Parse the resistors

Question

The aim is simple: given the string of resistance values, draw the part of "electrical circuit" of those resistors. Examples follow.

1. Input: 3. Output:

   --3--
   

2. Input: 1,2,3. Output:

   --1----2----3--
   

Next, the resistors can be joined parallel (by 2 or 3 resistors per join), but the required depth of parallel joins is only 1.

3. Input: 1|2. Output:

    |--1--|
   -|     |-
    |--2--|
   

4. Input: 1|2|3. Output:

    |--1--|
   -|--2--|-
    |--3--|
   

   Note, that --2-- resistor is now centered.

5. Input: 1|2,3|4|5. Output:

    |--1--|  |--3--|
   -|     |--|--4--|-
    |--2--|  |--5--|
   

Combined joins:

6. Input: 1,2,3|4,5,6|7|8,9. Output:

              |--3--|       |--6--|
   --1----2---|     |---5---|--7--|---9--
              |--4--|       |--8--|
   

If it would be more convenient, the input can be a two-dimensional array. I.e. the input for the last example would look like this: [1,2,[3,4],5,[6,7,8],9].

Some notes:

• No other forms of (in|out)puts are allowed.
• The resistance of each resistor can vary in the range from 1 to 9. Other values, like -, 42 or 0 are disallowed.
• The joins like 1|2|3|4 are invalid. As already mentioned, max 3 per parallel join. Empty joins, i.e. ...,|,... or ...,,... are invalid.
• Oh, and this is code golf :)

Answer 1

Vyxal, 64 bytes

ƛI⁼[h\-dpømð5*:‟W|ƛ`|--`pøm;:L2=[÷`|   `øm$W]];DvhṄ,ƛ1i;\-j,vtṄ,

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Answer 2

05AB1E, 45 42 bytes

ε€…--ÿDgið.ø.BëÁð3ת3£À'|ì… - SδìÅ\]€.ºøJ»

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Answer 3

Python 2, 195 bytes

a=b=c=''
B=' '*5
H='--'
S=' |'+H;E=H+'| '
for x in input():
 if x>9:a+=S+`x[0]`+E;c+=S+`x[-1]`+E;b+='-|'+[B,H+`x[1]`+H][len(x)>2]+'|-'
 else:b+=H+`x`+H;a+=B;c+=B
print[b,a+'\n'+b+'\n'+c]['|'in a]

Takes input as:

[1,2,[3,4],5,[6,7,8],9]

Output:

           |--3--|       |--6--|      
--1----2---|     |---5---|--7--|---9--
           |--4--|       |--8--|      

Answer 4

JavaScript (Node.js), 149 bytes

n=>(H=M=>n.map(e=>(T=e[M]||e[1])?` |${(G=e=>`--${e}--`)(T)}| `:'     ').join``)(0)+`
${n.map(e=>+e?G(e):`-|${e[2]?G(e[1]):'     '}|-`).join``}
`+H(2)

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No need to overcomplicate things, just keep it simple

Answer 5

Haskell, 132 bytes

foldr(zipWith(:))e.(>>=g)
a=" - "
b="|||"
s%c=[a,b,c,c,s,c,c,b,a]
g[x]=[a,a,' ':x:" ",a,a]
g[x,y]=[x,' ',y]%"- -"
g z=z%"---"
e=[]:e

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Takes input as a [[Char]] and outputs a list of lines.

Answer 6

Julia 1.5, 129 bytes

s=" "^5
+x="--$x--"
!x=map(*,[" |","-|"," |"],x,["| ","|-","| "])
-x=[[s,.+x,s],![+x[1],s,+x[(l=end;)]],!.+x][l]
~x=map(*,.-x...)

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output is a list of lines