With Python < 3 (e.g. 2.6 [see comments] or 2.7), there are two ways to do so.

# Option one
older_method_string = "%.9f" % numvar

# Option two
newer_method_string = "{:.9f}".format(numvar)

But note that for Python versions above 3 (e.g. 3.2 or 3.3), option two is preferred.

For more information on option two, I suggest this link on string formatting from the Python documentation.

And for more information on option one, this link will suffice and has info on the various flags.

Python 3.6 (officially released in December of 2016), added the f string literal, see more information here, which extends the str.format method (use of curly braces such that f"{numvar:.9f}" solves the original problem), that is,

# Option 3 (versions 3.6 and higher)
newest_method_string = f"{numvar:.9f}"

solves the problem. Check out @Or-Duan's answer for more info, but this method is fast.

but should I really force manually how many decimal numbers I want? Yes.

And even with specifying 10 decimal digits, you are still not printing all of them. Floating point numbers don't have that kind of precision anyway, they are mostly approximations of decimal numbers (they are really binary fractions added up). Try this:

>>> format(38.2551994324, '.32f')
'38.25519943239999776096738060005009'

there are many more decimals there that you didn't even specify.

When formatting a floating point number (be it with '%f' % number, '{:f}'.format(number) or format(number, 'f')), a default number of decimal places is displayed. This is no different from when using str() (or '%s' % number, '{}'.format(number) or format(number), which essentially use str() under the hood), only the number of decimals included by default differs; Python versions prior to 3.2 use 12 digits for the whole number when using str().

If you expect your rational number calculations to work with a specific, precise number of digits, then don't use floating point numbers. Use the decimal.Decimal type instead:

• Decimal “is based on a floating-point model which was designed with people in mind, and necessarily has a paramount guiding principle – computers must provide an arithmetic that works in the same way as the arithmetic that people learn at school.” – excerpt from the decimal arithmetic specification.

• Decimal numbers can be represented exactly. In contrast, numbers like 1.1 and 2.2 do not have exact representations in binary floating point. End users typically would not expect 1.1 + 2.2 to display as 3.3000000000000003 as it does with binary floating point.

You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.

With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).

Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.

For example,

>>> 125650429603636838/(2**53)
13.949999999999999

>>> 234042163/(2**24)
13.949999988079071

>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999

If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:

1. Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
2. Or use a fixed point number like decimal.

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}"