CODEWITHSUNDEEP
pythondictionary
dilot
dilot

Reputation: 67

Variable in python dictionary

I have this dictionary:

A = {"a": 1, "b":2}

And I want to write, instead of above initialization, a variable instead of 2 that is 4 times a's value that updates if a's value changes.

I cannot do this:

A={"a": 1, "b": 4*A["a"]}

How can I handle this?

‡: Here in my code,

flight = {
    "dest": "".join(random.choice(string.ascii_uppercase) for n in xrange(3)),
    "from": "".join(random.choice(string.ascii_uppercase) for n in xrange(3)),
    "not_available_seats": [],
    "available_seats": 50,
    "uid": uid,
    #"price": 10,
    "date": datetime.datetime.now()
}
flight["price"] = lambda: (51 - flight["available_seats"])*10

So when print flight(), I get error.

Upvotes: 1

Views: 88

Answers (2)

jpp
jpp

Reputation: 164663

The more I look at this problem the more I believe you want an object-oriented solution.

Here is one implementation:

import random, string
from datetime import datetime

class Flight(object):

    def __init__(self, dest, origin, not_available_seats, available_seats, uid, date):
        self.dest = dest
        self.origin = origin
        self.not_available_seats = not_available_seats
        self.available_seats = available_seats
        self.uid = uid
        self.date = date
        self.price = (51 - self.available_seats) * 10

    def set_available_seats(self, available_seats):
        self.available_seats = available_seats
        self.price = (51 - self.available_seats) * 10
        return None

You can create a class instance as easily as adding a dictionary item:

F1 = Flight("".join(random.choice(string.ascii_uppercase) for n in range(3)),
            "".join(random.choice(string.ascii_uppercase) for n in range(3)),
            [],
            50,
            123456789,
            datetime.now())

Updating available seats causes the price to update:

print(F1.price)              # 10
F1.set_available_seats(30)
print(F1.price)              # 210

Upvotes: 1

user3483203
user3483203

Reputation: 51175

Use a lambda function, then call A["b"] when you want the proper value:

>>> A={"a": 1}
>>> A["b"]= lambda: 4*A["a"]
>>> A["b"]()
4
>>> A["a"] = 5
>>> A["b"]()
20

Updated for the case where A["a"] is a list:

>>> A={"a": [1,2,3,4,5]}
>>> A["b"]= lambda: [4*i for i in A["a"]]
>>> A["b"]()
[4, 8, 12, 16, 20]

Upvotes: 3

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