Remove last element from list in Elixir? [duplicate]

Question

How would you simply remove 20 from the end of this example?

[46, 238, 64, 30, 105, 136, 98, 75, 23, 157, 11, 20]

Answer 1

fastest solution that I know of with @mudasobwa's one
solution 1:

x |> Enum.reverse() |> tl() |> Enum.reverse()

Some other solutions:
solution 2:

[k] = Enum.chunk(x, length(x)-1)

solution 3:

List.delete_at(x, length(x)-1)

solution 4:

x |> List.to_tuple() |> Tuple.delete_at(length(x)-1) |> Tuple.to_list

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An even more interesting question, is how to do that the fastest way ? let's benchmark it !

I got the following result using Benchee, each solution was computed 50000 times, and I used a list created with Enum.to_list(1..10_000)

Name                                  ips        average  deviation         median         99th %
mudasobwa's solution              14.80 K       67.57 μs    ±34.17%          79 μs          91 μs
solution 1                        14.79 K       67.59 μs    ±33.96%          79 μs          91 μs
solution 4                        10.71 K       93.38 μs    ±32.67%          81 μs         201 μs
Roman Rabinovich's solution        8.45 K      118.33 μs    ±18.27%         118 μs         171 μs
OneSneakyMofo's solution           5.07 K      197.34 μs    ±13.60%         193 μs         331 μs
dawner's solution                  4.57 K      219.00 μs    ±11.87%         216 μs      256.23 μs
solution 3                         3.41 K      292.91 μs    ±20.01%         290 μs      506.64 μs
solution 2                         0.83 K     1205.52 μs    ±22.25%        1105 μs     2061.77 μs

Comparison: 
mudasobwa's solution              14.80 K
solution 1                        14.79 K - 1.00x slower
solution 4                        10.71 K - 1.38x slower
Roman Rabinovich's solution        8.45 K - 1.75x slower
OneSneakyMofo's solution           5.07 K - 2.92x slower
dawner's solution                  4.57 K - 3.24x slower
solution 3                         3.41 K - 4.33x slower
solution 2                         0.83 K - 17.84x slower

Answer 2

You could do:

List.pop_at(x, -1)
{20, [46, 238, 64, 30, 105, 136, 98, 75, 23, 157, 11]}