Decimal numbers get round up while formatting python

Question

Here is what I have tried:

>>> pi = 3.14159265
>>> format(pi, '.3f') #print 3.142 # 3 precision after the decimal point
'3.142'
>>> format(pi, '.1f') #print 3.1
'3.1'
>>> format(pi, '.10f') #print 3.1415926500, more precision than the original
'3.1415926500'
>>> format(pi, '.5f') #print 3.14159, more precision than the original
'3.14159'
>>> format(pi, '.4f') 
'3.1416'

The concern part is this:

>>> format(pi, '.3f') #print 3.142 # 3 precision after the decimal point
'3.142'
>>> format(pi, '.4f') 
'3.1416'

Where I was expecting to have 3.1415, I am getting 3.1416. Please suggest me.
The SO is showing these 2 links:
http://stackoverflow.com/questions/21895756/why-are-floating-point-numbers-inaccurate
http://stackoverflow.com/questions/1089018/why-cant-decimal-numbers-be-represented-exactly-in-binary
But these are not the one I am looking for.

Answer 1

It's because it is rounding it, so it is just like the round function.

To fix it:

>>> l = str(pi).split('.')
>>> l[0] + '.' + l[1][:4]
'3.1415'
>>> float(l[0] + '.' + l[1][:4])
3.1415

A function version of it:

def first_n(a, b):
    l = str(a).split('.')
    return int(l[0] + '.' + l[1][:b])

And now:

print(first_n(pi, 4))

Gives:

3.1415

Answer 2

You could simply strip the last character:

pi = 3.14159

print(format(pi, '.5f')[:-1]) # 3.1415

Answer 3

I'm not sure what kind of suggestion you want, but here's a way to truncate a number to a given number of decimal places:

pi = 3.14159265

def truncate(v, places):
    return int(v * 10**places) / 10**places

print(truncate(pi, 3))  # -> 3.141
print(truncate(pi, 4))  # -> 3.1415

Answer 4

You may try this. May be exactly list the answer @U9-Forward, a little compact.

>>> str(pi)[:6]
'3.1415'

Hope this will help.

Answer 5

there is 3.1416 for format(pi, '.4f') for the same reason there is 3.142 for format(pi, '.3f') - the next digit after your desired number of digits in formatted output is at least five - it is rounded up.