如何将python的Decimal()类型转换为INT和指数 [英] How to Convert pythons Decimal() type into an INT and exponent

问题描述

我想在python中使用Decimal()数据类型并将其转换为整数和指数，这样我就可以将该数据发送到具有全精度和十进制控制的微控制器/plc. https://docs.python.org/2/library/decimal.html

I would like to use the Decimal() data type in python and convert it to an integer and exponent so I can send that data to a microcontroller/plc with full precision and decimal control. https://docs.python.org/2/library/decimal.html

我可以使用它，但是它有点破旧；有谁知道更好的方法?如果不是这样，我将采取什么方式自己编写一个较低级别的"as_int()"函数?

I have got it to work, but it is hackish; does anyone know a better way? If not what path would I take to write a lower level "as_int()" function myself?

示例代码:

from decimal import *
d=Decimal('3.14159')
t=d.as_tuple()
if t[0] == 0:
    sign=1
else:
    sign=-1

digits= t[1]
theExponent=t[2]
theInteger=sign * int(''.join(map(str,digits)))

theExponent
theInteger

对于那些没有编程PLC的人，我的替代选择是使用int并在两个系统中声明小数点，或者使用浮点数(仅某些PLC支持)，这是有损的.这样您就能知道为什么能够做到这一点了！

For those that havent programmed PLCs, my alternative to this is to use an int and declare the decimal point in both systems or use a floating point (that only some PLCs support) and is lossy. So you can see why being able to do this would be awesome!

提前谢谢！

推荐答案

您可以这样做:

[这比其他方法快三倍]

[ This is 3 times faster than the other methods ]

d=Decimal('3.14159')

list_d = str(d).split('.')   
# Converting the decimal to string and splitting it at the decimal point

# If decimal point exists => Negative exponent
# i.e   3.14159 => "3", "14159"
# exponent = -len("14159") = -5
# integer = int("3"+"14159") = 314159

if len(list_d) == 2:
    # Exponent is the negative of length of no of digits after decimal point
    exponent = -len(list_d[1])
    integer = int(list_d[0] + list_d[1])



# If the decimal point does not exist => Positive / Zero exponent
# 3400
# exponent = len("3400") - len("34") = 2
# integer = int("34") = 34

else:
    str_dec = list_d[0].rstrip('0')
    exponent = len(list_d[0]) - len(str_dec)
    integer = int(str_dec)

print integer, exponent

性能测试

def to_int_exp(decimal_instance):

    list_d = str(decimal_instance).split('.')

    if len(list_d) == 2:
        # Negative exponent
        exponent = -len(list_d[1])
        integer = int(list_d[0] + list_d[1])

    else:
        str_dec = list_d[0].rstrip('0')
        # Positive exponent
        exponent = len(list_d[0]) - len(str_dec)
        integer = int(str_dec)

    return integer, exponent

def to_int_exp1(decimal_instance):
    t=decimal_instance.as_tuple()

    if t[0] == 0:
        sign=1
    else:
        sign=-1

    digits= t[1]

    exponent = t[2]

    integer = sign * int(''.join(map(str,digits)))

    return integer, exponent

计算两种方法的100,000个循环所花费的时间:

Calculating the time taken for 100,000 loops for both methods :

ttaken = time.time()
for i in range(100000):
    d = Decimal(random.uniform(-3, +3))
    to_int_exp(d)    
ttaken = time.time() - ttaken
print ttaken

字符串解析方法花费的时间:1.56606507301

Time taken for string parsing method : 1.56606507301

ttaken = time.time()
for i in range(100000):
    d = Decimal(random.uniform(-3, +3))
    to_int_exp1(d)    
ttaken = time.time() - ttaken
print ttaken

转换为元组然后提取方法所用的时间:4.67159295082

Time taken for convertion to tuple then extract method : 4.67159295082

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