Python setting Decimal Place range without rounding?

You can do:

def truncate(f, n):
    return math.floor(f * 10 ** n) / 10 ** n

testing:

>>> f=1.923328437452
>>> [truncate(f, n) for n in range(7)]
[1.0, 1.9, 1.92, 1.923, 1.9233, 1.92332, 1.923328]

A super simple solution is to use strings

x = float (str (w)[:-1])
y = float (str (w)[:-2])
z = float (str (w)[:-3])

Any of the floating point library solutions would require you dodge some rounding, and using floor/powers of 10 to pick out the decimals can get a little hairy by comparison to the above.

Integers are faster to manipulate than floats/doubles which are faster than strings. In this case, I tried to get time with both approach :

  timeit.timeit(stmt = "float(str(math.pi)[:12])", setup = "import math", number = 1000000)

~1.1929605630000424

for :

timeit.timeit(stmt = "math.floor(math.pi * 10 ** 10) / 10 ** 10", setup = "import math", number = 1000000)

~0.3455968870000561

So it's safe to use math.floor rather than string operation on it.