问题
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?
回答1:
The easy way is replace! One simple example:
value=str('6,0865000000e-01')
value2=value.replace(',', '.')
float(value2)
0.60865000000000002
回答2:
The reason is the use of comma in 6,0865000000e-01. This won't work because float() is not locale-aware. See PEP 331 for details.
Try locale.atof(), or replace the comma with a dot.
回答3:
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)
回答4:
Simply by casting string into float:
new_val = float('9.81E7')
回答5:
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
回答6:
The float is correct, just use format to display it as you want, i.e.:
print format(the_float, '.8f')
来源:https://stackoverflow.com/questions/9195800/converting-exponential-to-float