list manipulation - How to repeat the nonblank string?

As the title, if I have a list:

{"", "", "", "2$70", ""}

I will expect:

{"", "", "", "2$70", "2$70"}

If I have

{"", "", "", "3$71", "", "2$72", ""}

then:

{"", "", "", "3$71", "3$71", "2$72", "2$72"}

And

{"", "", "", "3$71", "","", "2$72", ""}

should give

{"", "", "", "3$71", "3$71", "", "2$72", "2$72"}

This is my try:

{"", "", "", "2$70", ""} /. {p : Except["", String], ""} :> {p, p}

But I don't know why it doesn't work. Poor ability of pattern match. Can anybody give some advice?

Answer

As I presently interpret the question

(Now with refinements after considering Chris Degnen's simultaneous answer)

fn[list_] :=
  Partition[list, 2, 1, -1, ""] // Cases[{p_, ""} | {_, p_} :> p]

Test:

 {"", "x", "y", "", "z", ""} // fn

{"", "x", "y", "y", "z", "z"}

Patterns

Since you seem only to be interested in a pattern-matching solution here is my proposal to avoid the extremely slow use of ReplaceRepeated while still using pattern-matching as the core of the operation.

fn2[list_] :=
  list // ReplacePart @ 
   ReplaceList[list, {a___, p_, "", ___} :> (2 + Length@{a} -> p)]

Recursive replacement

I just realized that this is a perfect time to use a self-referential replacement:

fn3[list_] :=
  list /. {a___, p_, "", b___} :> {a, p, p, ##& @@ fn3@{b}}

Benchmark

All three methods are much faster than kglr's foo (note the log-log scale).

Now with Carl Woll's fc, the fastest method yet. ( but no patterns ;-) )

Needs["GeneralUtilities`"]

$RecursionLimit = 1*^5;

BenchmarkPlot[{foo, fn, fn2, fn3, fc},
  RandomChoice[{"", "", "", "a", "b"}, #] &

]

Comments

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