将指数转换为浮点数

发布于 2025-01-03 16:00:22 字数 789 浏览 2 评论 0原文

这是我的代码，尝试将该行的第二个字段从指数转换为浮点数。

outputrrd = processrrd.communicate()
(output, error) = outputrrd
output_lines = output.split('\n')
for line in output_lines:
    m = re.search(r"(.*): ", line)
    if m != None:
        felder = line.split(': ')
        epoch =  felder[0].strip(':')
        utc = epoch2normal(epoch).strip("\n")
        #print felder[1]
        data = float(felder[1])
        float_data = data * 10000000
        print float_data
        resultslist.append( utc + ' ' + hostname + ' ' +  float_data)

但是，程序因以下错误而停止：

File "/opt/omd/scripts/python/livestatus/rrdfetch-convert.py", line 156, in <module>
    data = float(felder[1])
ValueError: invalid literal for float(): 6,0865000000e-01

有人知道原因吗？

原文

This is my code, trying to convert the second field of the line from exponential into float.

outputrrd = processrrd.communicate()
(output, error) = outputrrd
output_lines = output.split('\n')
for line in output_lines:
    m = re.search(r"(.*): ", line)
    if m != None:
        felder = line.split(': ')
        epoch =  felder[0].strip(':')
        utc = epoch2normal(epoch).strip("\n")
        #print felder[1]
        data = float(felder[1])
        float_data = data * 10000000
        print float_data
        resultslist.append( utc + ' ' + hostname + ' ' +  float_data)

But, the program stops with this error:

File "/opt/omd/scripts/python/livestatus/rrdfetch-convert.py", line 156, in <module>
    data = float(felder[1])
ValueError: invalid literal for float(): 6,0865000000e-01

Does anyone know the reason?

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昇り龍 2025-01-10 16:00:23

最简单的方法就是更换！一个简单的例子：

value=str('6,0865000000e-01')
value2=value.replace(',', '.')
float(value2)
0.60865000000000002

The easy way is replace! One simple example:

value=str('6,0865000000e-01')
value2=value.replace(',', '.')
float(value2)
0.60865000000000002

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樱花落人离去 2025-01-10 16:00:23

原因是 6,0865000000e-01 中使用了逗号。这不起作用，因为 float() 不支持区域设置。有关详细信息，请参阅 PEP 331。

尝试 locale.atof()，或者将逗号替换为点。

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朮生 2025-01-10 16:00:23

float 值是正确的；您只需根据需要显示它，例如：

print(format(the_float, '.8f'))
# print(f'{the_float:.8f}') # For Python >= 3.6

The float value is correct; you simply need to display it as desired, e.g.:

print(format(the_float, '.8f'))
# print(f'{the_float:.8f}') # For Python >= 3.6

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伴我心暖 2025-01-10 16:00:23

这对我有用，试试吧。

def remove_exponent(value):
    decial = value.split('e')
    ret_val = format(((float(decial[0]))*(10**int(decial[1]))), '.8f')
    return ret_val

This work for me, try it out.

def remove_exponent(value):
    decial = value.split('e')
    ret_val = format(((float(decial[0]))*(10**int(decial[1]))), '.8f')
    return ret_val

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饮湿 2025-01-10 16:00:23

我认为这对你有用：

def remove_exponent(value):
    """
       >>>(Decimal('5E+3'))
       Decimal('5000.00000000')
    """
    decimal_places = 8
    max_digits = 16

    if isinstance(value, decimal.Decimal):
        context = decimal.getcontext().copy()
        context.prec = max_digits
        return "{0:f}".format(value.quantize(decimal.Decimal(".1") ** decimal_places, context=context))
    else:
        return "%.*f" % (decimal_places, value)

I think it is useful to you:

def remove_exponent(value):
    """
       >>>(Decimal('5E+3'))
       Decimal('5000.00000000')
    """
    decimal_places = 8
    max_digits = 16

    if isinstance(value, decimal.Decimal):
        context = decimal.getcontext().copy()
        context.prec = max_digits
        return "{0:f}".format(value.quantize(decimal.Decimal(".1") ** decimal_places, context=context))
    else:
        return "%.*f" % (decimal_places, value)

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浅浅 2025-01-10 16:00:23

只需将字符串转换为浮点数即可：

new_val = float('9.81E7')

Simply by casting string into float:

new_val = float('9.81E7')

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琴流音 2025-01-10 16:00:23

我在尝试从科学/指数表示法中的字符串转换为浮点数时遇到了类似的问题（如果太长，也可能导致指数表示法）

num = '-8e-05'

def scientific_to_float(exponential):
  split_word = 'e'
  e_index = exponential.index('e')
  base = float(exponential[:e_index])
  exponent = float(exponential[e_index + 1:])
  float_number = base * (10 ** exponent)
  return float_number

scientific_to_float(num) # return -8e-05 float number

I had a similar issue trying to convert from string in scientific/exponential notation to a float number (that can result also in exponential notation if too long)

num = '-8e-05'

def scientific_to_float(exponential):
  split_word = 'e'
  e_index = exponential.index('e')
  base = float(exponential[:e_index])
  exponent = float(exponential[e_index + 1:])
  float_number = base * (10 ** exponent)
  return float_number

scientific_to_float(num) # return -8e-05 float number

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