#
# Dictionaries 
#
# Practice 
#

# Preliminary notes: 
- Based on Chapter 9 
- Important! 
- Use only studied material 




# Q. 1
# Given is a dictionary in which each key 
# is a product code and each value is a list 
# of the prices of product in 5 stores 
# (the stores are the same for all products, 
# and the information about their prices is 
# arranged in the same order). 
# NA means unavailable data.
# Write a program that recieves an index of 
# a store and cacculates the average price 
# of all products in that store. The calculation 
# does not take the NAs into account.  

products = {'A': [2, 3, 1, "NA", 2],
            'B': [3, "NA", 4, 2, 4],
            'C': [1, 2, 3, "NA", 3],
            'D': [4, 1, 2, 1, 2]}
ind = 3
prices = [] 
for v in products.values():  
    if v[ind] != "NA": 
        prices.append(v[ind])
print(sum(prices) / len(prices))



# Q. 2
# Given are two dictionaries of prices 
# of three products sold in two different 
# stores. Produce a dictionary holding 
# the price difrerences for each product. 

store1 = {'a': 5, 'b': 7, 'c': 2}
store2 = {'b': 5, 'c': 7, 'a': 2}
diffs = {} 
for k in store1.keys(): 
    diffs[ k ] = store1[k] - store2[k]
print(diffs)











# Q. 3
# You have a dictionary of all cheapest products 
# in each of the three stores in the neighborhood. 
# Write a program that recevies, one after another, 
# a name of a store and a list of products sold in 
# that store and their prices, having this structure:
# [['a', 7],['b',9],['c',2]]
# The iteration stops when the string q is entered. 
# The program updates the dictionary. If the name of 
# store already exists in the dictionary, the program also 
# should update the minimal price, if needed. In that case 
# it issues a proper message. 

cheapest = {'store1': 5, 'store2': 7, 'store3': 2}
storeName = input("Please enter a store name: ")
while storeName != 'q': 
    prices = eval(input("Please enter a list of prices: "))
    minPrice = min(list(dict(prices).values()))
    if storeName not in cheapest: 
        cheapest[storeName] = minPrice 
    else: 
        if cheapest[storeName] > minPrice: 
            cheapest[storeName] = minPrice 
            print(storeName, "minimal price was updated.")
    storeName = input("Please enter a store name: ")
print(cheapest)

lst = [['a', 7],['b',9],['c',2]]
d = dict(lst)
print(d)
prices = d.values() 
print(min(list(prices)))

minValue = min(list(dict(lst).values()))
print(minValue)


cheapest = {'store1': 5, 'store2': 7, 'store3': 2}
storeName = input("Please enter a store name: ")
while storeName != 'q': 
    prices = eval(input("Please enter a list of prices: "))
    minPrice = min(list(dict(prices).values()))
    currentMinimal = cheapest.get(storeName, False)
    if currentMinimal == False : 
        cheapest[storeName] = minPrice 
    else: 
        if currentMinimal > minPrice: 
            cheapest[storeName] = minPrice 
            print(storeName, "minimal price was updated.")
    storeName = input("Please enter a store name: ")
print(cheapest)








# Q. 4
# We are 4 friends. We meet every week. 
# The host serves each of us with a plate 
# of 4 cookies. The meeting ends when at 
# least one of use ate all her/his cookies. 
# Write a program that receives, one by 
# another, a friend-s name who eats one 
# cookie. The program prints the name 
# of the friend who ate all his cookies first. 

cookies = {'A': 4, 'B': 4, 'C':4, 'D': 4}
proceed = True     
while proceed: 
    friend = input("Please enter friend's code: ")
    cookies[friend] = cookies[friend] - 1
    if cookies[friend] == 0: 
        proceed = False         
print(friend)
print(cookies)









# Q. 5
# Given is a dictionary of prices in a 
# certain store: keys are product codes, 
# values are product prices. Write a program 
# that receives a positive number and prints 
# codes of all pairs of products whose prices 
# sum is less or equal to the received number. 
# Each such pair should be printed only once. 
# Try to avoid checking a pair more than once. 

