比较两个列表的元素和顺序 python3

Question

import random

colors = ['blue', 'yellow', 'green', 'orange', 'black']
random_colors = (random.choices(colors, k=4))

print(f'\nWelcome to the color game! your color options are: \n{colors}\n')


userinput = []
for i in range(0, 4):
    print("Enter your color choice: {}: ".format(i+1))
    userchoice = str(input())
    userinput.append(userchoice)


def compare():
    set_colors = set(random_colors)
    set_userin = set(userinput)
    result = set_userin - set_colors
    print(result)


compare()

我想将 random_colors 集与用户输入集进行比较。如果用户在 random_colors 集中输入了错误的颜色或颜色位置，我想指定颜色是在错误的位置还是不在 random_colors 集中。我创建的比较函数不检查顺序。

例如。最终结果：
random_colors=橙蓝蓝黑
userinput = 橙蓝黄蓝

预期 - 'yellow is a wrong color' 和 'blue is wrong position'

我还没有开始打印，因为我不确定如何比较这些集合。

Answer 1

set 不保留顺序，因此您无法判断用户输入的位置是否正确；您可以改用 list。此外，您可以使用 zip 来判断两个元素是否处于同一位置。尝试以下操作：

import random

colors = ['blue', 'yellow', 'green', 'orange', 'black']
random_colors = random.choices(colors, k=4)

user_colors = [input(f"Color {i+1}: ") for i in range(4)]

for u, r in zip(user_colors, random_colors):
    if u not in random_colors:
        print(f"{u} is a wrong color.")
    elif u != r:
        print(f"{u} is in a wrong position.")

Answer 2

在 Python 中，set 不保留其元素的顺序。取而代之的是，您可以尝试将它们作为列表进行比较：

def compare(colors, userin):
  # Iterate over the index and value of each color
  # in the user input.
  for idx, color in enumerate(userin):

    # Check if the color is not the same as the color
    # at the same position in the random colors.
    if color != colors[idx]:

      # If the color exists at some other position in
      # random colors, then print that the color is
      # in the wrong position.
      if color in colors:
        print(f"{color} is wrong position")

      # Is the color is not present in the list of random
      # colors then print that the color is a wrong color
      else:
        print("{color} is a wrong color")

比较两个列表的元素和顺序 python3

Compare elements AND order of two lists python3

python

compare

list