On Tue, 30 Sep 2008 03:56:03 +0200, Ivan Reborin wrote:
> a = 2.000001
> b = 123456.789
> c = 1234.0001
> d = 98765.4321
> # same as above except for d
>
> print (3 * '%12.3f') % (a, b, c)
> #this works beautifully
>
> How to add d at the end but with a different format now, since I've
> "used" the "format part" ?
>
> Again, my weird wishful-thinking code: print (3*'%12.3f', '%5.3f')
> %(a,b,c),d
Maybe you should stop that wishful thinking and programming by accident
and start actually thinking about what the code does, then it's easy to
construct something working yourself.
The ``%`` operator on strings expects a string on the left with format
strings in it and a tuple with objects to replace the format strings
with. So you want
'%12.3f%12.3f%12.3f%5.3f' % (a, b, c, d)
But without repeating the '%12.3f' literally. So you must construct that
string dynamically by repeating the '%12.3f' and adding the '%5.3f':
In [27]: 3 * '%12.3f'
Out[27]: '%12.3f%12.3f%12.3f'
In [28]: 3 * '%12.3f' + '%5.3f'
Out[28]: '%12.3f%12.3f%12.3f%5.3f'
Now you can use the ``%`` operator on that string:
In [29]: (3 * '%12.3f' + '%5.3f') % (a, b, c, d)
Out[29]: ' 2.000 123456.789 1234.00098765.432'
(I guess there should be at least a space before the last format string.)
This time you *have* to put parenthesis around the construction of the
format string BTW because ``%`` has a higher priority than ``+``. So
implicit parentheses look like this:
3 * '%12.3f' + '%5.3f' % (a, b, c, d)
<=> 3 * '%12.3f' + ('%5.3f' % (a, b, c, d))
And there are of course not enough formatting place holders for four
objects in '%5.3f'.
It's also important to learn why your wrong codes fail. In your wishful
thinking example you will get a `TypeError` saying "unsupported operand
type(s) for %: 'tuple' and 'tuple'". That's because on the left side of
the ``%`` operator you wrote a tuple:
In [34]: (3 * '%12.3f', '%5.3f')
Out[34]: ('%12.3f%12.3f%12.3f', '%5.3f')
Ciao,
Marc 'BlackJack' Rintsch
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