Question

I need help understanding whats going on in the function, especially the return statements. I know what the return statements do but not how they do it. I know that they format the string but I just don’t understand how its being done. It would help if you guys take it step by step.

def intF(n, d, l=40):

    s=str(n*10**l / d) 
    if len(s) < l: 
        return '0.{:0>{width}}'.format(s,width=l) 
    if len(s) > l: 
        return s[0:len(s)-l]+'.'+s[len(s)-l:]  

    return '0.'+s

Answer 1

Here’s a line-by-line breakdown:

def intF(n, d, l=40):

Pretty obvious. n is a number, d is another number (the divisor) and l is the number of digits to print after the decimal point.

    s=str(n*10**l / d)

This does something a bit unusual. Rather than relying on floating point arithmetic, this multiplies n by 10 ** l, i.e. by a 1 followed by l digits. That way the final result won’t have any floating point error — assuming d is always an integer. (But of course any remaining digits get truncated. Also, replace / with // in Python 3 to get the same behavior.)

At this point, s will be a string representation of a whole number — again assuming d is an integer — but it will have the same digits as the result of float(n) / d. So now we just have to insert the decimal point in the right place.

    if len(s) < l: 
        return '0.{:0>{width}}'.format(s,width=l) 

If the length of s is less than l, then we need to pad it and prepend a 0.. That’s what this does. The {:0>{width}} field says to create a zero-padded field of width width, and insert a value into it on the right (>) side. Then s is passed in via format, and we have our result.

    if len(s) > l: 
        return s[0:len(s)-l]+'.'+s[len(s)-l:]

If the length of a is greater than l, then we need to insert the decimal point in the correct spot. That’s what this does. It removes the trailing l digits from s, appends a ., and then appends the remaining l digits.

    return '0.'+s

The final possibility is that s is exactly l digits long. In that case, we don’t need to do any padding; we can just prepend a 0 and a decimal point.

As a final note: if you pass anything but integers to this function, it will not work as expected. Consider this:

>>> intF(10, 10.1, 10)
'990.0990099.01'

Or this:

>>> intF(10.1, 10, 10)
'101.00000000.0'