如何用Python四舍五入到2个小数?

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

您可以使用四舍五入运算符最多为2小数

num = round(343.5544, 2)
print(num) // output is 343.55

2019-02-06 06:27:49

其他回答

到目前为止我发现的最简单的解决方案，不知道为什么人们不使用它。

# Make sure the number is a float
a = 2324.55555
# Round it according to your needs
# dPoints is the decimals after the point
dPoints = 2
# this will round the float to 2 digits
a = a.__round__(dPoints)
if len(str(a).split(".")[1]) < dPoints:
    # But it will only keep one 0 if there is nothing,
    # So we add the extra 0s we need
    print(str(a)+("0"*(dPoints-1)))
else:
    print(a)

2022-06-09 03:46:27

from decimal import Decimal, ROUND_HALF_UP

# Here are all your options for rounding:
# This one offers the most out of the box control
# ROUND_05UP       ROUND_DOWN       ROUND_HALF_DOWN  ROUND_HALF_UP
# ROUND_CEILING    ROUND_FLOOR      ROUND_HALF_EVEN  ROUND_UP

our_value = Decimal(16.0/7)
output = Decimal(our_value.quantize(Decimal('.01'), 
rounding=ROUND_HALF_UP))
print output

2021-08-29 18:19:39

因为你想要你的答案在十进制数字，所以你不需要类型转换你的答案变量str在printC()函数。

然后使用printf风格的字符串格式

2013-12-08 18:37:28

不知道为什么，但'{:0.2f}'.format(0.5357706)给我'0.54'。唯一适合我的解决方案(python 3.6)如下:

def ceil_floor(x):
    import math
    return math.ceil(x) if x < 0 else math.floor(x)

def round_n_digits(x, n):
    import math
    return ceil_floor(x * math.pow(10, n)) / math.pow(10, n)

round_n_digits(-0.5357706, 2) -> -0.53 
round_n_digits(0.5357706, 2) -> 0.53

2018-04-02 02:21:31

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

2021-03-21 03:31:39

如何用Python四舍五入到2个小数?

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