If you have Python 2.6 or newer, use format:
'{0:.3g}'.format(num)
For Python 2.5 or older:
'%.3g'%(num)
Explanation:
{0}tells format to print the first argument -- in this case, num.
Everything after the colon (:) specifies the format_spec.
.3 sets the precision to 3.
g removes insignificant zeros. See
http://en.wikipedia.org/wiki/Printf#fprintf
For example:
tests=[(1.00, '1'),
(1.2, '1.2'),
(1.23, '1.23'),
(1.234, '1.23'),
(1.2345, '1.23')]
for num, answer in tests:
result = '{0:.3g}'.format(num)
if result != answer:
print('Error: {0} --> {1} != {2}'.format(num, result, answer))
exit()
else:
print('{0} --> {1}'.format(num,result))
yields
1.0 --> 1
1.2 --> 1.2
1.23 --> 1.23
1.234 --> 1.23
1.2345 --> 1.23
Using Python 3.6 or newer, you could use f-strings:
In [40]: num = 1.234; f'{num:.3g}'
Out[40]: '1.23'
Answer from unutbu on Stack OverflowVideos
If you have Python 2.6 or newer, use format:
'{0:.3g}'.format(num)
For Python 2.5 or older:
'%.3g'%(num)
Explanation:
{0}tells format to print the first argument -- in this case, num.
Everything after the colon (:) specifies the format_spec.
.3 sets the precision to 3.
g removes insignificant zeros. See
http://en.wikipedia.org/wiki/Printf#fprintf
For example:
tests=[(1.00, '1'),
(1.2, '1.2'),
(1.23, '1.23'),
(1.234, '1.23'),
(1.2345, '1.23')]
for num, answer in tests:
result = '{0:.3g}'.format(num)
if result != answer:
print('Error: {0} --> {1} != {2}'.format(num, result, answer))
exit()
else:
print('{0} --> {1}'.format(num,result))
yields
1.0 --> 1
1.2 --> 1.2
1.23 --> 1.23
1.234 --> 1.23
1.2345 --> 1.23
Using Python 3.6 or newer, you could use f-strings:
In [40]: num = 1.234; f'{num:.3g}'
Out[40]: '1.23'
Only first part of Justin's answer is correct. Using "%.3g" will not work for all cases as .3 is not the precision, but total number of digits. Try it for numbers like 1000.123 and it breaks.
So, I would use what Justin is suggesting:
>>> ('%.4f' % 12340.123456).rstrip('0').rstrip('.')
'12340.1235'
>>> ('%.4f' % -400).rstrip('0').rstrip('.')
'-400'
>>> ('%.4f' % 0).rstrip('0').rstrip('.')
'0'
>>> ('%.4f' % .1).rstrip('0').rstrip('.')
'0.1'
I want to format a float so that it will round to two decimal places, but I am not sure how to do that. Can someone help me? I tried using round() but it doesn't work.
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
- Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
- Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True