Assume that we have an operation $f$ such that $$ f([a_0,\ldots, a_n])= [f(a_0),\ldots,f(a_n)] $$

Do we say that $f$ distributes over lits or do we say it commutes with lists? Which one is the correct way of expressing this property?

How about the following? What do we say for the relation between $f$ and $\vec{a} = a_0,\ldots,a_n$? $$ f(g(\vec{a})) = g(f(\vec{a})) = g(f(a_0),\ldots,f(a_n)) $$

  • $\begingroup$ I don't have a reference on hand, but a dim memory suggests "component-wise function". If I had to choose between distributive and commutative, I'd go with distributive. $\endgroup$ – Luke Mathieson Jul 11 '13 at 9:34
  • $\begingroup$ @Luke, now I have to decide between three in place of two, nice job. ;) $\endgroup$ – Kaveh Jul 11 '13 at 9:47
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    $\begingroup$ $f$ has a different type on the left and right hand side of the equation – unless it is polymorphic. I'd go with distributive, too. $\endgroup$ – Dave Clarke Jul 11 '13 at 10:58
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    $\begingroup$ As @DaveClarke points out: $f_{\mathrm{list}}([a_0,\dots, a_n]) = [f_{\mathrm{element}}(a_0),\ldots,f_{\mathrm{element}}(a_n)]$. So I'd say, "$f_{\mathrm{list}}$ maps $f_{\mathrm{element}}$ over lists." I think we also say that any function that maps over lists is linear with respect to the list concatenation (addition) operation because $f_{\mathrm{list}}(a+b) = f_{\mathrm{list}}(a) + f_{\mathrm{list}}(b)$. map is what it's actually called in most programming languages too, so bonus. $\endgroup$ – Wandering Logic Jul 11 '13 at 12:15
  • $\begingroup$ Suggestion "componentwise" by @Luke seems in line with suggestion by wikipedia (taking the difference between operations and functions for granted). Some people seem to say it is an elementwise function. $\endgroup$ – Hendrik Jan Jul 11 '13 at 21:24

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