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Groups > comp.lang.python > #17537
| Date | 2011-12-19 20:42 -0600 |
|---|---|
| From | Tim Chase <python.list@tim.thechases.com> |
| Subject | Re: Transform two tuples item by item |
| References | <21814930.66.1324346655123.JavaMail.geo-discussion-forums@prmw6> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.3840.1324348945.27778.python-list@python.org> (permalink) |
On 12/19/11 20:04, Gnarlodious wrote:
> What is the best way to operate on a tuple of values
> transforming them against a tuple of operations? Result can be
> a list or tuple:
>
> tup=(35, '34', 0, 1, 31, 0, '既濟')
>
> from cgi import escape
> [tup[0], "<span> class='H'>{}</span>".format(tup[1]), bool(tup[2]),
> bool(tup[3]), tup[4], bool(tup[5]), escape(tup[6])]
>
> -> [35, "<span class='H'>34</span>", False, True, 31, False,
> '&#26082;&#28639;']
>
> But I want to loop rather than subscripting.
Well, you can do something like
>>> from cgi import escape
>>> nop = lambda x: x
>>> tup = (35, '34', 0, 1, 31, 0, '既濟')
>>> ops = [nop, "<span class='H'>{0}</span>".format, bool, bool,
nop, bool, escape]
>>> [f(x) for f, x in zip(ops,tup)]
[35, "<span class='H'>34</span>", False, True, 31, False,
'&#26082;&#28639;']
Note #1: I had to change your format from "{}" to "{0}", at least
in 2.6 I've got here)
Note #2: it's spelled ".format" not ".format()" which puts the
function reference in the "ops" list, not the results of calling it.
-tkc
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Transform two tuples item by item Gnarlodious <gnarlodious@gmail.com> - 2011-12-19 18:04 -0800 Re: Transform two tuples item by item Tim Chase <python.list@tim.thechases.com> - 2011-12-19 20:42 -0600 Re: Transform two tuples item by item Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-12-20 03:40 +0000 Re: Transform two tuples item by item Gnarlodious <gnarlodious@gmail.com> - 2011-12-19 19:40 -0800
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