Number of digits in exponent
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-0051.00008709279394e+0051.1000e+01231.1000e-0123
You can use np.format_float_scientific
from numpy import format_float_scientificf = 0.0000870927939438012format_float_scientific(f, exp_digits=3) # prints '8.70927939438012e-005'format_float_scientific(f, exp_digits=5, precision=2) #prints '8.71e-00005'
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'