python查找具有多个列表的所有组合。 完整的编程新手

Question

So I'm 100% new to programming and although I'm a very fast learner with most things, I am in need of assistance.

I want to find all possible combinations using multiple lists on Python. I know there is an intertool for it, but I honestly don't even know where to begin, how to use it or even how to enter my data.

A basic example of what I'm trying to do:

Flavors        Sizes      Toppings         Syrups
==========     =======    =============    ==============
Chocolate      Small      Sprinkles        Hot fudge
Vanilla        Medium     Gummy bears      Caramel 
Strawberry     Large      Oreo             Strawberry
Coffee                    Cookie dough     White chocolate
                          Snickers         etc.
                          Brownies
                          etc.

SO for flavors and sizes there can only be ONE choice, but let's say for syrups I let them pick THREE choices, and for toppings I also let them pick THREE. And I want to find all combinations.

Is this hard to do? What is the exact code I need and how exactly do I enter my variables?

Thanks. Much appreciated.

Ps- Is there a limit to how many combinations python can take? How much can the cpu of an average macbook pro take?

Answer 1

I think what you're looking for is product :

Example:

import itertools

 a1 = [1,2,3] a2 = [4,5,6] a3 = [7,8,9] result = list(itertools.product(a1,a2,a3)) >>> print result [(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9)] 

Answer 2

from itertools import product, combinations, combinations_with_replacement

flavors  = ["chocolate", "vanilla", "strawberry", "coffee"]
sizes    = ["small", "medium", "large"]
toppings = ["sprinkles", "gummy bears", "oreos", "cookie dough", "snickers", "brownies"]
syrups   = ["hot fudge", "caramel", "strawberry", "white chocolate"]

#
# pick a flavor and a size
for flavor,size in product(flavors, sizes):
    #
    # pick three toppings, but no more than one of each
    for top_a, top_b, top_c in combinations(toppings, 3):
        #
        # pick three syrups, allowing repeats
        for syr_a, syr_b, syr_c in combinations_with_replacement(syrups, 3):
            #
            # now do something with the result:
            print(", ".join([flavor, size, top_a, top_b, top_c, syr_a, syr_b, syr_c]))

and output looks like

chocolate, small, sprinkles, gummy bears, oreos, hot fudge, hot fudge, hot fudge
chocolate, small, sprinkles, gummy bears, oreos, hot fudge, hot fudge, caramel
chocolate, small, sprinkles, gummy bears, oreos, hot fudge, hot fudge, strawberry
chocolate, small, sprinkles, gummy bears, oreos, hot fudge, hot fudge, white chocolate
chocolate, small, sprinkles, gummy bears, oreos, hot fudge, caramel, caramel
chocolate, small, sprinkles, gummy bears, oreos, hot fudge, caramel, strawberry
# ... etc
# (4800 combinations in total)

Edit:

The other thing to point out is that this presumes the order of toppings is unimportant - ie ["sprinkles", "oreos", "cookie dough"] is effectively identical to ["oreos", "sprinkles", "cookie dough"] .

If the order matters, you need to look at itertools.permutations(toppings, 3) instead (not allowing more than one of each) or itertools.product(toppings, repeat=3) (allowing multiples).

Be aware that taking order into account greatly increases the number of combinations - from 4800 to 92160 in this example.

Answer 3

from itertools import product, combinations, combinations_with_replacement

flavors  = ["chocolate", "vanilla", "strawberry", "coffee"]
sizes    = ["small", "medium", "large"]
syrups   = ["hot fudge", "caramel", "strawberry", "white chocolate"]
toppings = ["sprinkles", "gummy bears", "oreos", "cookie dough", "snickers", "brownies"]

all_combos = list(
    product(flavors, sizes, combinations(syrups, 3), combinations(toppings, 3))
)