Re^4: pattern matching from an array to an array of arrays

Ok. So my starting data look like this:

1) this is how each array in @AoA is structured

[someword firstvalue secondvalue ...]

[otherword first value secondvalue ...]

[....]

2) @other_array

[someword

otherword

...]

I need to match the words ad align the firstvalue of each array in @AoA to each of the words in @other_array to obtain something like this:

@other_array

[someword firstvalueofthefirstarrayin@AoA firstvalueofthesecondarrayin@AoA ...

otherword firstvalueofthefirstarrayin@AoA firstvalueofthesecondarrayin@AoA ...]

I thought of rendering @other_array an array of arrays, but I thought that a more economic way to do that would be just to add the values to each string.

Thanks immensely for your patience

Comment on Re^4: pattern matching from an array to an array of arrays

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Re^5: pattern matching from an array to an array of arrays by AnomalousMonk (Archbishop) on Oct 22, 2016 at 15:17 UTC
Further to Corion's reply: What you have provided in your reply is not a Short, Self Contained, Correct (Compilable), Example (SSCCE), but some kind of pseudo-code. Please see how hippo defined `@AoA` in the SSCCE given here: `my @AoA = ( ['first word', 'greek latin'], ['alpha omega', 'beta test'] );` [download] This is the sort of thing we'd like to see from you. Please remember to provide a similar compilable definition for `@other_array` and for your expected output. Again, please help us to help you. Please see How do I post a question effectively? and How (Not) To Ask A Question. Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]
Re^5: pattern matching from an array to an array of arrays by Corion (Patriarch) on Oct 22, 2016 at 13:48 UTC
Please try to convert the code you have posted above into a short, self-contained example that properly initializes `@AoA` and `@other_array` and then shows how this interacts with your code. Maybe, if you are just trying to find the words in `@AoA` that are in `@other_array`, a hash would make the lookup easier for you?	[reply] [d/l] [select]