CODEWITHSUNDEEP
python
user13273864
user13273864

Reputation:

Count the number of values in a dictionary?

I created a dictionary and want to create a function called count_type that will take that dictionary and return another dictionary that is called count. This function takes in a menu (dictionary) and returns a count (dictionary) that maps each item type to the count of how many items of that type exist in the menu. The new dictionary starts looking like:

count = {"Entrees": 0, "Salads": 0, "Sides": 0,
         "Kid's Meals": 0, "Desserts": 0, "Drinks": 0}

I want the end result to be the same dictionary but the 0's to be changed to the count of how many times there is a key:value with the value 'Entrees' etc in the original dictionary. an example of what I would like the output to look like: 

count = {"Entrees": 3, "Salads": 15, "Sides": 3,
         "Kid's Meals": 6, "Desserts": 4, "Drinks": 5}

So far I have the code:

def count_type(carte):

    count = {"Entrees": 0, "Salads": 0, "Sides": 0,
             "Kid's Meals": 0, "Desserts": 0, "Drinks": 0}
    menu = read_file("menu1.csv")
    ent = 0
    salad = 0
    side = 0
    kid = 0
    dessert = 0
    drink = 0
    for value in menu:
        if value == 'Entrees':
            ent += 1
        elif value =='Salads':
            salad += 1
        elif value =='Sides':
            side +=1
        elif value== "Kid's Meals":
            kid +=1
        elif value =='Desserts':
            dessert +=1
        else:
            drink +=1

This loop only gives me a count for drink of 46 which is all the values. How do I do this?

The read_file function that I have created already is:

def read_file(filename):

    file = open("menu1.csv", "r", encoding='utf-8-sig')
    file.readline()
    menu = {}
    for line in file:
        line = line.rstrip('\n')
        line = line.split(',')
        item = line[0]
        price = float(line[1])
        cal = int(line[2])
        typ = str(line[3])
        lst= (price, cal, typ)
        tup = tuple(lst)
        menu[item] = tup
    return menu

print(menu) returns a long dictionary:

{'Chic-fil-A Chicken Sandwich': (3.75, 440, 'Entrees'),
 'Chic-fil-A Deluxe Sandwich': (4.45, 500, 'Entrees'),
 'Spicy Chicken Sandwich': (3.99, 460, 'Entrees'),
 'Spicy Deluxe Sandwich': (4.69, 550, 'Entrees'),
 'Grilled Chicken Sandwich': (5.15, 320, 'Entrees'),
 'Grilled Chicken Club': (6.55, 460, 'Entrees')

Where some are 'Entrees', 'Salads', etc.

Upvotes: 0

Views: 189

Answers (3)

Blckknght
Blckknght

Reputation: 104852

Your code doesn't work because you're iterating over the keys of your menu dictionary, which is the names of the items (e.g. 'Spicy Chicken Sandwich'), but you're expecting to have an item type (e.g. "Entrees").

To fix this you need to change your loop. I'd suggest:

for _, _, value in menu.values():
    if value == 'Entrees':
        ...

I'd note that the name value is not very clear what kind of value it is, especially when you're getting a bunch of other data from the dictionary (and throwing them into the _ variable that will be ignored). Unpacking into more clearly named variables, like price, cal, typ again, might make the code clearer.

It might also be a good idea to match 'Drinks' explicitly, rather than letting any invalid types fall in there. You could have your code raise an exception if it gets an unknown type instead:

elif value == 'Drinks':
    drinks += 1
else:
    raise ValueError("Unknown menu item type: {}".format(typ))

Upvotes: 1

Sri
Sri

Reputation: 2328

In your for loop

for value in menu: instead of checking against value, check against the following

for value in menu:
    count[menu[value][2]] += 1

and as mentioned above, instead of creating seperate counters for each variables, you can make use of the already created count dictionary by using

count["Entrees"] += 1

Upvotes: 0

Phuong-Uyen Phan-Thi
Phuong-Uyen Phan-Thi

Reputation: 66

I recommend you try: count[value] += 1 in for value in menu loop

Update: you should edit your loop

for k,v in menu.items():
    count[v[2]] += 1

Upvotes: 1

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