如何打印项目以及它来自Python的列表？

Question

I am trying to create a roulette simulator and currently the program only prints the number. How do I get it to print the number and the list title it is in?

Example:

12 Red

22 Black

import random
red=[1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36]
black=[2,4,6,8,10,11,13,15,17,20,22,24,26,28,29,31,33,35]
spin=random.randint(0,36)
print(spin)

Answer 1

if spin in red:
    print('{} Red'.format(spin))
elif spin in black:
    print('{} Black'.format(spin))
else:
    print('{}'.format(spin))

Answer 2

You only need check two conditions, either it is 0 or in one of red or black, if it is not 0 and not in red in must be it black

import random

red = {1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36}
spin = random.randint(0, 36)
print("{} {}".format(spin, ["black","red"][spin in red] if spin != 0 else "green"))

Which written out explicitly becomes:

red = {1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36}

spin = random.randint(0, 36)
if spin == 0:
   print("green 0")
elif spin in red:
    print("red {}".format(spin))
else:
    print("black {}".format(spin))

The black list of numbers is irrelevant.

If you want to represent 00 you will need to use strings instead of integers as a leading 0 means octal in python 2 and is invalid syntax in python3

import random

red = {'25', '12', '14', '16', '19', '32', '30', '23', '36', '34', '1', '3', '27', '5', '7', '18', '9', '21'}

nums = ["00",'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26', '27', '28', '29', '30', '31', '32', '33', '34', '35']

spin = random.choice(nums)
if spin  in {"0","00"}:
   print("green {}".format(spin))
elif spin in red:
    print("red {}".format(spin))
else:
    print("black {}".format(spin))

you could also use a dict mapping numbers to colour and picking a random key:

d = {'00':"green",'0':"green",'24': 'black', '25': 'red', '26': 'black', '27': 'red', '20': 'black',
     '21': 'red', '22': 'black', '23': 'red', '28': 'black', '29': 'black', '1': 'red', '3': 'red',
     '2': 'black', '5': 'red', '4': 'black', '7': 'red', '6': 'black', '9': 'red', '8': 'black',
     '11': 'black', '10': 'black', '13': 'black', '12': 'red', '15': 'black', '14': 'red', '17':
    'black', '16': 'red', '19': 'red', '18': 'red', '31': 'black', '30': 'red', '36': 'red', '35':
    'black', '34': 'red', '33': 'black', '32': 'red'}

spin = random.choice(list(d))
print("{} {}".format(d[spin],spin))

If you want to get the percentages you can run n trials and use a Counter dict to keep count:

nums = list(d)

from collections import Counter

num_count = Counter()
colours = Counter()
for test in range(10000):
    spin = random.choice(nums)
    num_count[spin] += 1
    colours[d[spin]] += 1


for k,v in colours.most_common():
    print("{} rolled {} times which is {}% of the total spin".format(k,v, v / 10000.0 * 100))
print(" ")
for k,v in num_count.most_common():
    print("{} rolled {} times which is {}% of the total spins".format(k,v, v / 10000.0 * 100))

Output:

black rolled 4748 times which is 47.48% of the total spin
red rolled 4726 times which is 47.26% of the total spin
green rolled 526 times which is 5.26% of the total spin

20 rolled 301 times which is 3.01% of the total spins
36 rolled 296 times which is 2.96% of the total spins
23 rolled 292 times which is 2.92% of the total spins
3 rolled 285 times which is 2.85% of the total spins
00 rolled 284 times which is 2.84% of the total spins
13 rolled 282 times which is 2.82% of the total spins
12 rolled 280 times which is 2.8% of the total spins
27 rolled 278 times which is 2.78% of the total spins
15 rolled 273 times which is 2.73% of the total spins
14 rolled 272 times which is 2.72% of the total spins
6 rolled 270 times which is 2.7% of the total spins
8 rolled 270 times which is 2.7% of the total spins
2 rolled 268 times which is 2.68% of the total spins
5 rolled 267 times which is 2.67% of the total spins
35 rolled 267 times which is 2.67% of the total spins
19 rolled 266 times which is 2.66% of the total spins
34 rolled 265 times which is 2.65% of the total spins
33 rolled 265 times which is 2.65% of the total spins
21 rolled 264 times which is 2.64% of the total spins
4 rolled 264 times which is 2.64% of the total spins
24 rolled 263 times which is 2.63% of the total spins
17 rolled 262 times which is 2.62% of the total spins
31 rolled 261 times which is 2.61% of the total spins
11 rolled 260 times which is 2.6% of the total spins
18 rolled 257 times which is 2.57% of the total spins
29 rolled 255 times which is 2.55% of the total spins
30 rolled 253 times which is 2.53% of the total spins
10 rolled 251 times which is 2.51% of the total spins
22 rolled 248 times which is 2.48% of the total spins
28 rolled 247 times which is 2.47% of the total spins
1 rolled 245 times which is 2.45% of the total spins
25 rolled 243 times which is 2.43% of the total spins
32 rolled 243 times which is 2.43% of the total spins
0 rolled 242 times which is 2.42% of the total spins
9 rolled 242 times which is 2.42% of the total spins
26 rolled 241 times which is 2.41% of the total spins
7 rolled 240 times which is 2.4% of the total spins
16 rolled 238 times which is 2.38% of the total spins

Answer 3

Basically you can search using 'in' operator.

if spin in red:
    //do for red
elif spin in black:
    //do for black...
else:
    //take a spin again