Question

让我看看我的病情：

a = int(b) >= 1230 and int(b) not in [1300, 1305, 1250]

在此，列表可以有多个值[1300,1303,1306,1307]等。所以我想检查一下：

if int(b) >= 1230 and int(b) = 130* and int(b) != 1250:
    do something
else:
    so something

如何检查以130 *开头的数字？

Answer 1

如果您要覆盖的范围很广，则可以检查数字是否在range()内。不要多次转换为int，请存储int值：bAsInt = int(b)并使用它。

如果要检查特定的单个值，请使用set()（如果有4个或更多值）-列表查找会更快：

even = {1300,1302,1304,1306,1308}

for number in range(1299,1311):
    # print(number," is 130* :", number//10 == 130 ) # works too, integer division
    print(number," is 130* :", number in range(1300,1310), 
          " and ", "even" if number in even else "odd")

输出：

1299  is 130* : False  and  odd
1300  is 130* : True  and  even
1301  is 130* : True  and  odd
1302  is 130* : True  and  even
1303  is 130* : True  and  odd
1304  is 130* : True  and  even
1305  is 130* : True  and  odd
1306  is 130* : True  and  even
1307  is 130* : True  and  odd
1308  is 130* : True  and  even
1309  is 130* : True  and  odd
1310  is 130* : False  and  odd

从用户那里获取价值并进行比较：

可以像这样解决：

def inputNumber():
    # modified from other answer, link see below       
    while True:
        try:
            number = int(input("Please enter number: "))
        except ValueError:
            print("Sorry, I didn't understand that.")
            continue
        else:
            return number

b = inputNumber()

even = {1300,1302,1304,1306,1308}

if b > 1230 and b not in even and b in range(1300,1310):
    #  r > 1230 is redundant if also b in range(1300,1310)
    print("You won the lottery")

它将打印1301、1303、1305、1307、1309的smth（由于in even）。

其他答案：Asking the user for input until they give a valid response

Answer 2

怎么样

if int(b) >= 1230 and str(b).startswith('130') and int(b) != 1250:
    do something
else:
    do something

Answer 3

您也可以使用此

def start_with(nub, start, stop):
    return [x for x in range(start,stop) if str(x).startswith(str(nub))]

print(start_with(130, 1, 5000))

使用此功能，您可以找到以任何数字开头的任何列表

Python：检查数字中的条件，以列表中的130 *开头

3 个答案: