Is there a function in Mathematica which removes brackets from an expression?
RemoveBrackets[ {3} ] 3
Note: inspired by @garej's answer to this question :Brackets around each item in matrix
- further edits/comments/answers welcome
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- 1$\begingroup$ related: 21174 $\endgroup$Kuba– Kuba2016-02-10 09:09:17 +00:00Commented Feb 10, 2016 at 9:09
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2 Answers
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This question is closely related to:
- How to remove redundant {} from a nested list of lists?
- How to completely delete the head of a function expression
If you wish to strip the brackets from a single expression in a nontrivial case please consider Delete as described in my answer to the second referenced question above.
Unlike using
Apply(e.g.# & @@ {1}orSequence @@ {1}) it does not first replaceListwith something else which means it behaves better inside held expressions.Unlike
Part,First, etc. it works equally well with multiple arguments, allowing us to strip the{}from e.g.HoldComplete[{1, 2, 3}]using// Delete[{1, 0}].
If you wish to strip brackets throughout an expression you can use the methods from the first linked question, being mindful of the tradeoff between brevity and efficiency.
expr = {{0, {{1, 2}}, {3}}, {4}}; expr //. {x_} :> x {{0, {1, 2}, 3}, 4}
Replace[expr, {x_} :> x, {0, -2}] {{0, {1, 2}, 3}, 4}
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4 I learned from @Kuba:
First[{3}] 3
#& @@{3} 3
these do this task even more generally than I imagined.
given any Atomic expressions wrapped with any function >> the FullForm of almost everthing in Mathematica
ArbitraryHead[expr1, expr2 ]
First[ArbitraryHead[expr1, expr2]] expr1
#&@@ArbitraryHead[expr1,expr2] expr1
here are specific examples:
First[(1/a^2)] a
why?:
(1/a^2)//FullForm Power[a, -2]
First[List[3]] 3
P.S. I feel this community is building a tall cathedral of knowledge. I'm doing my little bit of brickwork.
{3}[[1]] 3
##& @@{3} 3
Seqence[{3}] Sequence[3]
- 1$\begingroup$ Don't forget
PartandSequence. $\endgroup$Sjoerd C. de Vries– Sjoerd C. de Vries2016-02-10 08:47:59 +00:00Commented Feb 10, 2016 at 8:47 - 1$\begingroup$ I agree, I've always been partial to using
a[[1]]instead ofFirst[a]$\endgroup$Jason B.– Jason B.2016-02-10 08:50:23 +00:00Commented Feb 10, 2016 at 8:50 - 2$\begingroup$
Seqence[{3}]evaluates toSeqence[{3}]-- did you forget something? $\endgroup$Mr.Wizard– Mr.Wizard2016-02-10 09:58:08 +00:00Commented Feb 10, 2016 at 9:58 - 2