如何使用用戶輸入和 1 到 100 之間的值對 python 中的列表進行排序，並且沒有重復值？

Question

我想制作一個程序，以升序對數字列表進行排序，值在 1 到 100 之間。這就是我所做的，如果您運行代碼，列表會打印多次並且沒有正確指出錯誤：

# This is the user input
a = int(input("Enter the first number: "))
b = int(input("Enter the second number: "))
c = int(input("Enter the third number: "))
d = int(input("Enter the fourth number: "))
e = int(input("Enter the fifth number: "))
f = int(input("Enter the sixth number: "))
# This the list to sort
array = [a, b, c, d, e, f]
#This will count how many time each value is repeated
aa = int(array.count(a))
bb = int(array.count(b))
cc = int(array.count(c))
dd = int(array.count(d))
ee = int(array.count(e))
ff = int(array.count(f))
# This will loop through the list
for i in array:
    # This is checking if any values are repeated or not between 1 and 100
    if i > 100 or i < 1 or aa and bb and cc and dd and ee and ff > 1:
        print("Number should be smaller than 100 and larger than 1 and numbers can't be repeated")
    # This is to print the sorted list if you get everything right
    else:
        array.sort()
        print(array)

Answer 1

我認為您想用if...aa and bb and cc and dd and ee and ff > 1:的or語句替換您的and語句。 否則，僅當所有數字都重復時，條件才會為True 。

if i > 100 or i < 1 or aa or bb or cc or dd or ee or ff > 1:

注意：這將打印Number should be smaller than 100 and larger than 1 and numbers can't be repeated 6 次，這是您真正想要的嗎？

否則，有一種更短的方法可以做您（可能）想做的事情：

• 檢查 max(array) > 100
• 檢查 min(array) < 1
• 將array轉換為一個set （只保留唯一值）並比較它們的長度

嘗試這個 ：

# This is the user input
a = int(input("Enter the first number: "))
b = int(input("Enter the second number: "))
c = int(input("Enter the third number: "))
d = int(input("Enter the fourth number: "))
e = int(input("Enter the fifth number: "))
f = int(input("Enter the sixth number: "))

# This the list to sort
array = [a, b, c, d, e, f]

if max(array)>100 or min(array)<1 or len(set(array)) != len(array):
    print("Number should be smaller than 100 and larger than 1 and numbers can't be repeated")
else:
    array.sort()
    print(array)

Answer 2

替代方法：使用集合來收集您正在接受的號碼，並使用它來拒絕已經存在的號碼

我知道 StackOverflow 的本意不是給別人寫代碼，但是我想推薦給你的很多東西，我只能用代碼來表達。 在這里，我認為你可以通過分析來學習一些東西。

（它需要 Python 3.8 或更高版本）：

accepted_numbers = set() # Begin with an empty bag


## Main Job ##
def main():
  for i in range(1, 6+1):
      while not satisfy_conditions(number := get_ith_number(i)):
          notify_error()
      accepted_numbers.add(number)

  report()
## END Main Job ##


## Support ##

def satisfy_conditions(number):
    # Returns True if the number is not in the bag and is between 1 and 100
    return number not in accepted_numbers and (1 <= number <= 100)

def get_ith_number(i):
    word_number = {1:'first', 2:'second', 3:'third', 4:'forth', 5:'five', 6:'sixth'}
    return int(input(f'Enter the {word_number[i]} number: '))

def notify_error():
    print("Number should be smaller than 100 and larger than 1 and numbers can't be repeated")

def report():
    print(sorted(accepted_numbers))

## END Support ##

if __name__ == '__main__':
    main()