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?

Sennight

Answer 1

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BTW, in the recent Python where .format() has superseded %, you could use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

Cornett

Answer 2

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)

Amphithecium

Answer 3

This should work too

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

where i is the precision required.

Cornelius

Answer 4

Tested in Python 3.8 and 3.9:

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

Thornberry

This solution has already been given. – Judijudicable