Michael Spencer wrote:
> Neal Becker wrote:
>
>>Like a puzzle? I need to interface python output to some strange old
>>program. It wants to see numbers formatted as:
>>
>>e.g.: 0.23456789E01
>>
>>That is, the leading digit is always 0, instead of the first significant
>>digit. It is fixed width. I can almost get it with '% 16.9E', but not
>>quite.
>>
>>My solution is to print to a string with the '% 16.9E' format, then parse it
>>with re to pick off the pieces and fix it up. Pretty ugly. Any better
>>ideas?
>>
>>
>
> Does this do what you want?
Not, if the input is 0 or 1. Here's a correction, with a more comprehensive
test
from math import log10, modf, fabs
def format(n, mantplaces = 9, expplaces = 2):
"""Formats n as '0.mmmmmmmmmEee'"""
if n:
sign, absn = n/fabs(n), fabs(n)
f, i = modf(log10(absn))
mant, exp = sign * 10** (f - (f>=0)), i + (f>=0)
else:
mant, exp = 0, 0
return "%.*fE%0*d" % (mantplaces, mant, expplaces, exp)
def test_format(N = 10000, step = 1):
"""Verifies format(n) and format(1/n) for -N < n < N"""
assert format(0,9) == '0.000000000E00'
assert format(0, 7, 3) == '0.0000000E000'
def verify(n):
DIGITS = '123456789'
try:
f = format(n)
assert round(float(format(n)),6) == round(n, 6)
assert f[0] == "-" and f[3] in DIGITS or f[2] in DIGITS
except AssertionError:
raise AssertionError("Failed on: %f, formatted as %s" % (n, f))
for n in xrange(-N, N, step):
if n:
verify(n)
verify(1.0/n)
Michael
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