当用户必须输入随机数的值时程序显示错误

Question

我被要求让程序生成一个由 15 个随机整数组成的数组，然后要求用户输入数组中的数字并让它显示一条消息，说明它在数组中，但是，我得到一个错误。

import numpy as ny
randnums = ny.random.randint(1,101,15)

print(randnums)

target = int(input("Please pick a random number: "))

for counter in range(0,15):
  while target != randnums:
    print("This number is not in the list")
    target = int(input("Please pick a random number: "))
  else:
   if target == randnums:
      print("The number" , target , "has been found in the list.")

输出：

Traceback (most recent call last):
  File "python", line 9, in <module>
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Answer 1

问题在于 != 和 == 运算符。
“randnums”是一个元素列表。您不能将单个值与整个列表进行比较。相反，您想检查该值是否在列表中。
您可以通过使用“in”和“not in”运算符来做到这一点，因此您的代码将如下所示：

import numpy as ny
randnums = ny.random.randint(1,101,15)

print(randnums)

target = int(input("Please pick a random number: "))

for counter in range(0,15):
  while target not in randnums:
    print("This number is not in the list")
    target = int(input("Please pick a random number: "))
  else:
   if target in randnums:
      print("The number" , target , "has been found in the list.")

Answer 2

问题是您的代码的第 9 行，正如所提供的输出所解释的那样。

while target != randnums: 这将检查target 变量是否不等于 randnums 变量，后者是一个 numpy 数组。

你真正想要的是这个

while target not in randnums: 如果target 变量的值是randnums numpy 数组中的值之一，它将遍历randnums 变量并返回一个布尔值。

Answer 3

短版：

import numpy as ny

randnums = ny.random.randint(1, 101, 15)

while True:
    target = int(input('Please pick a random number: '))
    if target in randnums:
        print(f'The number {target} has been found in the list')
        break
    else:
        print('This number is not in the list')