Display a decimal in scientific notation

Question:

How can I display Decimal('40800000000.00000000000000') as '4.08E+10'?

I’ve tried this:

>>> '%E' % Decimal('40800000000.00000000000000')
'4.080000E+10'

But it has those extra 0’s.

Asked By: Greg

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Answers:

from decimal import Decimal

'%.2E' % Decimal('40800000000.00000000000000')

# returns '4.08E+10'

In your ‘40800000000.00000000000000’ there are many more significant zeros that have the same meaning as any other digit. That’s why you have to tell explicitly where you want to stop.

If you want to remove all trailing zeros automatically, you can try:

def format_e(n):
    a = '%E' % n
    return a.split('E')[0].rstrip('0').rstrip('.') + 'E' + a.split('E')[1]

format_e(Decimal('40800000000.00000000000000'))
# '4.08E+10'

format_e(Decimal('40000000000.00000000000000'))
# '4E+10'

format_e(Decimal('40812300000.00000000000000'))
# '4.08123E+10'
Answered By: eumiro

See tables from Python string formatting to select the proper format layout. In your case it’s %.2E.

Answered By: Mihai Maruseac

My decimals are too big for %E so I had to improvize:

def format_decimal(x, prec=2):
    tup = x.as_tuple()
    digits = list(tup.digits[:prec + 1])
    sign = '-' if tup.sign else ''
    dec = ''.join(str(i) for i in digits[1:])
    exp = x.adjusted()
    return '{sign}{int}.{dec}e{exp}'.format(sign=sign, int=digits[0], dec=dec, exp=exp)

Here’s an example usage:

>>> n = decimal.Decimal(4.3) ** 12314
>>> print format_decimal(n)
3.39e7800
>>> print '%e' % n
inf
Answered By: ubershmekel

Here’s an example using the format() function:

>>> "{:.2E}".format(Decimal('40800000000.00000000000000'))
'4.08E+10'

Instead of format, you can also use f-strings:

>>> f"{Decimal('40800000000.00000000000000'):.2E}"
'4.08E+10'
Answered By: Cees Timmerman

To convert a Decimal to scientific notation without needing to specify the precision in the format string, and without including trailing zeros, I’m currently using

def sci_str(dec):
    return ('{:.' + str(len(dec.normalize().as_tuple().digits) - 1) + 'E}').format(dec)

print( sci_str( Decimal('123.456000') ) )    # 1.23456E+2

To keep any trailing zeros, just remove the normalize().

Answered By: MikeM
def formatE_decimal(x, prec=2):
    """ Examples:
    >>> formatE_decimal('0.1613965',10)
    '1.6139650000E-01'
    >>> formatE_decimal('0.1613965',5)
    '1.61397E-01'
    >>> formatE_decimal('0.9995',2)
    '1.00E+00'
    """
    xx=decimal.Decimal(x) if type(x)==type("") else x 
    tup = xx.as_tuple()
    xx=xx.quantize( decimal.Decimal("1E{0}".format(len(tup[1])+tup[2]-prec-1)), decimal.ROUND_HALF_UP )
    tup = xx.as_tuple()
    exp = xx.adjusted()
    sign = '-' if tup.sign else ''
    dec = ''.join(str(i) for i in tup[1][1:prec+1])   
    if prec>0:
        return '{sign}{int}.{dec}E{exp:+03d}'.format(sign=sign, int=tup[1][0], dec=dec, exp=exp)
    elif prec==0:
        return '{sign}{int}E{exp:+03d}'.format(sign=sign, int=tup[1][0], exp=exp)
    else:
        return None
Answered By: Andrej

This worked best for me:

import decimal
'%.2E' % decimal.Decimal('40800000000.00000000000000')
# 4.08E+10
Answered By: Matthew Fitch

Given your number

x = Decimal('40800000000.00000000000000')

Starting from Python 3,

'{:.2e}'.format(x)

is the recommended way to do it.

e means you want scientific notation, and .2 means you want 2 digits after the dot. So you will get x.xxE±n

Answered By: patapouf_ai

No one mentioned the short form of the .format method:

Needs at least Python 3.6

f"{Decimal('40800000000.00000000000000'):.2E}"

(I believe it’s the same as Cees Timmerman, just a bit shorter)

Answered By: Eulenfuchswiesel

I prefer Python 3.x way.

cal = 123.4567
print(f"result {cal:.4E}")

4 indicates how many digits are shown shown in the floating part.

cal = 123.4567
totalDigitInFloatingPArt = 4
print(f"result {cal:.{totalDigitInFloatingPArt}E} ")
Answered By: snr

This is a consolidated list of the “Simple” Answers & Comments.

PYTHON 3

from decimal import Decimal

x = '40800000000.00000000000000'
# Converted to Float
x = Decimal(x)

# ===================================== # `Dot Format`
print("{0:.2E}".format(x))
# ===================================== # `%` Format
print("%.2E" % x)
# ===================================== # `f` Format
print(f"{x:.2E}")
# =====================================
# ALL Return: 4.08E+10
print((f"{x:.2E}") == ("%.2E" % x) == ("{0:.2E}".format(x)))
# True
print(type(f"{x:.2E}") == type("%.2E" % x) == type("{0:.2E}".format(x)))
# True
# =====================================

OR Without IMPORT‘s

# NO IMPORT NEEDED FOR BASIC FLOATS
y = '40800000000.00000000000000'
y = float(y)

# ===================================== # `Dot Format`
print("{0:.2E}".format(y))
# ===================================== # `%` Format
print("%.2E" % y)
# ===================================== # `f` Format
print(f"{y:.2E}")
# =====================================
# ALL Return: 4.08E+10
print((f"{y:.2E}") == ("%.2E" % y) == ("{0:.2E}".format(y)))
# True
print(type(f"{y:.2E}") == type("%.2E" % y) == type("{0:.2E}".format(y)))
# True
# =====================================

Comparing

# =====================================
x
# Decimal('40800000000.00000000000000')
y
# 40800000000.0

type(x)
# <class 'decimal.Decimal'>
type(y)
# <class 'float'>

x == y
# True
type(x) == type(y)
# False

x
# Decimal('40800000000.00000000000000')
y
# 40800000000.0

So for Python 3, you can switch between any of the three for now.

My Fav:

print("{0:.2E}".format(y))
Answered By: JayRizzo

Here is the simplest one I could find.

format(40800000000.00000000000000, '.2E')
#'4.08E+10'

(‘E’ is not case sensitive. You can also use ‘.2e’)

Answered By: Ibrahim Khalil

Adding an updated answer to show how to apply e notation to small numbers only

value = 0.1
a = "{:,}".format(value) if value >= 0.001 else "{:,.3e}".format(value)
print(a) # 0.1

value = 0.00002488
a = "{:,}".format(value) if value >= 0.001 else "{:,.3e}".format(value)
print(a) # 2.488e-05
Answered By: Beppe C