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
下面是一个解析时不会产生错误的函数。它处理明显的情况,失败时返回None(在CPython上默认处理最多2000个'-/+'符号!):
#!/usr/bin/env python
def get_int(number):
splits = number.split('.')
if len(splits) > 2:
# too many splits
return None
if len(splits) == 2 and splits[1]:
# handle decimal part recursively :-)
if get_int(splits[1]) != 0:
return None
int_part = splits[0].lstrip("+")
if int_part.startswith('-'):
# handle minus sign recursively :-)
return get_int(int_part[1:]) * -1
# successful 'and' returns last truth-y value (cast is always valid)
return int_part.isdigit() and int(int_part)
一些测试:
tests = ["0", "0.0", "0.1", "1", "1.1", "1.0", "-1", "-1.1", "-1.0", "-0", "--0", "---3", '.3', '--3.', "+13", "+-1.00", "--+123", "-0.000"]
for t in tests:
print "get_int(%s) = %s" % (t, get_int(str(t)))
结果:
get_int(0) = 0
get_int(0.0) = 0
get_int(0.1) = None
get_int(1) = 1
get_int(1.1) = None
get_int(1.0) = 1
get_int(-1) = -1
get_int(-1.1) = None
get_int(-1.0) = -1
get_int(-0) = 0
get_int(--0) = 0
get_int(---3) = -3
get_int(.3) = None
get_int(--3.) = 3
get_int(+13) = 13
get_int(+-1.00) = -1
get_int(--+123) = 123
get_int(-0.000) = 0
如有需要,可使用:
def int_predicate(number):
return get_int(number) is not None
其他回答
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
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()
可以用下面的方法进行检查。
def check_if_string_is_int(string1):
for character in string1:
if not character.isdigit():
return "Not a number"
else:
return "Is a number"
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。
下面是一个解析时不会产生错误的函数。它处理明显的情况,失败时返回None(在CPython上默认处理最多2000个'-/+'符号!):
#!/usr/bin/env python
def get_int(number):
splits = number.split('.')
if len(splits) > 2:
# too many splits
return None
if len(splits) == 2 and splits[1]:
# handle decimal part recursively :-)
if get_int(splits[1]) != 0:
return None
int_part = splits[0].lstrip("+")
if int_part.startswith('-'):
# handle minus sign recursively :-)
return get_int(int_part[1:]) * -1
# successful 'and' returns last truth-y value (cast is always valid)
return int_part.isdigit() and int(int_part)
一些测试:
tests = ["0", "0.0", "0.1", "1", "1.1", "1.0", "-1", "-1.1", "-1.0", "-0", "--0", "---3", '.3', '--3.', "+13", "+-1.00", "--+123", "-0.000"]
for t in tests:
print "get_int(%s) = %s" % (t, get_int(str(t)))
结果:
get_int(0) = 0
get_int(0.0) = 0
get_int(0.1) = None
get_int(1) = 1
get_int(1.1) = None
get_int(1.0) = 1
get_int(-1) = -1
get_int(-1.1) = None
get_int(-1.0) = -1
get_int(-0) = 0
get_int(--0) = 0
get_int(---3) = -3
get_int(.3) = None
get_int(--3.) = 3
get_int(+13) = 13
get_int(+-1.00) = -1
get_int(--+123) = 123
get_int(-0.000) = 0
如有需要,可使用:
def int_predicate(number):
return get_int(number) is not None