问题
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
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?
Example code:
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
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!
Thanks in advance!
回答1:
You could do this :
[ 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
Performance testing
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
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
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
Time taken for convertion to tuple then extract method : 4.67159295082
回答2:
from functools import reduce # Only in Python 3, omit this in Python 2.x
from decimal import *
d = Decimal('3.14159')
t = d.as_tuple()
theInteger = reduce(lambda rst, x: rst * 10 + x, t.digits)
theExponent = t.exponent
回答3:
Get the exponent directly from the tuple as you were:
exponent = d.as_tuple()[2]
Then multiply by the proper power of 10:
i = int(d * Decimal('10')**-exponent)
Putting it all together:
from decimal import Decimal
_ten = Decimal('10')
def int_exponent(d):
exponent = d.as_tuple()[2]
int_part = int(d * (_ten ** -exponent))
return int_part, exponent
回答4:
from decimal import *
d=Decimal('3.14159')
t=d.as_tuple()
digits=t.digits
theInteger=0
for x in range(len(digits)):
theInteger=theInteger+digits[x]*10**(len(digits)-x)
来源:https://stackoverflow.com/questions/24944863/how-to-convert-pythons-decimal-type-into-an-int-and-exponent