Question

I am having difficulty converting a float to string in the following manner:

20.02  --> 20.02
20.016 --> 20.02
20.0   --> 20

It seems that%g format is the best for that, but I am getting strange results:

In [30]: "%.2g" % 20.03
Out[30]:  20 

In [31]: "%.2g" % 20.1
Out[31]:  20 

In [32]: "%.2g" % 20.3
Out[32]:  20 

In [33]: "%.2g" % 1.2
Out[33]:  1.2 

In [34]: "%.2g" % 1.0
Out[34]:  1 

In [35]: "%.2g" % 2.0
Out[35]:  2 

In [36]: "%.2g" % 2.2
Out[36]:  2.2 

In [37]: "%.2g" % 2.25
Out[37]:  2.2 

In [38]: "%.2g" % 2.26
Out[38]:  2.3 

In [39]: "%.3g" % 2.26
Out[39]:  2.26 

In [40]: "%.3g" % 2.25
Out[40]:  2.25 

In [41]: "%.3g" % 20.02
Out[41]:  20 

In [42]: "%.3g" % 20.016
Out[42]:  20 

In [43]: "%.20g" % 20.016
Out[43]:  20.015999999999998238

The only solution I know at the moment is checking whether the number is an int and applying %d instead of %f formatting - this is too complicated, I think.

Does anybody know why the things above are hapenning? How to do this in a simpler way?

Thanks.

Answer 1

Using %f format specifier:

( %.2f  % (value,)).rstrip( 0 ).rstrip( . )

Using round() function:

str(round(value)).rstrip( 0 ).rstrip( . )

Answer 2

Use round together with %g -- you want at most 2 digits shown, so round to two digits, then use %g to print it as short as possible:

>>> "%g" % round(20.016, 2)
 20.02 
>>> "%g" % round(20, 2)
 20

Answer 3

I m not sure what exactly you are finding complicated here -- you re getting exactly the results specified e.g. here. E.g.:

In [32]: "%.2g" % 20.3
Out[32]:  20 

In [33]: "%.2g" % 1.2
Out[33]:  1.2

In each case, you have requested that 2 digits be shown in all, and that s what is happening (both digits come before the trailing dot in one case, one before and one after in the other case, but that s an obvious consequence of the numbers respective magnitudes).

When you ask for 20 digits, you re shown 20 digits -- most aren t meaningful of course (a double-precision IEEE floating point is good for only about 16 digits of precision), so it s more sensible to ask for fewer. You do know of course that floats are represented in binary, as explained here, right? Use decimal (much slower, of course, since what your computer s hardware supplies is binary floating point, the decimal version must all be synthesized in software) is what you need are decimal-represented floats (e.g., for monetary computations).

Answer 4

Try something a little less than 20 digits. If you use 20 digits, it will give you the 20 digit representation of the floating point value, which might look a bit odd.

Answer 5

use built-in function round

Answer 6

You can use .format instead of %.

>>>  {:.2f} .format(3.0101)
 3.01 
>>> type( {:.2f} .format(3.0101))
str

There is lot more stuff which is not possible using % but can be done using .format.

For more formatting details, take a look at pyformat

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