round does take negative ndigits parameter!

>>> round(46,-1)
50

may solve your case.

def round_down(num, divisor):
    return num - (num%divisor)

In [2]: round_down(19,10)
Out[2]: 10

In [3]: round_down(19,5)
Out[3]: 15

In [4]: round_down(10,10)
Out[4]: 10

Here's how to round a number to the nearest tenth:

>>> round(21.3331, 1)
21.3

Here's how to print out a floating point number with one decimal point:

>>> print "%.1f" % (21.3331,)
21.3

Note that if you do it using %d it will get printed as rounded to the nearest integer:

>>> print "%d" % (21.3331,)
21

See the String Formatting Operations docs for more on how the % formatting works.

If you always round to the largest previous power of 10, use $$r=10^{\lfloor\log_{10}(n)\rfloor},$$ where $\lfloor x\rfloor$ is the floor function.

• If you want to ceil, then you want to use math.ceil. This is as easy as:

  \$10 \lceil \frac{value}{10} \rceil\$

• You may want to make a generic function to ceil within a bound.

• You can set your somewhat weird limits to the function.

import math


def ceil(number, bound=1):
    return bound * math.ceil(number / bound)


def get_bin_pos(value):
    if value > 70:
        return None
    return ceil(max(value, 1), 10)

Actually, you could still use the round function:

>>> print round(1123.456789, -1)
1120.0

This would round to the closest multiple of 10. To 100 would be -2 as the second argument and so forth.