Question

我有一个文本文件，其中包含一个多个点的数字字符串，如'1.2.3'。我想将此数字转换为float或int，如'1.2.3'，以便稍后可以将此数字与其他数字进行比较。我在linux下使用python 3.5。要解释这个问题，请考虑这个简单的例子： -

$ python3.5
Python 3.5.2 (default, Sep 10 2016, 08:21:44) 
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> _str = '1.2.3'
>>> float(_str)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: could not convert string to float: '1.2.3'
>>> _str = '1.2'
>>> float(_str)
1.2

很明显，python不会将包含多个点的数字视为float。我做了很多搜索，但似乎没有人有类似的问题。

我如何克服这个问题？

Answer 1

我创建了一个名为WrongFloat的类，因为你要求的不是浮点数

from itertools import izip_longest

class WrongFloat:
    def __init__(self, val):
        if isinstance(val, WrongFloat):
            self.parts = val.parts[:]
        elif isinstance(val, str):
            self.parts=[int(v) for v in val.split('.')]
        elif isinstance(val, float):
            self.parts=str(val).split('.')
        elif isinstance(val, int):
            self.parts=[val]

    def __str__(self):
        return ".".join([str(v) for v in self.parts])


    def __lt__(self, val):
        o = WrongFloat(val)
        parts = list(izip_longest(self.parts, o.parts, fillvalue=0))
        for p in parts:
            if p[0] > p[1]:
                return False

        if parts[-1][0] == parts[-1][1]:
            return False
        return True

    def __le__(self, val):
        o = WrongFloat(val)
        parts = list(izip_longest(self.parts, o.parts, fillvalue=0))
        for p in parts:
            if p[0] > p[1]:
                return False
        return True        

    def __eq__(self, val):
        o = WrongFloat(val)
        first, second = tuple(zip(*list(izip_longest(self.parts, o.parts, fillvalue=0))))
        return first == second

    def __ne__(self, val):
        o = WrongFloat(val)
        first, second = tuple(zip(*list(izip_longest(self.parts, o.parts, fillvalue=0))))
        return first != second

    def __gt__(self, val):
        o = WrongFloat(val)
        parts = list(izip_longest(self.parts, o.parts, fillvalue=0))
        for p in parts:
            if p[0] < p[1]:
                return False

        if parts[-1][0] == parts[-1][1]:
            return False
        return True

    def __ge__(self, val):
        o = WrongFloat(val)
        parts = list(izip_longest(self.parts, o.parts, fillvalue=0))
        for p in parts:
            if p[0] < p[1]:
                return False
        return True

init 将从string，float，int或其他WrongFloat创建一个实例

所以这可以用于施工

a = WrongFloat("1.2.3")
print a

b = WrongFloat(1.2)
print b

c = WrongFloat(1)
print c

d = WrongFloat(a)
print d

try:
    e = WrongFloat("d.r.f")
except ValueError:
    print "ValueError"

添加了方法 gt ， ge ， le ， lt ， eq 和 ne 所以你可以比较两个值

print a < "1.2.4"
print a < "1.2.3"
print a < "1.2.2"
print a < "1.2.3.1"
print a < "1.2.3.0"
print a < "1.2.2.9"

print a <= "1.2.4"
print a <= "1.2.3"
print a <= "1.2.2"
print a <= "1.2.3.1"
print a <= "1.2.3.0"
print a <= "1.2.2.9"

print a == "1.2.4"
print a == "1.2.3"
print a == "1.2.2"

print a != "1.2.4"
print a != "1.2.3"
print a != "1.2.2"

print a <= "1.2.2"
print a <= "1.2.3"
print a <= "1.2.4"

print a < "1.2.2"
print a < "1.2.3"
print a < "1.2.4"

您还可以使用添加和 sub 方法进行加法和减法，我不认为乘法和除法对于这个带有多个点的浮点类型是合理的< / p>

如何将多点串转换为浮点值

1 个答案: