Python float formatting with no exponent and max precision not fixed precision

Question

I have a long list of floating point numbers to be formatted as follows:

Examples:

case 1) 6.0 -> 6.0 (No trailing zeros)

case 2) 1.23456789 -> 1.234567 (or 1.234568) (Max precision of 6)

case 3) 0.000004 -> 0.000004 (No exponent)

I can use

'{}'.format(round(x, 6))

for cases 1 & 2 but 3 gives 4e-06

If I use

'{:6f}'.format(6.0)

I get 6.000000 for case 1)

Is there a clean way to get the formatting I want?

Well, considering you want two different formats, you'll need two different format strings. — Peter Wood, Jan 30 '17 at 13:26
I don't want two different formats, just no trailing zeros, no exponent and max precision of 6 — Paul Hester, Jan 30 '17 at 14:22

score -1 · Answer 1 · answered Jan 30 '17 at 13:20

-1

Use %.6f%(data)

Hope it works !

answered Jan 30 '17 at 13:20

Jay Parikh

score -1 · Accepted Answer · edited May 23 '17 at 11:45

-1

Perhaps you can consider this solution.

Your case:

print ('%.6f' % data).rstrip('0').rstrip('.')

edited May 23 '17 at 11:45

Community

answered Jan 30 '17 at 13:33

NAmorim

I just knew it would have been asked before (or similar). This works. – Paul Hester Jan 30 '17 at 14:21

2 Answers2