("3.", letters, "-", get_text_valuation(letter_value_dict, letters)) prints: 1. BLAH 7 2: tbou mbeh rowth as the end of the money - 70 3. too much month at the end of the money def get_text_valuation(1etter_worth dict, text) return 0 # word-len-dict() get Define the get_word len dict() function which is passed a string of text as a parameter. The function returns a dictionary with keys which are integers and corresponding values which are lists of unique words. The list of words corresponding to a key contains all the unique words from the text that have a length equal to the key value. The corresponding lists of unique words should be in sorted alphabetical order. Note: the testing code makes use of the print_dict_in_key_order(a dict) which prints the dictionary pairs in sorted key order. For example, the following code: text "May your coffee be strong and your Monday be short" the dict -get_word_len_ dict (text) print_dict_in key_order(the dict) print() text why does someone believe you when you say there are four billion stars but they have to check when you say the paint is wet' the dict get_word_len_dict(text) print dict_in_key_order (the dict) prints: 2 ['be'] 3 I'May' 'and' 5 : ['short'1 6 : I'Monday, 'coffee, 'strong'] 2: 'is', 'to' 3 ['are', 'but', 'say,'the', 'wet', 'why, 4 : I'does','four', 'have, 'they, 'when' 1 5 : ['check', 'paint', 'stars', 'there'] 7: I'believe', 'billion', 'someone' 1 'why''you' def get_word_len_dict (text): return (th # 4444444444444444444444444444444444444444444444444 # get-triples-dict() Define the get triples diet) funetion which is passed a string of text as a The function first converts the parameter string to lower case turns a dictionary with keys which are all the unique consecutive tic characters from the text, and the corresponding values are characters appear in parameter. and then re three alphabe the number of times the three consecutive alphabetic text. Use the isalpha() method to check if a character is alphabetic or the not. The dictionary After your dietionary has been created and populated, you need to remove any key-value pairs which have a corresponding value of 1. For example, text is "Super, duper" the algorithm proceeds as follows: should only contain entries which occur more than once. Character 's' String is s,Dictionary is Character u': String is "su", Dictionary is ( Character 'p': String is "sup", change string to "up" Dictionary is ('sup': 1) Character e: string is "upe", change string to "pe", Dictionary is ('sup': 1, upe: 1) Character 'r': string is "per" change string t er Dictionary is ('sup 1, upe 11, per': 1) upe' 1, per CharacterString is er, Dict tonary is 'sup Character String is erDictionary is 'sup1, upe': 1, per: Character d': String is "erd", change string to rdDictionary is E'sup upe1, per'1, erd': 1 Character 'u':Stringis "rdu change string to "du Dictionary is Dictionary is Character 'p': String is "dup change string to "up Character 'e': String is "upe change string to pe" Dictionary is Character 'r: String is "per", change string to erDictionary is Remove all entries with a value of 1: Dictionary is ('upe': 2, per 2) For example, executing the following code: print("1.*) print dict in key order(get_triples dict( super, duper)) print("In2.") print dict-in key-order( get-triples, dict( "ABC ABC ABC- print( n3.) print dict in key order(get triples dict( "Sometimes the smallest things make more room in your heart")) print("An4.) print dict in key order(get triples dict("My favourite painting is the painting i did of my dog in that painting in my den)) 3. est -2 sma -2 prints: (output is shown here in four ain3 abc-3 bca -2 per -2 upe 2 gin 2 ing -3 int -4 myd 2 ngi - 3 nti3 cab -2 separate columns) tin 3 def get triples dict (text): return t)

Answer 1

EXPLANATION IN CODE COMMENTS :

def remove_less_than_2(dic):
all_keys = list(dic.keys())
for key in all_keys:
if dic[key]==1:
del dic[key]
return dic
def print_dic_in_key_order(dic):
#print dictionary content in sorted key order
for key in sorted(dic):
print(key ,":", dic[key])
print()
  
def draw_rows(dic):
#run two loops and print the value key times
for key in sorted(dic):
print(key ,":",end =" ")
for i in range(key):
print(dic[key],end ="")
print()
print()
  
def get_text_valuation(dic,text):
sumA=0
#loop over text , if character is alphabet and it exits in dictionary then add value to sum
#else add 1 to the sum
for i in text.lower():
if i.isalpha()==True:
if i in dic.keys():
sumA+=dic[i]
else:
sumA+=1
return sumA

"""
33333333333333333333333333333333333333333333333333

"""
def get_word_len_dict(text):
l = text.split(" ")
dic = {}
#for each word in the list, get its length
#if len exists as key in dictionary , append the word to it
#else create a new entry in the dictionary
for i in l:
key = len(i)
if key in dic.keys():
if i not in dic[key]:
dic[key].append(i)
else:
dic[key]=[i]
for value in dic.values():
value.sort()
return dic

"""
444444444444444444444444444444444444444444444444444

"""
def get_triples(text):
dic = {}
s=""
#loop over text as lower() , take string as s
#in each iteration add each character if it is alphabet to s
#when length of s reaches 3, if it exists in the dictionary , add one to it
#else create a new entry
for i in text.lower():
if i.isalpha() == True:
s += i
if len(s)==3:
if s in dic.keys():
dic[s]+=1
else:
dic[s]=1
s = s[1:]
#remove all the entries in the dictionary whose value is == 1
return remove_less_than_2(dic)

dic = {4:'a',7:'c',4:'b'}
draw_rows(dic)

dic = {"r":2,"s":2,"h":4,"t":3,"m":7,"g":4,"v":8}
letters = "BLAH"
print("1. ",letters," - ",get_text_valuation(dic,letters),"\n")

text = "May your coffee be strong and your monday be short"
dic = get_word_len_dict(text)
print_dic_in_key_order(dic)

text = "super , duper"
print_dic_in_key_order(get_triples(text))