Question

因此，此函数接收来自用户的输入并将其输入转换为值。例如，如果他们输入1k，输出将是1000.我希望能够倒退。所以说如果我的值为325000，我想将其改为325k。关于如何实现这一目标的任何想法？

class Parsing:

    def __init__(self, instring):
        self.instring = instring

    def valueParsing(self):
        self.instring = self.instring.strip()
        self.parsedString = ''
        self.scalerDict = {'K': 1000, 'MEG': 1000000, 'G': 1000000000, 'M': 0.001, 'U': 0.000001, 'N':  0.000000001, 'P': 0.000000000001}
        self.scaler = 1.0
        self.stringCounter = 0
        self.errorflag = False
        self.Parsedvalue = 0.0
        self.inStringLength = len(self.instring)

        for self.stringCounter in range (self.inStringLength):
            if ((self.instring[self.stringCounter].upper()) == 'K'):
                self.scaler = self.scalerDict['K']
            elif ((self.instring[self.stringCounter].upper()) == 'G'):
                self.scaler = self.scalerDict['G']
            elif ((self.instring[self.stringCounter].upper()) == 'U'):
                self.scaler = self.scalerDict['U']
            elif ((self.instring[self.stringCounter].upper()) == 'N'):
                self.scaler = self.scalerDict['N']
            elif ((self.instring[self.stringCounter].upper()) == 'P'):
                self.scaler = self.scalerDict['P']
            elif ((self.instring[self.stringCounter].upper()) == 'M'):
                if (((self.instring.upper()).count('MEG'))):
                    self.scaler = self.scalerDict['MEG']
                else:
                    self.scaler = self.scalerDict['M']

            elif (( self.instring[ self.stringCounter ].upper() ) == 'F' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'W' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'S' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == '%' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'V' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'A' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'H' ):
                break
            elif (( self.instring[ self.stringCounter ].upper() ) == 'Z' ):
                break
            elif (( self.instring[ self.stringCounter ]) == '.' ):
                self.parsedString = self.parsedString + self.instring[ self.stringCounter ]

            elif (self.instring[self.stringCounter].isdigit()):
                if(int(self.instring[self.stringCounter]) >= 0):
                    if(int(self.instring[self.stringCounter]) <= 9):
                        self.parsedString = self.parsedString + self.instring[self.stringCounter]
                    else:
                        self.errorflag = True
                        break
            else:
                self.errorflag = True
                print('Invalid input, Try again.')

        if (self.errorflag):
            self.parsedvalue = -1

        else:
            self.parsedvalue = long(self.parsedString)*self.scaler

        return self.parsedvalue

print '1. Resistors in series\n',\
      '2. Resistors in Parallel\n',\
      '3. Voltage Divider\n'

iput = int(input("Enter your choice: "))

if iput == 1:
    r1 = raw_input("Enter first resistor:")
    r2 = raw_input("Enter second resistor:")
    R1 = Parsing(r1)
    R2 = Parsing(r2)
    req = R1.valueParsing() + R2.valueParsing()
    print "The value of the series resistors is %s." % req

Answer 1

试试这个：

import math

def fmtnum(num):
    k = (1e3, 'k')
    M = (1e6, 'M')
    G = (1e9, 'G')
    table = {3: k, 4: k, 5: k, 6: M, 7: M, 8: M, 9: G, 10: G, 11: G}
    num = float(num)
    exp = math.log10(num)
    if num < 0:
        exp = int(exp)-1
    else:
        exp = int(exp)
    try:
        denum, suffix = table[exp]
        return '{:g} {}'.format(num/denum, suffix)
    except KeyError:
        return '{:g}'.format(num)

将此解决方案扩展为10的负幂已经留给读者练习。： - ）

示例：

In [50]: fmtnum(3250)
Out[50]: '3.25 k'

In [51]: fmtnum(32500)
Out[51]: '32.5 k'

In [52]: fmtnum(325000)
Out[52]: '325 k'

In [53]: fmtnum(3250000)
Out[53]: '3.25 M'

In [54]: fmtnum(32500000)
Out[54]: '32.5 M'

In [55]: fmtnum(325000000)
Out[55]: '325 M'

In [56]: fmtnum(3250000000)
Out[56]: '3.25 G'

Answer 2

怎么样：

n = 1000
for exp, name in zip(range(9, -13, -3), ('GMk1munp')):
    if exp == 0:
        continue
    if isinstance(n, int):
        if n % 10**exp == 0:
            n = '{0:d}{1}'.format(n / 10**exp, name)
            break
    elif isinstance(n, basestring):
        if n[-exp:] == '0' * exp:
            n = '{0}{1}'.format(n[:-exp], name)
            break
        elif n[-1] == name:
            n = n[:-1] + '0' * exp
            break

向后运行指数，确保找到正确的匹配。

在Python

2 个答案: