使用 isdigit 表示浮点数？

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? ')

仅当用户输入整数时才有效，但我希望即使他们输入浮点也能工作，但当他们输入字符串时则不起作用>。

因此，用户应该能够输入 9 和 9.2，但不能输入 abc。

我该怎么做呢？

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，因为更 Pythonic 的方式通常是 try... except （即 EAFP，“请求宽恕比请求许可更容易”）。

但是，如果您确实想要进行验证，则可以在使用 isdigit() 之前精确删除 1 位小数点。

>>> "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，这并没有将浮点数视为与整数有任何不同。如果您确实需要的话，您可以添加该检查。

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

基于 dan04 的答案：

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))

请注意，这也适用于负浮点数。

Answer 6

我认为@dan04有正确的方法（EAFP），但不幸的是现实世界通常是一个特殊情况，并且确实需要一些额外的代码来管理事物 - 所以下面是一个更详细的，但也更务实（和现实） ：

import sys

while True:
    try:
        a = raw_input('How much is 1 share in that company? ')
        x = float(a)
        # validity check(s)
        if x < 0: raise ValueError('share price must be positive')
    except ValueError, e:
        print("ValueError: '{}'".format(e))
        print("Please try entering it again...")
    except KeyboardInterrupt:
        sys.exit("\n<terminated by user>")
    except:
        exc_value = sys.exc_info()[1]
        exc_class = exc_value.__class__.__name__
        print("{} exception: '{}'".format(exc_class, exc_value))
        sys.exit("<fatal error encountered>")
    else:
        break  # no exceptions occurred, terminate loop

print("Share price entered: {}".format(x))

使用示例：

> python numeric_input.py
How much is 1 share in that company? abc
ValueError: 'could not convert string to float: abc'
Please try entering it again...
How much is 1 share in that company? -1
ValueError: 'share price must be positive'
Please try entering it again...
How much is 1 share in that company? 9
Share price entered: 9.0

> python numeric_input.py
How much is 1 share in that company? 9.2
Share price entered: 9.2

Answer 7

import re

string1 = "0.5"
string2 = "0.5a"
string3 = "a0.5"
string4 = "a0.5a"

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

if p.match(string1):
    print(string1 + " float or int")
else:
    print(string1 + " not float or int")

if p.match(string2):
    print(string2 + " float or int")
else:
    print(string2 + " not float or int")

if p.match(string3):
    print(string3 + " float or int")
else:
    print(string3 + " not float or int")

if p.match(string4):
    print(string4 + " float or int")
else:
    print(string4 + " not float or int")

output:
0.5 float or int
0.5a not float or int
a0.5 not float or int
a0.5a not float or int

Answer 8

如果字符串包含一些特殊字符，例如下划线（例如“1_1”），则提供的答案将失败。在我测试的所有情况下，以下函数都会返回正确的答案。

def IfStringRepresentsFloat(s):
try:
    float(s)
    return str(float(s)) == s
except ValueError:
    return False

Answer 9

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

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