I want to create a List of Associations from data in a two-dimensional List by using ReplaceAll, but did not manage to produce the expected result.

So here is a short example, the original lists are stored in the variable values:

values = {{1, a}, {2, b}, {3, c}}
{{1, a}, {2, b}, {3, c}}

and I want to transform this to a List of Associations with Keys "N" and "C" for the numbers and the characters

{<|"N" -> 1, "C" -> a|>, <|"N" -> 2, "C" -> b|>, <|"N" -> 3, "C" -> c|>}

If I use the ReplaceAll command to create an Association I do not get the desired result

values /. {num_, char_} -> <|"N" -> num, "C" -> char|>
{<|"N" -> num, "C" -> char|>, <|"N" -> num, "C" -> char|>, <|"N" -> num, "C" -> char|>}

So the values for num and char are not considered correctly in ReplaceAll. However, If I would replace the Association by a List of Rules this works:

values /. {num_, char_} -> {"N" -> num, "C" -> char}
{{"N" -> 1, "C" -> a}, {"N" -> 2, "C" -> b}, {"N" -> 3, "C" -> c}}

Is there any explanation why the behaviour is different for the Association? What is the usual way to create Associations from Lists?

  • $\begingroup$ Okay, I just found out that using :> (RuleDelayed) instead of -> (Rule) in ReplaceAll does the trick. Is there any explanation why? So the working command is values /. {num_, char_} :> <|"N" -> num, "C" -> char|> $\endgroup$
    – Mathias
    Jan 21, 2022 at 13:29

1 Answer 1


If you use "Rule" the right side is evaluated at once resulting in:

{<|"N" -> num, "C" -> char|>, <|"N" -> num, "C" -> char|>, <|"N" -> num, "C" -> char|>}

On the other hand, if you use RuleDelayed, the right side is only evaluated when the pattern has matched:

values = {{1, a}, {2, b}, {3, c}};
values /. {num_, char_} :> <|"N" -> num, "C" -> char|>
(*{<|"N" -> 1, "C" -> a|>, <|"N" -> 2, "C" -> b|>, <|"N" -> 3, "C" -> c|>} *)
  • 1
    $\begingroup$ But why is there no problem when using a List instead of an Association on the right hand side? $\endgroup$
    – Mathias
    Jan 21, 2022 at 13:44
  • 2
    $\begingroup$ The reason is that Association has the attribute HoldAllComplete. $\endgroup$ Jan 21, 2022 at 13:52
  • $\begingroup$ I don't think that is true, after all HoldComplete also has that attribute yet n is substituted in 1 /. n_ -> HoldComplete[n]. I believe the cause is the fact that association constructor expressions evaluate to atomic association objects (see Evaluated vs. unevaluated Association). $\endgroup$
    – WReach
    Jan 21, 2022 at 21:46
  • $\begingroup$ Thank's. There is always new to learn.. $\endgroup$ Jan 22, 2022 at 9:42

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