有没有办法告诉一个字符串是否代表一个整数(例如,'3','-17'但不是'3.14'或'asfasfas')而不使用try/except机制?
is_int('3.14') == False
is_int('-7') == True
有没有办法告诉一个字符串是否代表一个整数(例如,'3','-17'但不是'3.14'或'asfasfas')而不使用try/except机制?
is_int('3.14') == False
is_int('-7') == True
当前回答
正确的RegEx解决方案应该结合Greg Hewgill和Nowell的思想,但不使用全局变量。可以通过将属性附加到方法来实现这一点。另外,我知道在方法中导入是不受欢迎的,但我想要的是像http://peak.telecommunity.com/DevCenter/Importing#lazy-imports这样的“惰性模块”效果
edit:到目前为止,我最喜欢的技术是使用String对象的独占方法。
#!/usr/bin/env python
# Uses exclusively methods of the String object
def isInteger(i):
i = str(i)
return i=='0' or (i if i.find('..') > -1 else i.lstrip('-+').rstrip('0').rstrip('.')).isdigit()
# Uses re module for regex
def isIntegre(i):
import re
if not hasattr(isIntegre, '_re'):
print("I compile only once. Remove this line when you are confident in that.")
isIntegre._re = re.compile(r"[-+]?\d+(\.0*)?$")
return isIntegre._re.match(str(i)) is not None
# When executed directly run Unit Tests
if __name__ == '__main__':
for obj in [
# integers
0, 1, -1, 1.0, -1.0,
'0', '0.','0.0', '1', '-1', '+1', '1.0', '-1.0', '+1.0',
# non-integers
1.1, -1.1, '1.1', '-1.1', '+1.1',
'1.1.1', '1.1.0', '1.0.1', '1.0.0',
'1.0.', '1..0', '1..',
'0.0.', '0..0', '0..',
'one', object(), (1,2,3), [1,2,3], {'one':'two'}
]:
# Notice the integre uses 're' (intended to be humorous)
integer = ('an integer' if isInteger(obj) else 'NOT an integer')
integre = ('an integre' if isIntegre(obj) else 'NOT an integre')
# Make strings look like strings in the output
if isinstance(obj, str):
obj = ("'%s'" % (obj,))
print("%30s is %14s is %14s" % (obj, integer, integre))
对于那些不太喜欢冒险的同学,输出如下:
I compile only once. Remove this line when you are confident in that.
0 is an integer is an integre
1 is an integer is an integre
-1 is an integer is an integre
1.0 is an integer is an integre
-1.0 is an integer is an integre
'0' is an integer is an integre
'0.' is an integer is an integre
'0.0' is an integer is an integre
'1' is an integer is an integre
'-1' is an integer is an integre
'+1' is an integer is an integre
'1.0' is an integer is an integre
'-1.0' is an integer is an integre
'+1.0' is an integer is an integre
1.1 is NOT an integer is NOT an integre
-1.1 is NOT an integer is NOT an integre
'1.1' is NOT an integer is NOT an integre
'-1.1' is NOT an integer is NOT an integre
'+1.1' is NOT an integer is NOT an integre
'1.1.1' is NOT an integer is NOT an integre
'1.1.0' is NOT an integer is NOT an integre
'1.0.1' is NOT an integer is NOT an integre
'1.0.0' is NOT an integer is NOT an integre
'1.0.' is NOT an integer is NOT an integre
'1..0' is NOT an integer is NOT an integre
'1..' is NOT an integer is NOT an integre
'0.0.' is NOT an integer is NOT an integre
'0..0' is NOT an integer is NOT an integre
'0..' is NOT an integer is NOT an integre
'one' is NOT an integer is NOT an integre
<object object at 0x103b7d0a0> is NOT an integer is NOT an integre
(1, 2, 3) is NOT an integer is NOT an integre
[1, 2, 3] is NOT an integer is NOT an integre
{'one': 'two'} is NOT an integer is NOT an integre
其他回答
对于正整数,可以使用.isdigit:
>>> '16'.isdigit()
True
但它对负整数不起作用。假设您可以尝试以下方法:
>>> s = '-17'
>>> s.startswith('-') and s[1:].isdigit()
True
它不适用于'16.0'格式,在这个意义上,'16.0'格式类似于int类型强制转换。
编辑:
def check_int(s):
if s[0] in ('-', '+'):
return s[1:].isdigit()
return s.isdigit()
使用正则表达式:
import re
def RepresentsInt(s):
return re.match(r"[-+]?\d+$", s) is not None
如果你必须接受小数也:
def RepresentsInt(s):
return re.match(r"[-+]?\d+(\.0*)?$", s) is not None
如果经常这样做,为了提高性能,请使用re.compile()只编译正则表达式一次。
如果你想只接受低ascii码的数字,这里有这样做的测试:
Python 3.7+:(u.isdecimal() and u.isascii()))
Python <= 3.6:(u.isdecimal() and u == str(int(u)))
其他答案建议使用.isdigit()或.isdecimal(),但它们都包含一些上unicode字符,如'缌' (u'\u0662'):
u = u'\u0662' # '٢'
u.isdigit() # True
u.isdecimal() # True
u.isascii() # False (Python 3.7+ only)
u == str(int(u)) # False
Greg Hewgill的方法缺少了几个组件:前导的“^”只匹配字符串的开头,并且预先编译re。但是这种方法可以让你避免尝试:
import re
INT_RE = re.compile(r"^[-]?\d+$")
def RepresentsInt(s):
return INT_RE.match(str(s)) is not None
我很感兴趣为什么你试图避免尝试:除了?
如果你真的不喜欢到处使用try/except,请写一个helper函数:
def represents_int(s):
try:
int(s)
except ValueError:
return False
else:
return True
>>> print(represents_int("+123"))
True
>>> print(represents_int("10.0"))
False
它将需要更多的代码来精确覆盖Python认为是整数的所有字符串。要我说,你就用蟒语吧。