generate association based on list

Question

a is a list {1,2,3,4}

b is an association <|1->a, 2->b, 3->c, 4->d|>

I am trying to write a single line of code to get b from a

b=<|#->myFunction[#] &|> /@ a

Here is the output:

Association[#1->myFunction[#1]&][1]

Can someone help find out a correct way to do it?

btw, I can do it using for loop, of course, just trying to see if I can use only one line to do it :-)

Try b = <|# -> myFunction[#]|> & /@ a do not use () for function parameters Mathematica uses [] — Vahagn Tumanyan, Apr 29, 2017 at 14:14
Also try a = Range[4]; b = CharacterRange["a", "d"]; Association@((#[[1]] -> #[[2]] &) /@ Transpose[{a, b}]) — foxcode64, Apr 29, 2017 at 14:28
@VahagnTumanyan I did use [#] in my code, just a typo in my original post, now just corrected it. — user1470393, Apr 29, 2017 at 16:13

jjc385 · Accepted Answer · 2017-04-29 17:01:57Z

2

Just found a function AssociationMap

listOfKeys = {key1, key2, ...};

AssociationMap[f, listOfKeys] 
->  <|key1->f[key1], key2->f[key2], ...|>

jjc385

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answered Apr 29, 2017 at 16:17

user1470393

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$\begingroup$ xs = Range[4]; Association @@ Map[Rule[#, f[#]] &]@xs $\endgroup$
– webcpu
Apr 29, 2017 at 18:31
$\begingroup$ @tomd just realized you had done it before me :) there is also a function AssociationThread that merges what you just did. $\endgroup$
– Ali Hashmi
Apr 29, 2017 at 18:51

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Ali Hashmi · Accepted Answer · 2017-04-29 18:44:45Z

2

AssociationThread[# -> FromLetterNumber[#]] &@Range[4]

answered Apr 29, 2017 at 18:44

Ali Hashmi

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1

$\begingroup$ This is what I was trying to do :-) (Variant: AssociationMap[FromLetterNumber, a]). $\endgroup$
– user1066
Apr 29, 2017 at 18:54
$\begingroup$ @tomd thanks ! yours is more compact. $\endgroup$
– Ali Hashmi
Apr 29, 2017 at 19:03

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2 Answers 2