如何检查号码是否与列表中的号码相同

Question

我想检查用户输入的数字是否与列表中的数字相同，以及我 select 列表中的项目使用 for 循环与数字进行比较的方式不起作用。 就像它只是遍历列表中的第一项。 即使我输入 43（示例），它仍然会打印（列表中有 43），我认为第一项是 65，这与 43 不同。

ARR = [65,75,85,95,105]

def check_number ():
    NUM = int (input ("Please enter the number you want to check with"))
    for x in range (len (arr)):   
        check == ARR [x]
    if check == ARR [x]:
        print (NUM,"is available in the list")      
    else:
        print (NUM,"is not available is the list")
check_number ()

Answer 1

检查一个值（包括数字和字符串的任何可散列值）是否是列表成员的更好方法是使用 Python 的内置set类型：

arr = set([65, 75, 85, 95, 105])

def check_number():
    num = int (input ("Please enter the number you want to check with"))
    if num in arr:
        print (num, "is available in the list")      
    else:
        print (num, "is not available is the list")

check_number ()

对于更大的数据集，它既更省时，也更惯用。

Answer 2

你可以简单地做

ARR = [65,75,85,95,105]

def check_number ():
    NUM = int (input ("Please enter the number you want to check with"))

    if NUM in ARR:
        print (NUM,"is available in the list")      
    else:
        print (NUM,"is not available is the list")
check_number ()

使用in检查数组是否包含元素是“ pythonic ”的方法。

Answer 3

您可以使用in关键字进行比较。

ARR = [65,75,85,95,105]

def check_number ():
    NUM = int (input ("Please enter the number you want to check with"))
    if NUM in ARR:
        print(NUM, ' is available in this list')
    else:
        print(NUM, ' is not available in this list')
check_number()

Answer 4

无需使用loop进行比较，因为它会traverse并将number与list进行比较。 这是一个有点冗长的过程，而不是你可以使用in来比较number和list 。 相同场景的代码如下所述：-

# Declared List named 'ARR'
ARR = [65,75,85,95,105]

# Declaration of 'check_number()' function for checking whether Number is Present in list or not
def check_number ():

    # User imput of 'Number' for Comparision
    NUM = int (input ("Please enter the number you want to check with:- "))

    # If 'Number (NUM)' is present in 'List (ARR)'
    if NUM in ARR:
        print (NUM, "is available in the list") 

    # If 'Number (NUM)' is not present in 'List (ARR)'
    else:
        print (NUM, "is not available is the list")

check_number()

# Output_1 for the above code is given below:-
Please enter the number you want to check with:- 45
45 is not available is the list

# Output_2 for the above code is given below:-
Please enter the number you want to check with:- 65
65 is available in the list