Variable in python dictionary

Question

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.

If I'm not mistaken, within the {} A doesn't exist yet, so A["a"] has no meaning. — Drise
– Drise, Commented Mar 2, 2018 at 21:40
Possible duplicate of Can you set a dictionary value dependant on another dictionary entry? — jpp
– jpp, Commented Mar 2, 2018 at 21:41
You edited the question, but now it's confusing. If you initialize A with static values, and then try to re-initialize it with a dynamic value (as reflected in your question), it will work fine. If you instead ONLY try to initialize it self-referentially, then it won't work (see @jpp's link). — Jim Stewart
– Jim Stewart, Commented Mar 2, 2018 at 21:42
Could you do two assignments? Ie A={"a": 1}; A["b"]=4*A["a"];. At this point, A exists, and contains a "a" key with a value. — Drise
– Drise, Commented Mar 2, 2018 at 21:43

user3483203 · Accepted Answer · 2018-03-02 22:07:09Z

3

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]

edited Mar 2, 2018 at 22:07

answered Mar 2, 2018 at 21:49

user3483203

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2 Comments

jpp Over a year ago

The issue I have with this implementation is that while the syntax is nice ("just add () after a dictionary call"), in reality the dictionary isn't storing the value itself. It must recalculate each time. I posted an object-oriented solution which I believe better suits the problem.

user3483203 Over a year ago

@jpp Oh I agree, the overarching issue is best solved with an OOP solution. Definitely an XY problem. +1 to your approach!

jpp · Accepted Answer · 2018-03-02 22:53:26Z

1

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

answered Mar 2, 2018 at 22:53

jpp

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1 Comment

dilot Over a year ago

Thanks, this is great and what I want!

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