string formatting: engineering notation

[Darren Dale]

Does anyone know if it is possible to represent a number as a string with
engineering notation (like scientific notation, but with 10 raised to
multiples of 3: 120e3, 12e-6, etc.). I know this is possible with the
decimal.Decimal class, but repeatedly instantiating Decimals is inefficient
for my application (matplotlib plotting library). If it is not currently
possible, do you think the python devs would be receptive to including
support for engineering notation in future releases?

Thanks,
Darren

 

[Steve Holden]

Does anyone know if it is possible to represent a number as a string with
engineering notation (like scientific notation, but with 10 raised to
multiples of 3: 120e3, 12e-6, etc.). I know this is possible with the
decimal.Decimal class, but repeatedly instantiating Decimals is inefficient
for my application (matplotlib plotting library). If it is not currently
possible, do you think the python devs would be receptive to including
support for engineering notation in future releases?

How close is this:
'3.142e+00'

regards
Steve
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[Darren Dale]

How close is this:

'3.142e+04'

What I am looking for is '31.4159e+03'

Darren

 

[Jorge Godoy]

How close is this:

Not close when you have the exponent.

 

[Grant Edwards]

How close is this:

'3.142e+00'

Not close at all. It should be "3.14159"
'3.142e+01'

should be 31.4159
'3.142e+02'

should be 314.159
'3.142e+04'

should be 31.4159e3

 

[attn.steven.kuo]

Does anyone know if it is possible to represent a number as a string with
engineering notation (like scientific notation, but with 10 raised to
multiples of 3: 120e3, 12e-6, etc.). I know this is possible with the
decimal.Decimal class, but repeatedly instantiating Decimals is inefficient
for my application (matplotlib plotting library). If it is not currently
possible, do you think the python devs would be receptive to including
support for engineering notation in future releases?

Do you also consider this to be too inefficient?


import math

for exponent in xrange(-10, 11):
flt = 1.23 * math.pow(10, exponent)
l = math.log10(flt)
if l < 0:
l = l - 3
p3 = int(l / 3) * 3
multiplier = flt / pow(10, p3)
print '%e => %fe%d' % (flt, multiplier, p3)

 

[Darren Dale]

Do you also consider this to be too inefficient?


import math

for exponent in xrange(-10, 11):
flt = 1.23 * math.pow(10, exponent)
l = math.log10(flt)
if l < 0:
l = l - 3
p3 = int(l / 3) * 3
multiplier = flt / pow(10, p3)
print '%e => %fe%d' % (flt, multiplier, p3)

That's a good suggestion. It's probably fast enough. I was hoping that
something like '%n'%my_number already existed.