Display a decimal in scientific notation

How can I display this:

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.

eumiro

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'

Cees Timmerman

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

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

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

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)

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

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

This worked best for me:

import decimal
'%.2E' % decimal.Decimal('40800000000.00000000000000')
# 4.08E+10

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().

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

来源：https://stackoverflow.com/questions/6913532/display-a-decimal-in-scientific-notation

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string-formatting