部隊蟒蛇不輸出標準形式/科學計數法浮點值/指數形式

Question

所以這工作得很好：

>>> float(1.0e-1) 
0.10000000000000001 

但具有較大數量的問題時，它不會打印：

>>> float(1.0e-9) 
1.0000000000000001e-09 

有沒有辦法強制這個？也許使用numpy或其他東西。

Answer 1

print '{0:.10f}'.format(1.0e-9) 

String formatting在文檔中。

Answer 2

這是非常標準的打印格式，專爲浮動：

print "%.9f" % 1.0e-9 

Answer 3

>>> a 
1.0000000000000001e-09 
>>> print "heres is a small number %1.9f" %a 
heres is a small number 0.000000001 
>>> print "heres is a small number %1.13f" %a 
heres is a small number 0.0000000010000 
>>> b 
11232310000000.0 
>>> print "heres is a big number %1.9f" %b 
heres is a big number 11232310000000.000000000 
>>> print "heres is a big number %1.1f" %b 
heres is a big number 11232310000000.0 

Answer 4

每個人都建議使用f字符串格式代碼都隱含假設它沒關係後修復的位數小數點。這對我來說似乎是一個非常不穩定的假設。但是，如果你沒有做出這樣的假設，就沒有內置的機制去做你想做的事情。這是我遇到類似問題時遇到的最好的黑客（在PDF生成器中 - PDF中的數字不能使用指數表示法）。你可能想要把所有的b關掉，並且在這裏可能還有其他的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 

Answer 5

使用％E打印您的號碼時：

>>> a = 0.1234567 
>>> print 'My number is %.7e'%a 
My number 1.2345670e-01 

如果你使用％G它會自動選擇最佳的可視化您：

>>> print 'My number is %.7g'%a 
My number is 0.1234567 

Answer 6

這裏的zwol的答案簡化和轉換到標準的python格式：

import re 
def format_float_in_standard_form(f): 
    s = str(f) 
    m = re.fullmatch(r'(-?)(\d)(?:\.(\d+))?e([+-]\d+)', s) 
    if not m: 
     return s 
    sign, intpart, fractpart, exponent = m.groups('') 
    exponent = int(exponent) + 1 
    digits = intpart + fractpart 
    if exponent < 0: 
     return sign + '0.' + '0'*(-exponent) + digits 
    exponent -= len(digits) 
    return sign + digits + '0'*exponent + '.0'