如何将建筑格式的测量值转换为浮点数？

Question

我有一个数据库，是一个建筑公司创建并使用的。所有的测量数据都以这样的格式存储：15-3/4" 和 12' 6-3/4"。

有没有办法在Python中把这些测量值转换成浮点数？或者有没有什么库可以提供这样的功能？

同样的，怎样才能把浮点数转换成上面的格式呢？

Answer 1

从建筑单位转换到浮点数：

import re

regex = re.compile('(\d+\' )*(\d+)-(\d+)\/(\d+)"')
regex.sub(lambda m: str((int((m.group(1) or '0').split("'")[0]) * 12)
  + int(m.group(2)))
  + ('%.2f' % (int(m.group(3)) / float(m.group(4))))[1:], measurement)

这真是太糟糕了，不过我有一段时间没用Python了；我相信肯定有更简单的方法来做到这一点，但这个方法确实很好地处理了没有英尺的情况。不过，它总是期待输入为英寸，所以像12'这样的测量值必须写成12' 0"才能正确解析。

Answer 2

考虑一下下面这段自我注释的代码。我尽量让它简单易懂。

>>> from fractions import Fraction
>>> def Arch2Float(num):
    #First Partition from Right so that the Feet and Unit always
    #Remains aligned even if one of them is absent
    ft,x,inch=num.rpartition("\'")
    #Convert the inch to a real and frac part after stripping the
    #inch (") identifier. Note it is assumed that the real and frac
    #parts are delimited by '-'
    real,x,frac=inch.strip("\"").rpartition("-")
    #Now Convert every thing in terms of feet which can then be converted
    #to float. Note to trap Error's like missing or invalid items, its better
    #to convert each items seperately
    result=0
    try:
        result = int(ft.strip("\'"))
    except ValueError:
        None
    #Convert the real inch part as a fraction of feet
    try:
        result +=  Fraction(int(real),12)
    except ValueError:
        None
    #Now finally convert the Fractional part using the fractions module and convert to feet
    try:
        result+=Fraction(frac)/12
    except ValueError:
        None
    return float(result)    

酸性测试

>>> print Arch2Float('15-3/4"')     # 15-3/4" (without ft)
1.3125
>>> print Arch2Float('12\' 6-3/4"') #12' 6-3/4"
12.5625
>>> print Arch2Float('12\'6-3/4"')  #12'6-3/4" (without space)
12.5625
>>> print Arch2Float('3/4"')        #3/4" (just the inch)
0.0625
>>> print Arch2Float('15\'')        #15' (just ft)
15.0
>>> print Arch2Float('15')          #15 (without any ascent considered as inch)
1.25

把浮点数转换成架构会很简单，因为你不需要费心去解析。

>>> def Float2Arch(num):
    num=Fraction(num)
    ft,inch=Fraction(num.numerator/num.denominator),Fraction(num.numerator%num.denominator)/num.denominator*12
    real,frac=inch.numerator/inch.denominator,Fraction(inch.numerator%inch.denominator,inch.denominator)
    return '{0}\' {1}-{2}"'.format(ft,real,frac)

酸性测试

>>> print Float2Arch(Arch2Float('12\' 6-3/4"'))
12' 6-3/4"
>>> print Float2Arch(Arch2Float('15-3/4"'))
1' 3-3/4"
>>> print Float2Arch(Arch2Float('12\'6-3/4"'))
12' 6-3/4"
>>> print Float2Arch(Arch2Float('3/4"'))
0' 0-3/4"
>>> print Float2Arch(Arch2Float('15\''))
15' 0-0"
>>> print Float2Arch(Arch2Float('15'))
1' 3-0"
>>> 

注意*** 保持浮点数的表示很重要，要么用最小的单位（英寸），要么用最大的单位（英尺）。在这个例子中，我选择了最大的单位，英尺。如果你想用更小的单位，可以把它乘以12。

---

更新以满足四舍五入的请求 （不确定这是否优雅，但能完成任务）

def Float2Arch(num):
    num=Fraction(num)
    ft,inch=Fraction(num.numerator/num.denominator),Fraction(num.numerator%num.denominator)/num.denominator*12
    real,frac=inch.numerator/inch.denominator,Fraction(inch.numerator%inch.denominator,inch.denominator)
    for i in xrange(1,17):
        if Fraction(frac) < Fraction(1.0/16*i): break
    frac=Fraction(1.0/16*i)
    if frac>= 1:
        real+=1
        frac=0
    return '{0}\' {1}-{2}"'.format(ft,real,frac)

Answer 3

根据模式的规律性，你可以使用 str.partition 来进行解析：

def architectural_to_float(text):
    ''' Convert architectural measurements to inches.

        >>> for text in """15-3/4",12' 6-3/4",3/4",3/4',15',15",15.5'""".split(','):
        ...     print text.ljust(10), '-->', architectural_to_float(text)
        ...
        15-3/4"    --> 15.75
        12' 6-3/4" --> 150.75
        3/4"       --> 0.75
        3/4'       --> 9.0
        15'        --> 180.0
        15"        --> 15.0
        15.5'      --> 186.0

    '''
    # See http://stackoverflow.com/questions/8675714
    text = text.replace('"', '').replace(' ', '')
    feet, sep, inches = text.rpartition("'")
    floatfeet, sep, fracfeet = feet.rpartition('-')
    feetnum, sep, feetdenom = fracfeet.partition('/')
    feet = float(floatfeet or 0) + float(feetnum or 0) / float(feetdenom or 1)
    floatinches, sep, fracinches = inches.rpartition('-')
    inchesnum, sep, inchesdenom = fracinches.partition('/')
    inches = float(floatinches or 0) + float(inchesnum or 0) / float(inchesdenom or 1)
    return feet * 12.0 + inches