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I have a list where some elements are strings or sublists:

list = {"DTLCIGYHANNSTDT", "LCLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKI", \ "CLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKIC", {"H25", " H45", " I361", " D362"}, "N197", {"H25", " H45", " S46", " V47", " T332", " V361", " D362", " G363", " W364", " Q381", " K382", " T384", " Q385", " I388", " N389", " V395", " N396", " I399"}, {"H25", " H45", " V47", " N48", " L49", " T332", " D362", " G363", " W364", " Q381", " T384", " Q385", " I388", " N389", " T392", " V395", " N396", " I399", " E400"}, "NSTDTVDTVLEKNVT", {"D31", " S46"}, "S145", ... }

I want to target the 1st level strings that look like "N197" or "S145" and replace them with {N197} and {S145}. Transform from String to List, essentially.

I've tried with 

list /. x_String /; StringMatchQ[x, RegularExpression["^[A-Z]\\d{1,3}$"]] :> List[x]

This command affects the desired elements, PLUS the first string element of every sublist. I know I'm missing something obvious, but how can I keep the changes to the 1st-level of the list?

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Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.

pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\\d{1,3}$"]]; Replace[list, pattern :> List[x], {1}]
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    $\begingroup$ Thanks for the tip about ReplaceAll! $\endgroup$ Commented Oct 7, 2018 at 20:32

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