How to put array into variables

Hi All

I tried to put an array into variables like this

s, h = b.scan(/^(\w+)\s(\w+)$/)

But for some reason ‘s’ becomes an array and ‘h’ is undefined!!

Any suggestions how to put the first result in ‘s’ and de second in ‘h’
?

Thnx a lot
LuCa

Luca S. wrote:

Hi All

I tried to put an array into variables like this

b = “aaa bbb”

s, h = b.scan(/^(\w+)\s(\w+)$/)

But for some reason ‘s’ becomes an array and ‘h’ is undefined!!

The problem is that scan returns an array of arrays:
[ # first match:
[ #first group :
“aaa”,
# second group:
“bbb”
]
]

So you might want in your case:

s, h = b.scan(/^(\w+)\s(\w+)$/)[0]
s, h = b.scan(/^(\w+)\s(\w+)$/).flatten

Vince

thanks a lot!!!

LuCa

a, b = *[1, 2, 3]
a => 1
b => 2
3 has been rejected.

[email protected] a écrit :

a,*b = [1,2,3,4,5,6,7,8,9]
a => 1
b => [2, 3, 4, 5, 6, 7, 8, 9]

Ken Allen wrote:

This also works:
b = “word1 word2”
((a,b)) = b.scan(/^(\w+)\s(\w+)$/)
p a => “word1”
p b => “word2”

The outer parens matches the outer array, and the inner parens matches
the inner array, allowing a and b to decompose it.
It’s a pretty useful trick when decomposing nested arrays.

Ken

Seems like the long way around…

foo = [“one”, “two”]
a, b = foo
a #=> “one”
b #=> “two”

El Gato wrote:

Ken Allen wrote:

This also works:
b = “word1 word2”
((a,b)) = b.scan(/^(\w+)\s(\w+)$/)
p a => “word1”
p b => “word2”

The outer parens matches the outer array, and the inner parens matches
the inner array, allowing a and b to decompose it.
It’s a pretty useful trick when decomposing nested arrays.

Ken

Seems like the long way around…

foo = [“one”, “two”]
a, b = foo
a #=> “one”
b #=> “two”

Actually, to be more clear

foo = “hello world”
a,b = foo.split
a #=> “hello”
b #=> “world”

This also works:
b = “word1 word2”
((a,b)) = b.scan(/^(\w+)\s(\w+)$/)
p a => “word1”
p b => “word2”

The outer parens matches the outer array, and the inner parens matches
the inner array, allowing a and b to decompose it.
It’s a pretty useful trick when decomposing nested arrays.

Ken

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