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Python Python Collections (Retired) Dictionaries String Formatting with Dictionaries

Andrew Forte
Andrew Forte
1,296 Points
Posted by Andrew Forte
Andrew Forte
1,296 Points

Challenge: string.py - String Factory

Hi Treehouse,

Seems like I have found my kryptonite with the course. I understand the principles of the lists and dictionaries but when it comes to writing the code logic I am useless :)

I am stumped on this challenge and this is as far as I got. My apologies as I know that this is a poor attempt but I am not sure what to do next. I know that there is some form of a for loop that comes into play that iterates through the dicts list of dictionaries and add's the values to my new list using the string.

Anyone's help would be much appreciated!

Andrew

strings.py 
dicts = [
    {'name': 'Michelangelo',
     'food': 'PIZZA'},
    {'name': 'Garfield',
     'food': 'lasanga'},
    {'name': 'Walter',
     'food': 'pancakes'},
    {'name': 'Galactus',
     'food': 'worlds'}
]

string = "Hi, I'm {name} and I love to eat {food}!"

# creating a function named string_factory that accepts dicts and strings above
def string_factory(dicts, string):

  #creating a new list called new_list
  new_list = []

1 Answer

Fenimore Cooper
Fenimore Cooper
3,591 Points
Fenimore Cooper
Fenimore Cooper
3,591 Points

Hi Andrew,

The question asks that you convert data from a dictionary format into a list -- which isn't too hard -- but dictionaries can be tricky to get the hang of at first. If I were you I'd mess around with dictionaries on your own computer -- I find its a lot easier to debug problems when you can see what your program is returning -- which you can't on the 'work space'.

Iterating over dictionaries is similar to iterating over lists -- except each iteration is a new dictionary, rather than a new list item. In our example, this is our first tiny dictionary: {'name': 'Michelangelo', 'food': 'PIZZA'}. So, if you do a for loop over our list of dictionaries -- the name of the key will act as, well... a key to the stored value. In the case of the code below, entry['name'] gets you each name i.e. 'Michelangelo', and entry['food'] gets each food item i.e. 'PIZZA.' The rest is just a sting substitution &c.

Hope that helps.

def string_factory(dicts, string):
    holding_list = []
    for entry in dicts:
        holding_list.append(string.format(name=entry['name'], food=entry['food']))
    return holding_list
Andrew Forte
Andrew Forte
1,296 Points
Andrew Forte
Andrew Forte
1,296 Points

Hi Fenimore,

Thank you for this, its great! Much appreciated for helping me.

I am certainly going to take your advice and mess around with both lists and dictionaries on my computer. As you say much easier to debug vs using the workspace. I think I just need to play more until the penny drops :)

Thanks again, I really appreciate it!

Thanks,

Andrew

Chris Freeman
Chris Freeman
Treehouse Moderator 68,457 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,457 Points

There is a shortcut available when you want to turn a dict into a list of keyword arguments.

If in the first iteration, entry were {'name': 'Michelangelo', 'food': 'PIZZA'}, then

**entry is equivalent to name='Michelangelo', food='PIZZA'

This means the loop statement can be written as:

holding_list.append(string.format(**entry))