How to specify floating point decimal precision from variable?

Question

I have the following repetitive simple code repeated several times that I would like to make a function for:

for i in range(10):
    id  = "some id string looked up in dict"
    val = 63.4568900932840928 # some floating point number in dict corresponding to "id"
    tabStr += '%-15s = %6.1f\n' % (id,val)

I want to be able to call this function: def printStr(precision)
Where it preforms the code above and returns tabStr with val to precision decimal points.

For example: printStr(3)
would return 63.457 for val in tabStr.

Any ideas how to accomplish this kind of functionality?

kennytm · Accepted Answer · 2017-08-31 06:16:11Z

50

tabStr += '%-15s = %6.*f\n' % (id, i, val)

where i is the number of decimal places.

BTW, in the recent Python where .format() has superseded %, you could use

"{0:<15} = {2:6.{1}f}".format(id, i, val)

for the same task.

Or, with field names for clarity:

"{id:<15} = {val:6.{i}f}".format(id=id, i=i, val=val)

If you are using Python 3.6+, you could simply use f-strings:

f"{id:<15} = {val:6.{i}f}"

edited Aug 31, 2017 at 6:16

answered Apr 6, 2011 at 22:36

kennytm

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mcragun · Accepted Answer · 2013-09-03 19:40:50Z

15

I know this an old thread, but there is a much simpler way to do this:

Try this:

def printStr(FloatNumber, Precision):
    return "%0.*f" % (Precision, FloatNumber)

answered Sep 3, 2013 at 19:40

mcragun

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LW001 · Accepted Answer · 2023-08-09 20:54:05Z

1

Tested in Python 3.8 and 3.9:

>>> val = 1.123456789
>>> decimals = 3
>>> f"{val:0.{decimals}f}"
'1.123'

edited Aug 9, 2023 at 20:54

LW001

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answered Aug 7, 2023 at 8:56

Felipe

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1 Comment

Itération 122442 Over a year ago

This solution has already been given.

nkarthiks · Accepted Answer · 2011-04-07 01:15:08Z

0

This should work too

tabStr += '%-15s = ' % id + str(round(val, i))

where i is the precision required.

edited Apr 7, 2011 at 1:15

answered Apr 7, 2011 at 1:08

nkarthiks

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