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I want my Python (2.4.3) output numbers to have a certain format. Specifically, if the number is a terminating decimal with <= 6 significant digits, show it all. However, if it has > 6 significant digits, then output only 6 significant digits.

"A" shows how Python is writing the floats. "B" shows how I want them written. How can I make Python format my numbers in that way?

A: 10188469102.605597 5.5657188485 3.539 22.1522612479 0 15.9638450858 0.284024 7.58096703786 24.3469152383 B: 1.01885e+10 5.56572 3.539 22.1523 0 15.9638 0.284024 7.58097 24.3469
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6
  • Floats lack the precision required to make this possible. Commented Sep 11, 2014 at 5:56
  • "Format decimal" - I think you mean "format float", because Decimal is a library in Python. Commented Sep 11, 2014 at 6:15
  • @BurhanKhalid yes, you're correct. Commented Sep 11, 2014 at 6:20
  • possible duplicate of Nicely representing a floating-point number in python Commented Sep 11, 2014 at 6:20
  • You really need to accept joachim's answer, not mine. Commented Sep 11, 2014 at 6:23

3 Answers 3

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74

You'll want the g modifier for format that drops insignificant zeroes;

>>> "{0:.6g}".format(5.5657188485) '5.56572' >>> "{0:.6g}".format(3.539) '3.539'

Sorry, my update also includes the fact that I am restricted to using Python 2.4.3, which does not have format() function.

The format specifiers work even without the .format() function:

>>> for i in a: ... print '%.6g' % (i,) ... 1.01885e+10 5.56572 3.539 22.1523 0 15.9638 0.284024 7.58097 24.3469
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3 Comments

Updated my question with a large number in example as well
@user1145925 As far as I can see, the format gives your exact B output.
Sorry, my update also includes the fact that I am restricted to using Python 2.4.3, which does not have format() function.
20

There is a way to retain trailing zeros so that it consistently shows the number of significant digits. Not exactly what OP wanted, but probably useful to many.

a = [10188469102.605597,5.5657188485,3.539,22.1522612479,0,15.9638450858,0.284024,7.58096703786,24.3469152383] for i in a: print("{:#.6g}".format(i))

Output

1.01885e+10 5.56572 3.53900 22.1523 0.00000 15.9638 0.284024 7.58097 24.3469

Note that this will only work with the format function and not with % operator.

According to the docs:

The '#' option causes the “alternate form” to be used for the conversion. The alternate form is defined differently for different types. This option is only valid for integer, float, complex and Decimal types.

'g': General format ... insignificant trailing zeros are removed from the significand, and the decimal point is also removed if there are no remaining digits following it, unless the '#' option is used.

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thanks; according to the docs, the '#' goes btw sign and the optional '0': docs.python.org/3.4/library/string.html
4

try this way 

a=[10188469102.605597,5.5657188485,3.539,22.1522612479,0,15.9638450858,0.284024,7.58096703786,24.3469152383] for i in a: if i >100: print '{:.6e}'.format(i) else: print '{:.6f}'.format(i)

for lower version of python

for i in a: if i >100: print '%6e'%i else: print '%6f'%i

output

1.018847e+10 5.565719 3.539000 22.152261 0.000000 15.963845 0.284024 7.580967 24.346915
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1 Comment

I cannot use format() function. That is not available for Python 2.4.3.

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