Ruby: ¿Cuál es la forma más fácil de eliminar el primer elemento de un array?

Digamos que tengo una matriz 

[0, 132, 432, 342, 234]

¿Cuál es la forma más fácil de deshacerse del primer elemento? (0)

Author: Sergio Tulentsev, 2010-09-01
Source

11 answers

"pop"ing el primer elemento de un Array se llama" shift "("unshift" siendo la operación de agregar un elemento delante de la matriz).

 250
Author: Sjoerd,
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2014-05-26 11:15:33 
a = [0,1,2,3]

a.drop(1)
# => [1, 2, 3] 

a
# => [0,1,2,3]

Y adicionalmente:

[0,1,2,3].drop(2)
=> [2, 3]

[0,1,2,3].drop(3)
=> [3]
 302
Author: scaryguy,
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2015-06-23 01:48:34

Utilice el método shift en array

>> x = [4,5,6]
=> [4, 5, 6]                                                            
>> x.shift 
=> 4
>> x                                                                    
=> [5, 6]

Si desea eliminar n elementos iniciales, puede usar x.shift(n)

 291
Author: bragboy,
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2010-09-01 07:07:58 
[0, 132, 432, 342, 234][1..-1]
=> [132, 432, 342, 234]

Así que a diferencia de shift o slice esto devuelve el array modificado (útil para un liners).

 122
Author: vise,
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2010-09-01 08:33:00

Esto es bastante limpio:

head, *tail = [1, 2, 3, 4, 5]
#==> head = 1, tail = [2, 3, 4, 5]

Como está escrito en los comentarios, hay una ventaja de no mutar la lista original.

 92
Author: hurikhan77,
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2015-06-09 08:22:50

O a.delete_at 0

 18
Author: zzzhc,
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2010-09-01 12:25:51

Use el método shift

array.shift(n) => Remove first n elements from array 
array.shift(1) => Remove first element

Https://ruby-doc.org/core-2.2.0/Array.html#method-i-shift

 11
Author: Rahul Patel,
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2017-05-19 11:16:22

Puedes usar: 

a.slice!(0)

¡Slice! se generaliza a cualquier índice o rango.

 10
Author: Matthew Flaschen,
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2010-09-01 07:06:51

Puede usar Array.delete_at (0) método que eliminará el primer elemento.

 x = [2,3,4,11,0]
 x.delete_at(0) unless x.empty? # [3,4,11,0]
 4
Author: lalit.sethi143,
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2015-10-06 11:13:38

Puedes usar:
arr - [arr[0]] o de lo contrario arr - [arr.shift()] o simplemente arr.shift(1)

 2
Author: Avijit Majhi,
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2017-07-07 07:32:09

Puedes usar: 

 a.delete(a[0])   
 a.delete_at 0

Ambos pueden funcionar

 0
Author: sam,
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2018-03-14 07:07:39