制定决策树

Question

I have ten systems, A, B, C, ..., J. Each system can be up or down. If, for example, systems A, B, D and J are down, with the remainder being up then I want to take action X. If systems C, D, and H are down, with the remainder being up then I want to take action Y. If systems A, E, F, H and I are down, with the remainder being up then I want to take action Z.

I am wanting to write a program that will print the various combinations (I believe with 10 systems and each can be up or down there are 100 combinations).

I have this so far:

import itertools
status_list = (
    "Up",
    "Down",
)
component_list = (
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "G",
    "G",
    "I",
    "J",
)
combinations = itertools.product(component_list, status_list)

Answer 1

If you don't mind using a binary representation, we could use 0 to represent Up and 1 to represent Down

Our values would range from 0 , to 10^2-1 ,

which is 0 to 1023 , or

0000000000 to 1111111111 .

So we could just loop from 0 to 1023, printing the numbers out in a 10-digit binary format:

for x in range(2**10):
    print '{0:010b}'.format(x)

Outputs:

0000000000
0000000001
0000000010
0000000011
0000000100
0000000101
0000000110
...etc

---

If you wanted to get fancier and print up and down instead of 0 and 1 , you could examine the digits one-by-one and convert:

for x in range(2**10):

    #Iterate over each character c in the binary number x
    #Convert c to an int
    #Look up the value in status_list by taking c mod length of status list
    print(" ".join(status_list[int(c)%len(status_list)] for c in '{0:010b}'.format(x)))

Outputs:

Up Up Up Up Up Up Up Up Up Up
Up Up Up Up Up Up Up Up Up Down
Up Up Up Up Up Up Up Up Down Up
Up Up Up Up Up Up Up Up Down Down

Answer 2

You can print out every possible combination of Up and Down using itertools.product :

import itertools
components = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J']
statuses = ['Up', 'Down']

for state in itertools.product(statuses, repeat=len(components)):
    for c,s in zip(components, state):
        print '{}:{}'.format(c,s),
    print

This produces output like the following:

A:Up B:Up C:Up D:Up E:Up F:Up G:Up H:Up I:Up J:Up
A:Up B:Up C:Up D:Up E:Up F:Up G:Up H:Up I:Up J:Down
...
A:Up B:Down C:Down D:Down E:Down F:Down G:Down H:Down I:Down J:Up
A:Up B:Down C:Down D:Down E:Down F:Down G:Down H:Down I:Down J:Down
A:Down B:Up C:Up D:Up E:Up F:Up G:Up H:Up I:Up J:Up
A:Down B:Up C:Up D:Up E:Up F:Up G:Up H:Up I:Up J:Down
...
A:Down B:Down C:Down D:Down E:Down F:Down G:Down H:Down I:Down J:Up
A:Down B:Down C:Down D:Down E:Down F:Down G:Down H:Down I:Down J:Down

It deals nicely with a variable number of components or states (for example, you could make states = ['Up', 'Down', 'Unknown'] and you would get 3^10 = 59049 outputs).