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How to partially evaluate elements in a hash-table
Posted: Fri Jun 14, 2019 6:50 am
by lyl
I construct a hash-table like this:
Code: Select all
(setq mytable '(
((char "a") "a: choiceA" (+ 0 101))
((char "b") "b: choiceB" (+ 0 102))
))
In this example, I want the first element of each sublist to be evaluated while other elements not. That is to say, (char "a") should be 97, (char "b") should be 98.
Or, in common, how to partially evaluate elements in a hash-table? Is there an easy to achieve this?
Re: How to partially evaluate elements in a hash-table
Posted: Fri Jun 14, 2019 7:14 am
by cameyo
Try this:
Code: Select all
(map (fn (x) (eval (first x))) mytable)
;-> (97 98)
Re: How to partially evaluate elements in a hash-table
Posted: Fri Jun 14, 2019 7:42 am
by lyl
Sorry, I didn't make my question clear.
What I want is not the result of each first element evaluation. I want to get a easy way to construct "mytable" from
Code: Select all
'(((char "a") "a: choiceA" (+ 0 101)) ((char "b") "b: choiceB" (+ 0 102)))
to
Code: Select all
'((97 "a: choiceA" (+ 0 101)) (98 "b: choiceB" (+ 0 102)))
Re: How to partially evaluate elements in a hash-table
Posted: Fri Jun 14, 2019 5:16 pm
by cameyo
Maybe this:
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(setq a '(((+ 6 2) (a) 2) ((- 2 5) (b) 5)))
;-> (((+ 6 2) (a) 2) ((- 2 5) (b) 5))
(dolist (el a)
(setf (first (a $idx)) (eval (first el)))
)
a
;-> ((8 (a) 2) (-3 (b) 5))
Re: How to partially evaluate elements in a hash-table
Posted: Sun Jun 23, 2019 2:06 am
by lyl
Yet, is there a better way to evaluate elements in hash-table when constructing hash-table, as like the comma expression in common lisp?
Re: How to partially evaluate elements in a hash-table
Posted: Sat Jun 29, 2019 8:31 am
by al1ranger
This can be one way of doing it. The
quote function can also be used if the list is being built at runtime.
Code: Select all
(setq mytable (list (list (char "a") "a: choiceA" '(+ 0 101)) (list (char "b") "b: choiceB" '(+ 0 102))))