prices = {'A':5, 'B':7, 'C':1, 'D':2}
num = 4
codes = list(prices.keys())
while len(codes) > 1: 
    curCode = codes.pop() 
    for code in codes: 
        if prices[curCode] + prices[code] <= num: 
            print(curCode, code)




# Q. 6
# Your store has four departments, 
# A, B, C and D. The dictionary stock
# holds the number of products sold in 
# each departement. The keys are letters
# representing the department, and the values 
# are lists of lists: each inner lists includes 
# a name of a products and the number of items 
# of this product in stock. E.g. 
# store = {'A':{'p1':5, 'p2':3, 'p3':2},
#          'B':{'p6':2},
#          'C':{'p1':1, 'p7':1, 'p8':3},
#          'D':{'p9':1, 'p10':1, 'p3':1}}
# Now you went bankrupt. 
# Write a program that receives a name a 
# department and number of items of a product 
# sold in that department. The program updates 
# the number of items. If it is 0 or below, 
# it deletes the inner list of that product. 
# If all products of this department are no 
# longer in stock, the department is "deleted" 
# from the dictionary. 

store = {'A':{'p1':5, 'p2':3, 'p3':2},
         'B':{'p6':2},
         'C':{'p1':1, 'p7':1, 'p8':3},
         'D':{'p9':1, 'p10':1, 'p3':1}}
department = 'A'
code = 'p1'
items = 6
store[department][code] = store[department][code] - items 
if store[department][code] <= 0:
    del store[department][code]
    if not store[department]: 
        del store[department]
print(store)    



tuple 

lst = [3, 4, 5]
lst[1] = 999 
lst.sort() 

tup = (3, 4, 5)
tup[1] = 990

# Q. 7
# Given is a dictionary of products: each key 
# is the product code and the store in which 
# it is sold, and the value is the product's 
# price and that store. E.g. 
# { ('A','store1') :6, 
#   ('A','store2') :5, 
#   ('B','store3') :3, 
#   ('C','store2') :1}
# Write a program that: 
#     (a) Receives a price and prints a list 
#      of all product codes sold in prices 
#      smaller than the given price in any store, 
#      with no duplicates.  
# version 1
products = {('A','store1'):6, ('A','store2'):5, ('B','store3'):3, ('C','store2'):1}
price = 7
['A', 'B', 'C']
codes = set() 
for k in products: 
    if products[k] < price: 
        codes.add(k[0])
print(list(codes))

# version 2 
products = {('A','store1'):6, ('A','store2'):5, ('B','store3'):3, ('C','store2'):1}
price = 7
['A', 'B', 'C']
codes = set() 
for k, v in products.items():   
    if v < price: 
        codes.add(k[0])
print(list(codes))


  
# (b) Receives a product code and finds its 
# cheapest price. 

products = {('A','store1'):6, 
            ('A','store2'):5, 
            ('B','store3'):3, 
            ('C','store2'):1}
myCode = 'C'
prices = [] 
for code, store in products: 
    if code == myCode: 
        prices.append(products[ (code, store) ])
print(min(prices)) 


# (c) Produces a new dictionary, in which each key 
#      is a product code and each value is a list 
#      of two values: the number of stores in which 
#      it is sold, and the product-s highest price, 
#      in any store. 
products = {('A','store1'):6, 
            ('A','store2'):5, 
            ('B','store3'):3, 
            ('C','store2'):1}
myDict = {}
for k, v in products.items(): 
    code = k[0]
    if code not in myDict: 
        myDict[code] = [1,v]
    else: 
        myDict[code][0] += 1
        if myDict[code][1] < v: 
            myDict[code][1] = v 
print(myDict)        

products = {('A','store1'):6, 
            ('A','store2'):5, 
            ('B','store3'):3, 
            ('C','store2'):1}
myDict = {}
for code, store in products: 
    myDict[code] = [0, 0]
for k, v in products.items(): 
    code = k[0] 
    myDict[code][0] += 1
    if myDict[code][1] < v: 
        myDict[code][1] = v 
print(myDict)