对于"get_roll“函数，我如何拆分用户的输入，例如，"1d3"，意思是一个3面骰子的1次滚动，作为一个随机整数？

Question

import random

def get_damage(attack, defense):
  if get_roll(attack) > get_roll(defense):
    return get_roll(attack) - get_roll(defense)
  else:
    return 0

def get_roll(rollstring):

def main_menu():
  x = int(input("How many rolls do you want to take?: "))
  z = 0
  list = []
  for roll in range(0,x):
    print("Input attack and defense roll", roll + 1, ": ", end="")
    y = input()
    list = list + y.split(",")
  for element in range(0, x):
    attack = list[z]
    defense = list[z+1]
    print ("Attack: ", attack, "Defense: ", defense, "Damage: ", get_damage(attack, defense))
    z = z + 2

main_menu()

样本输出

How many rolls do you want to take? 3
Input attack and defense roll 1: 1d3,1d3
Input attack and defense roll 2: 1d6,1d3
Input attack and defense roll 3: 1d6,1d10
Attack:1d3, Defense:1d3 : Damage: 0
Attack:1d6, Defense:1d3 : Damage: 2
Attack:1d6, Defense:1d10 : Damage: 0

Answer 1

您可以像这样使用random.randint：

import random

def get_damage(attack, defense):
  if attack > defense:
    return attack - defense
  else:
    return 0

def get_roll(rollstring):
    repeat = int(rollstring.split('d')[0])
    torandom = int(rollstring.split('d')[1])
    for i in range(repeat):
        result = random.randint(repeat, torandom)
    return result
def main_menu():
  x = int(input("How many rolls do you want to take?: "))
  z = 0
  list = []
  for roll in range(0,x):
    print("Input attack and defense roll", roll + 1, ": ", end="")
    y = input()
    list = list + y.split(",")
  for element in range(0, x):
    attack = list[z]
    defense = list[z+1]
    attack = get_roll(attack)
    defense = get_roll(defense)
    print ("Attack: ", attack, "Defense: ", defense, "Damage: ", get_damage(attack, defense))
    z = z + 2

main_menu()

我修复了你的一些代码。

Answer 2

import random
rollstring = '5d5'
num_rolls, sides = rollstring.split('d')
rolls = []
for roll in range(int(num_rolls)):
    rolls.append(random.choice(range(int(sides))))
print(rolls)
print(sum(rolls))