[英]Python - Input Validation
我正在寻找创建代码,该代码要求用户在继续之前输入大于 2 的整数。 我正在使用 python 3.3。 这是我到目前为止所拥有的:
def is_integer(x):
try:
int(x)
return False
except ValueError:
print('Please enter an integer above 2')
return True
maximum_number_input = input("Maximum Number: ")
while is_integer(maximum_number_input):
maximum_number_input = input("Maximum Number: ")
print('You have successfully entered a valid number')
我不确定的是如何最好地设置整数必须大于 2 的条件。我刚刚开始学习 python,但想养成良好的习惯。
这应该可以完成这项工作:
def valid_user_input(x):
try:
return int(x) > 2
except ValueError:
return False
maximum_number_input = input("Maximum Number: ")
while valid_user_input(maximum_number_input):
maximum_number_input = input("Maximum Number: ")
print("You have successfully entered a valid number")
或者更短:
def valid_user_input():
try:
return int(input("Maximum Number: ")) > 2
except ValueError:
return False
while valid_user_input():
print('You have successfully entered a valid number')
def take_user_in():
try:
return int(raw_input("Enter a value greater than 2 -> ")) # Taking user input and converting to string
except ValueError as e: # Catching the exception, that possibly, a inconvertible string could be given
print "Please enter a number as" + str(e) + " as a number"
return None
if __name__ == '__main__': # Somethign akin to having a main function in Python
# Structure like a do-whole loop
# func()
# while()
# func()
var = take_user_in() # Taking user data
while not isinstance(var, int) or var < 2: # Making sure that data is an int and more than 2
var = take_user_in() # Taking user input again for invalid input
print "Thank you" # Success
我的看法:
from itertools import dropwhile
from numbers import Integral
from functools import partial
from ast import literal_eval
def value_if_type(obj, of_type=(Integral,)):
try:
value = literal_eval(obj)
if isinstance(value, of_type):
return value
except ValueError:
return None
inputs = map(partial(value_if_type), iter(lambda: input('Input int > 2'), object()))
gt2 = next(dropwhile(lambda L: L <= 2, inputs))
def check_value(some_value):
try:
y = int(some_value)
except ValueError:
return False
return y > 2
这验证输入是整数,但确实拒绝看起来像整数的值(如3.0
):
def is_valid(x):
return isinstance(x,int) and x > 2
x = 0
while not is_valid(x):
# In Python 2.x, use raw_input() instead of input()
x = input("Please enter an integer greater than 2: ")
try:
x = int(x)
except ValueError:
continue
希望这有帮助
import str
def validate(s):
return str.isdigit(s) and int(s) > 2
使用其他答案中显示的内置int()
问题是它会将浮点数和布尔值转换为整数,因此它并不是真正检查您的参数是否为整数。
单独使用内置的isinstance(value, int)
方法很诱人,但不幸的是,如果传递一个布尔值,它将返回 True。 所以如果你想要严格的类型检查,这是我简短而甜蜜的 Python 3.7 解决方案:
def is_integer(value):
if isinstance(value, bool):
return False
else:
return isinstance(value, int)
结果:
is_integer(True) --> False
is_integer(False) --> False
is_integer(0.0) --> False
is_integer(0) --> True
is_integer((12)) --> True
is_integer((12,)) --> False
is_integer([0]) --> False
等等...
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.