Question

Challenge is to improve a previous challenge, “Who’s Your Daddy” (see my successful code: http://pastebin.com/AU2aRWHk) by adding a choice that let’s the user enter a name and get back a grandfather. Program should still only use one dictionary of son-father pairs.

I cannot get this to work. My entire code so far can be seen at: http://pastebin.com/33KrEMhT

I’ve obviously made this WAY more difficult than it need to be and am now trapped in a world of complexity. Here’s code I’ve F’d up:

# create dictionary
paternal_pairs ={"a": "b",
                 "b": "c",
                 "c": "d"}

# initialize variables
choice = None

# program's user interface
while choice != 0:
print(
"""
   Who's Yo Daddy:

   2 - Look Up Grandfather of a Son
   """
)

choice = input("What would you like to do?: ")
print() 

    # look up grandfather of a son
    if choice == "2":
        son = input("What is the son's name?: ")
        # verify input exists in dictionary
        if son in paternal_pairs.values():
            for dad, boy in paternal_pairs.items():
                if dad == son:
                    temp_son = dad
                    for ol_man, kid in paternal_pairs.items():
                        if temp_son == kid:
                            print("\nThe grandfather of", son, "is", ol_man)
                        else:
                            print("\nNo grandfather listed for", son)
                else:
                    print("\nNo grandfather listed for", son)
        # if input does not exist in dictionary:
        else:
            print("Sorry, that son is not listed. Try adding a father-son pair.")

After choosing “2”, my output:

What is the son's name?: d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

No grandfather listed for d

Obviously temporarily trapped in a small loop and it doesn’t work. All other code works as expected. Help!

Answer 1

You loop over each entry in the dict, and match the value, and if it doesn’t match, then for each key-value pair you print that it doesn’t match.

It is the equivalent of the following simplified loop:

>>> for i in range(3):
...     if i == 5:
...         print(i)
...     else:
...         print('Not 5')
... 
Not 5
Not 5
Not 5

Use the else: clause of the for loop instead, it’ll only be invoked if you completed going through all the values; use a break if you find a match:

for ol_man, kid in paternal_pairs.items():
    if temp_son == kid:
        print("\nThe grandfather of", son, "is", ol_man)
        break
else:
    print("\nNo grandfather listed for", son)

A small demonstration of how the else: clause works when used with a for loop:

>>> for i in range(3):
...     if i == 1:
...         print(i)
...         break
... else:
...     print('Through')
... 
1
>>> for i in range(3):
...     if i == 5:
...         print(i)
...         break
... else:
...     print('Through')
... 
Through

In the first example, we broke out of the loop with a break, but in the second example we never reached the break statement (i never was equal to 5) so the else: clause is reached and Through is printed.