Python 从用户输入检查列表中的重复项

所以我写了一段代码，生成一个随机列表，其中包含随机数目的值。然后询问用户要查找的号码，如果该号码在列表中，它将告诉用户该号码在列表中的位置

import random

a = [random.randint(1, 20) for i in range(random.randint(8, 30))]

a.sort()
print(a)


def askUser():

    n = input("What number are you looking for?")
    while not n.isdigit():
        n = input("What number are you looking for?")

    n = int(n)
    s = 0

    for numbers in a:
        if numbers == n:
            s += 1
            print("Number", n, "is located in the list and the position is:", (a.index(n)+1))
            # Something here to skip this index next time it goes through the loop
        else:
            pass
    if s == 0:
        print("Your number could not be found")


askUser()

我想添加一些东西，将跳过它第一次找到的索引，然后寻找重复的索引（如果有）

当前结果

[2, 4, 8, 9, 10, 10, 16, 19, 20, 20]
What number are you looking for?20
Number 20 is located in the list and the position is: 9
Number 20 is located in the list and the position is: 9

期望结果

[2, 4, 8, 9, 10, 10, 16, 19, 20, 20]
What number are you looking for?20
Number 20 is located in the list and the position is: 9
Number 20 is located in the list and the position is: 10

更改此行：

for numbers in a:

致：

然后更改打印索引的方式：

print("Number", n, "is located in the list and the position is:", (i+1))

样本输出：

[1, 2, 2, 5, 5, 5, 6, 7, 8, 8, 8, 10, 10, 10, 10, 10, 11, 11, 16, 17, 17, 19, 19]
What number are you looking for? 8
Number 8 is located in the list and the position is: 9
Number 8 is located in the list and the position is: 10
Number 8 is located in the list and the position is: 11

---

您可以使用

numpy

简化此代码以删除循环

a = np.array([random.randint(1,20) for i in range(random.randint(8,30))])

然后，您可以使用

np。其中

确定用户是否在随机值数组

a

中选择了一个值：

idx_selections = np.where(a == n)[0]

然后，您可以处理用户是否匹配答案：

if len(idx_selections) == 0:
    print("Your number could not be found")
else:
    for i in idx_selections:
        print("Number", n, "is located in the list and the position is:", i)

---

如果您喜欢，可以将一些循环转换为列表理解：

def askUser():

    n = input("What number are you looking for?")
    while not n.isdigit():
        n = input("What number are you looking for?")

    n = int(n)

    # get a list of all indexes that match the number
    foundAt = [p+1 for p,num in enumerate(a) if num == n]

    if foundAt:
        # print text by creating a list of texts to print and decompose them
        # printing with a seperator of linefeed
        print( *[f"Number {n} is located in the list and the position is: {q}" for 
                 q in foundAt], sep="\n")
    else: 
        print("Your number could not be found")

编辑：正如Chrisz指出的那样，

f”“

格式字符串是为Python3.6提供的（不知道：o/）-因此对于早期的3.x，Python必须使用

print( *["Number {} is located in the list and the position is: {}".format(n,q) for 
                 q in foundAt], sep="\n")

---

感谢它工作得很完美，这么小的一件事却产生了这么大的差异很高兴能提供帮助，通过使用这里的

enumerate

，您可以在O（n）时间内检查所有索引，而不是使用

arr.index

，它只找到第一个匹配项。

a=random.choices（范围（1，20），k=random.randint（8，30））

可能比您的列表组件更好。这是一个python 2.x答案-问题带有标记-并且您没有完全生成所需的输出；）@chrisz PEP498新增：）

def askUser():

    n = input("What number are you looking for?")
    while not n.isdigit():
        n = input("What number are you looking for?")

    n = int(n)

    # get a list of all indexes that match the number
    foundAt = [p+1 for p,num in enumerate(a) if num == n]

    if foundAt:
        # print text by creating a list of texts to print and decompose them
        # printing with a seperator of linefeed
        print( *[f"Number {n} is located in the list and the position is: {q}" for 
                 q in foundAt], sep="\n")
    else: 
        print("Your number could not be found")

print( *["Number {} is located in the list and the position is: {}".format(n,q) for 
                 q in foundAt], sep="\n")