Re: Python arrays and sting formatting options

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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