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week8tut.py

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+# q1
+# the following program takes in a users number
+# the number is set as a key and is value is the result of the key multiplied by itself
+dict_1 = {2: ''}
+num = dict_1.keys()
+num2 = num * num
+dict_1[2] = num2
+
+print(dict_1)
+
+
+
+
+
+

week8tut2.py

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+# the following program will combine to two dictionaries into one
+
+dict_a = {'a': 100, 'b': 200, 'c': 300}
+dict_b = {'a': 300, 'b': 200, 'd': 400}
+dict_c = {}
+
+for key in dict_a:
+    dict_c[key] = dict_a[key]
+
+
+for key in dict_b:
+    if key in dict_a.keys():
+        dict_c[key] += dict_b[key]
+    else:
+        dict_c[key] = dict_b[key]
+
+print(dict_c)
+
+

week8tut3.py

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+dict_b = {'1':'one', '2':'two', '3':'three', '4':'four', '5':'five', '6':'six', '7':'seven','8':'eight', '9':'nine', '0':'zer0'}
+
+num = input('enter a number\n')
+dict_keys = dict_b.keys()
+
+for char in num:
+    if char in dict_keys:
+        print(dict_b[char],end='')
+    else:
+        print('you must enter a number')

week8tut4.py

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+# the following program prompts a user to enter a country and its capital
+# takes a users input and puts it in a dictionary
+# and iterates this again
+
+# function checks if the user has entered yes or no, repeating the program
+def check_input(input):
+    if input == 'y' or input == 'n':
+        return  True
+    return False
+
+capitals = {}
+
+country_str = input('enter a country\n')
+country_cap = input('enter the capital of the country you just entered\n')
+capitals[country_str] = country_cap
+
+# here, the loop variable. holding the user decision is passed to the function
+loop = input('do you want to enter another country (y/n) : ')
+while(check_input(loop) == False):
+    loop = input('do ou want to enter another country(y/n) :')
+
+    if loop == 'n':
+        break
+
+# list is made to hold the country and capital
+value_key_list = []
+for key, val in capitals.items():
+    value_key_list.append((key,val))
+value_key_list.sort()
+
+for val, key in value_key_list:
+    print(key, ': ', val)
+
+

week8ut5.py

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+# this program checks if a sentence entered is a heterogram or not
+def heterogram(sentence):
+    sentence_list = list(sentence)
+
+    # ord function returns ascii version of character
+    alphabet = [ch for ch in sentence_list if(ord(ch) >= ord('a') and ord(ch) <= ord('z'))]
+    sentence_set = set(sentence_list)
+
+    if len(sentence_set) == len(alphabet):
+        print('yes, your word is a hectogram')
+    else:
+        print('no, your word is not a hectogram')
+
+
+print('enter a word that does not use a letter more than twice(hectogram)')