Truncate to three decimals in Python

Question

How do I get 1324343032.324?

As you can see below, the following do not work:

>>1324343032.324325235 * 1000 / 1000
1324343032.3243253
>>i         


        

Answer 1

'%.3f'%(1324343032.324325235)

It's OK just in this particular case.

Simply change the number a little bit:

1324343032.324725235

And then:

'%.3f'%(1324343032.324725235)

gives you 1324343032.325

Try this instead:

def trun_n_d(n,d):
    s=repr(n).split('.')
    if (len(s)==1):
        return int(s[0])
    return float(s[0]+'.'+s[1][:d])

Another option for trun_n_d:

def trun_n_d(n,d):
    dp = repr(n).find('.') #dot position
    if dp == -1:  
        return int(n) 
    return float(repr(n)[:dp+d+1])

Yet another option ( a oneliner one) for trun_n_d [this, assumes 'n' is a str and 'd' is an int]:

def trun_n_d(n,d):
    return (  n if not n.find('.')+1 else n[:n.find('.')+d+1]  )

trun_n_d gives you the desired output in both, Python 2.7 and Python 3.6

trun_n_d(1324343032.324325235,3) returns 1324343032.324

Likewise, trun_n_d(1324343032.324725235,3) returns 1324343032.324

---

Note 1 In Python 3.6 (and, probably, in Python 3.x) something like this, works just fine:

def trun_n_d(n,d):
    return int(n*10**d)/10**d

But, this way, the rounding ghost is always lurking around.

Note 2 In situations like this, due to python's number internals, like rounding and lack of precision, working with n as a str is way much better than using its int counterpart; you can always cast your number to a float at the end.