How to Convert pythons Decimal() type into an INT and exponent

Question

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!

Awesome! Thanks guys! I took rvraghav93's performance code and added everyone that replied here is the code(http://paste.ubuntu.com/7856669/) and here is the breakdown: rvraghav93: 1.11100006104 ericnutsch: 4.22099995613 smesseim: 4.70300006866 Mark_Ransom: 7.80999994278 Ehsan_Abd: 10.6519999504 — ericnutsch, Jul 25 '14 at 15:32

Raghav RV · Accepted Answer · 2014-07-25T00:00:36.987

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

You managed to bypass that as_tuple() function and for a major performance increase! Thanks for the benchmark too! — ericnutsch, Jul 25 '14 at 15:36
@ericnutsch I would never use `time.time` for code timing, it doesn't have sufficient precision. With `timeit.timeit` I get 0.31839284114539623 for to_int_exp and 1.5341496635228395 for to_int_exp1. I don't know why my results are so radically different, maybe because it's Python 3.8.6. — Mark Ransom, Jan 22 '22 at 05:35

smesseim · Answer 2 · 2014-07-24T23:01:45.633

5

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

edited Jul 24 '14 at 23:01

answered Jul 24 '14 at 22:29

smesseim

79
6

Thanks, that exactly what I'm looking for – Artem Baskakov Jun 16 '20 at 13:03

score 3 · Answer 3 · answered Jul 24 '14 at 23:02

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

score 0 · Answer 4 · answered Jul 24 '14 at 22:39

0

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)

answered Jul 24 '14 at 22:39

Ehsan88

3,569
5
29
52

How to Convert pythons Decimal() type into an INT and exponent

4 Answers4

Performance testing

Linked