使用isdigit浮动？

Question

a = raw_input('How much is 1 share in that company? ')

while not a.isdigit():
    print("You need to write a number!\n")
    a = raw_input('How much is 1 share in that company? ')

这仅在用户输入时才有效integer,但我希望即使他们输入了一个,但我希望它能够工作float,而不是在他们输入时string.

因此,用户应该能够同时输入9和9.2,但不会abc.

我该怎么办？

Answer 1

EAFP

try:
    x = float(a)
except ValueError:
    print("You must enter a number")

Answer 2

现有的答案是正确的,因为通常更多的Pythonic方式try...except(即EAFP).

但是,如果您确实想要进行验证,则可以在使用前删除正好1个小数点isdigit().

>>> "124".replace(".", "", 1).isdigit()
True
>>> "12.4".replace(".", "", 1).isdigit()
True
>>> "12..4".replace(".", "", 1).isdigit()
False
>>> "192.168.1.1".replace(".", "", 1).isdigit()
False

请注意,这并不会处理与int不同的浮点数.如果你真的需要它,你可以添加该检查.

Answer 3

使用正则表达式.

import re

p = re.compile('\d+(\.\d+)?')

a = raw_input('How much is 1 share in that company? ')

while p.match(a) == None:
    print "You need to write a number!\n"
    a = raw_input('How much is 1 share in that company? ')

Answer 4

以dan04的答案为基础:

def isDigit(x):
    try:
        float(x)
        return True
    except ValueError:
        return False

用法:

isDigit(3)     # True
isDigit(3.1)   # True
isDigit("3")   # True
isDigit("3.1") # True
isDigit("hi")  # False

Answer 5

s = '12.32'
if s.replace('.', '').replace('-', '').isdigit():
    print(float(s))

float请注意，这也适用于负数。