CODEWITHSUNDEEP
pythonfloating-pointexponent
nmadzharov
nmadzharov

Reputation: 502

Number of digits in exponent

Is it possible to set the number of digits to be used for printing the exponent of a floating-point number? I want to set it to 3.

Currently,

f = 0.0000870927939438012
>>> "%.14e"%f
'8.70927939438012e-05'
>>> "%0.14e"%f
'8.709279e-005'

What I want to print is: '8.70927939438012e-005'

Upvotes: 27

Views: 22918

Answers (4)

Anurag Uniyal
Anurag Uniyal

Reputation: 88777

There is a no way to control that, best way is to write a function for this e.g.

def eformat(f, prec, exp_digits):
    s = "%.*e"%(prec, f)
    mantissa, exp = s.split('e')
    # add 1 to digits as 1 is taken by sign +/-
    return "%se%+0*d"%(mantissa, exp_digits+1, int(exp))

print(eformat(0.0000870927939438012, 14, 3))
print(eformat(1.0000870927939438012e5, 14, 3))
print(eformat(1.1e123, 4, 4))
print(eformat(1.1e-123, 4, 4))

Output:

8.70927939438012e-005
1.00008709279394e+005
1.1000e+0123
1.1000e-0123

Upvotes: 23

sancho.s ReinstateMonicaCellio
sancho.s ReinstateMonicaCellio

Reputation: 15561

Similar to @Anurag Uniyal answer, but optionally removing the sign in the exponent (useful if one is tight in space).

def expformat(f, prec, exp_digits, sign='on'):
    """Scientific-format a number with a given number of digits in the exponent.
    Optionally remove the sign in the exponent"""
    s = "%.*e"%(prec, f)
    mantissa, exp = s.split('e')
    if (sign=='on') :
        # add 1 to digits as 1 is taken by sign +/-
        return "%se%+0*d"%(mantissa, exp_digits+1, int(exp))
    else :
        return "%se%0*d"%(mantissa, exp_digits, int(exp))

Upvotes: 0

Khalil Al Hooti
Khalil Al Hooti

Reputation: 4516

You can use np.format_float_scientific

from numpy import format_float_scientific

f = 0.0000870927939438012

format_float_scientific(f, exp_digits=3) # prints '8.70927939438012e-005'

format_float_scientific(f, exp_digits=5, precision=2) #prints '8.71e-00005'

Upvotes: 20

Ana Nimbus
Ana Nimbus

Reputation: 735

Here is a slightly more flexible answer (does either 'e' or 'E' to separate mantissa and exponent, is forgiving of missing/bad arguments). But I am upvoting the answer from @AnuragUniyal because that answer is so compact.

def efmte(x, n=6, m=2, e='e', nmax=16):
    def _expc(s): # return exponent character of e-formatted float
        return next(filter(lambda character: character in {'E', 'e'}, s))
    def _pad0(s, n): # return string padded to length n
        return ('{0:0>' + str(n) + '}').format(s)
    def _efmtes(s, n): # reformat e-formatted float: n e-digits
        m, e, p = s.partition(_expc(s)) # mantissa, exponent, +/-power
        return m + e + p[0] + _pad0(p[1:], n)
    def _efmt(x, n, e): # returns formatted float x: n decimals, 'e'/'E'
        return ('{0:.' + str(n) + e + '}').format(x)
    x = x if isinstance(x, float) else float('nan')
    nmax = 16 if not isinstance(nmax, int) else max(0, nmax)
    n = 6 if not isinstance(n, int) else min(max(0, n), nmax)
    m = 2 if not isinstance(m, int) else max(0, m)
    e = 'e' if e not in {'E', 'e'} else e
    return _efmtes(_efmt(x, n, e), m)

Examples:

>>> efmte(42., n=1, m=5)
'4.2e+00001'
>>> efmte('a')
'-1.#IND00e+00'
>>> # Yuck: I was expecting 'nan'. Anyone else?
>>> from math import pi
>>> efmte(pi)
'3.141593e+00'
>>> efmte(pi, m=3)
'3.141593e+000'

Upvotes: 0

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