我在这段代码的输出中得到了很多小数(华氏到摄氏度转换器)。

我的代码目前看起来是这样的:

def main():
    printC(formeln(typeHere()))

def typeHere():
    global Fahrenheit
    try:
        Fahrenheit = int(raw_input("Hi! Enter Fahrenheit value, and get it in Celsius!\n"))
    except ValueError:
        print "\nYour insertion was not a digit!"
        print "We've put your Fahrenheit value to 50!"
        Fahrenheit = 50
    return Fahrenheit

def formeln(c):
    Celsius = (Fahrenheit - 32.00) * 5.00/9.00
    return Celsius

def printC(answer):
    answer = str(answer)
    print "\nYour Celsius value is " + answer + " C.\n"



main()

我的问题是,如何让程序把每个答案四舍五入到小数点后2位?


当前回答

如果你不仅需要四舍五入的结果,而且还需要用四舍五入的结果进行数学运算,那么你可以使用十进制。十进制https://docs.python.org/2/library/decimal.html

from decimal import Decimal, ROUND_DOWN

Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32') 

其他回答

截2指炎:

somefloat = 2.23134133
truncated = int( somefloat * 100 ) / 100  # 2.23
float(str(round(answer, 2)))
float(str(round(0.0556781255, 2)))

可以使用python“%”的字符串格式化操作符。 “%。“2f”表示小数点后2位。

def typeHere():
    try:
        Fahrenheit = int(raw_input("Hi! Enter Fahrenheit value, and get it in Celsius!\n"))
    except ValueError:
        print "\nYour insertion was not a digit!"
        print "We've put your Fahrenheit value to 50!"
        Fahrenheit = 50
    return Fahrenheit

def formeln(Fahrenheit):
    Celsius = (Fahrenheit - 32.0) * 5.0/9.0
    return Celsius

def printC(answer):
    print "\nYour Celsius value is %.2f C.\n" % answer

def main():
    printC(formeln(typeHere()))

main()

http://docs.python.org/2/library/stdtypes.html#string-formatting

只需使用%格式即可。2f可以四舍五入到2个小数。

def printC(answer):
    print "\nYour Celsius value is %.2f C.\n" % answer

为了避免round()的惊人值,以下是我的方法:

Round = lambda x, n: eval('"%.'+str(int(n))+'f" % '+repr(int(x)+round(float('.'+str(float(x)).split('.')[1]),n)))

print(Round(2, 2))       # 2.00
print(Round(2.675, 2))   # 2.68