Python: Float to string with variable number of positions after the decimal point

Question

I need to represent float values as strings with different number of positions after the decimal point, which should be set by variable , not hardcoded like "%.2f" .

Now I'm using round approach:

def float_to_str(value, digits):
    return str(round(value, digits))

This function cuts extra digits, so the result has no more that specified number of digits. But it doesn't add trailing zeros.

float_to_str(1.3573, 2)   =>  "1.35"  - OK 
float_to_str(1.3573, 3)   =>  "1.357" - OK 
float_to_str(1.3000, 2)   =>  "1.3"   - NOT OK (needed "1.30")

Is there a nicer way to implement this function?

Answer 1

You should use the format function, which can use syntax similar to "%.2f" , and format that as well:

def float_to_str(value, digits):
    return "{value:.{digits}f}".format(value=value, digits=digits)

Answer 2

The formatting mini language also allows for nested replacement:

def float_to_str(value, digits):
    return f"{value:.{digits}f}"

Answer 3

You can use * to denote a variable width in your format string:

def float_to_str(value, digits):
    return "%.*f" % (digits, value)

Answer 4

Try this code.

def float_to_str(value, digits):
    formatting = '{:.' + str(digits) + 'f}'
    return formatting.format(round(value, digits))

For more information about rounding a number but keeping ending zeros in python, see this

Answer 5

Using f-strings is a possible way to go:

def float_to_str(x, n):
    return f"{x:.{n}f}"