所以这很好用:
>>> float(1.0e-1)
0.10000000000000001
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但是当处理更大的数字时,它将无法打印:
>>> float(1.0e-9)
1.0000000000000001e-09
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有没有办法强迫这个?也许使用numpy或其他东西.
zwo*_*wol 11
每个人都建议使用f字符串格式代码隐含地假设可以修复小数点后的位数.这对我来说似乎是一个非常不稳定的假设.但是,如果您没有做出这样的假设,那么就没有内置机制来做您想做的事情.这是我遇到类似问题时遇到的最好的黑客(在PDF生成器中 - PDF中的数字不能使用指数表示法).您可能希望将所有bs从字符串中删除,并且此处可能还有其他Python3-isms.
_ftod_r = re.compile(
br'^(-?)([0-9]*)(?:\.([0-9]*))?(?:[eE]([+-][0-9]+))?$')
def ftod(f):
"""Print a floating-point number in the format expected by PDF:
as short as possible, no exponential notation."""
s = bytes(str(f), 'ascii')
m = _ftod_r.match(s)
if not m:
raise RuntimeError("unexpected floating point number format: {!a}"
.format(s))
sign = m.group(1)
intpart = m.group(2)
fractpart = m.group(3)
exponent = m.group(4)
if ((intpart is None or intpart == b'') and
(fractpart is None or fractpart == b'')):
raise RuntimeError("unexpected floating point number format: {!a}"
.format(s))
# strip leading and trailing zeros
if intpart is None: intpart = b''
else: intpart = intpart.lstrip(b'0')
if fractpart is None: fractpart = b''
else: fractpart = fractpart.rstrip(b'0')
if intpart == b'' and fractpart == b'':
# zero or negative zero; negative zero is not useful in PDF
# we can ignore the exponent in this case
return b'0'
# convert exponent to a decimal point shift
elif exponent is not None:
exponent = int(exponent)
exponent += len(intpart)
digits = intpart + fractpart
if exponent <= 0:
return sign + b'.' + b'0'*(-exponent) + digits
elif exponent >= len(digits):
return sign + digits + b'0'*(exponent - len(digits))
else:
return sign + digits[:exponent] + b'.' + digits[exponent:]
# no exponent, just reassemble the number
elif fractpart == b'':
return sign + intpart # no need for trailing dot
else:
return sign + intpart + b'.' + fractpart
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