Display decimal in scientific notation

Question

How can I display this:

Decimal ('40800000000.00000000000000') as '4.08E + 10'?

I tried this:

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

But he has an extra 0.

Answer 1

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

Your “40800000000.00000000000000” has many more significant zeros that have the same meaning as any other digit. That is why you should clearly indicate where you want to stay.

If you want to automatically delete all trailing zeros, 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' 

Answer 2

Here is an example using the format() function:

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

• official documentation

• original format proposal ()

Answer 3

Given your number

 x = Decimal('40800000000.00000000000000') 

Starting with Python 3,

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

is the recommended way to do this.

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

Answer 4

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

Requires at least Python 3.6

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

(I believe it just like Siz Timmerman, just a little shorter)

Answer 5

See tables from Python String Formatting to choose the correct format. In your case, this is %.2E .

Answer 6

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

 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 a usage example:

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

Answer 7

This worked better for me:

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

Answer 8

To convert decimal to scientific notation without having to specify precision in the format string and without including trailing zeros, I currently use

 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 preserve any trailing zeros, simply remove normalize() .

Answer 9

 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 

Answer 10

Or just (python 3.6+):

 f"{40800000000.00000000000000:.2E}" output: 4.08E+10 

or hard_to_read_number = 40800000000.00000000000000

 f"{difficult_to_read_number:.2E}" output: 4.08E+10