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Re: new functional operator
Posted:
Mar 21, 2012 6:48 AM
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Here SIX several equivalent expressions from (IMHO) most intuitive or
readable to least:
Composition[g, f] /@ {1, 2, 3, 4}
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
g /@ f /@ {1, 2, 3, 4}
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
Apply[Composition, {g, f}] /@ {1, 2, 3, 4}
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
g@f@# & /@ {1, 2, 3, 4}
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
Compose[g, f@#] & /@ {1, 2, 3, 4}
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
{1, 2, 3, 4} // f /@ # & // g /@ # &
{g[f[1]], g[f[2]], g[f[3]], g[f[4]]}
The last is truly awful.
Bobby
On Tue, 20 Mar 2012 02:18:47 -0500, roby <roby.nowak@gmail.com> wrote:
>> That creates a information fog that makes *all* Mathematica code harder
>> to understand, and Mathematica much harder to learn than it used to be.
>
> {1, 2, 3, 4} /// f///g
>
>
>> {1, 2, 3, 4} // f /@ # & // g /@ # &
>
> sorry but I absolutly can't agree with your opinion in this case, the
> former expression is more or less fogless and would be much easier to
> understand.
> The latter expression bears a lot of clutter.
>
> Robert
>
>
>
--
DrMajorBob@yahoo.com
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