How do I remove the first array inside of a nested array with ruby on rails?

Question

let's say I have a nested array like so

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

How would I remove the very first array in the nested array so it could end up like so

array = [[4,5,6],[7,8,9]]

Do you need to first detect if it is an array or not, or is it always an array? — Rockwell Rice
– Rockwell Rice, Commented Mar 18, 2022 at 2:07
while mechnicov and Cary Swoveland point out many possible solutions, I think Farhad Ajaz's solution is the most strait forward. That being said, a simple google search of "ruby remove first element from array" would yield many solutions, including those found on this site, which might result in a the question being marked as a duplicate — nPn
– nPn, Commented Mar 18, 2022 at 3:59

Farhad Ajaz · Accepted Answer · 2022-03-17 22:04:04Z

3

you can try the following method

array.drop(1)

answered Mar 17, 2022 at 22:04

Farhad Ajaz

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1 Comment

Note that drop doesn't modify array. (as requested by the OP)

mechnicov · Accepted Answer · 2022-03-17 23:00:42Z

3

There are many-many ways to do it

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

You can mutate array in place with such methods

# Array#delete_at

array.delete_at(0)
array # => [[4, 5, 6], [7, 8, 9]]

# Array#slice!

array.slice!(0)
array # => [[4, 5, 6], [7, 8, 9]]

# Array#shift

array.shift # or array.shift(1)
array # => [[4, 5, 6], [7, 8, 9]]

or you can assign new value

# Array#drop

array = array.drop(1)

# Array#[]

array = array[1..]

or replace content to save link to object

# Array#replace

array.replace(array[1..-1]) # new array as argument of replace

or more difficult ways

# Array#reject

array = array.reject!.with_index { |_, index| index.zero? }

# Array#reject!

array.reject!.with_index { |_, index| index.zero? }

# Array#select

array = array.select.with_index { |_, index| index.positive? }

# Array#select!

array.select!.with_index { |_, index| index.positive? }

I'm sure there are many other methods

Please look in the docs to explore it:

answered Mar 17, 2022 at 22:53

mechnicov

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Two more: array.replace(array[1..-1]) (mutating) and _,*a = array, a being the new array (non-mutating).