同じ要素数の配列(リスト)を要素ごとにまとめて配列の配列を作るには (zip)

Scala 2013/09/17

Signature:

def scala.collection.Iterable[A]#zip(that: Iterable[B]): Iterable[(A, B)]
def scala.collection.Seq[A]#zip(that: Iterable[B]): Seq[(A, B)]

Python 2014/04/12

zip関数を使う。

例

lst1 = [1, 2, 3]
lst2 = ["abc", "def", "ghi"]

zlst = zip(lst1, lst2)

print(zlst)
# Python2の場合
# [(1, 'abc'), (2, 'def'), (3, 'ghi')]
# Python3の場合
# <zip object at 0x7fb3dfbc43b0>

for e in zlst:
    print(e)
# (1, 'abc')
# (2, 'def')
# (3, 'ghi')

長さが一致しないリストを与えると短い方に揃えられる。

Ruby / JRuby 2015/01/04

Array#zip を使う。

p ['a', 'b'].zip([1, 2])         # => [["a", 1], ["b", 2]]
p ['a', 'b'].zip([1, 2], [3, 4]) # => [["a", 1, 3], ["b", 2, 4]]

Array#zip | Ruby 2.1 リファレンスマニュアル
http://docs.ruby-lang.org/ja/2.1.0/method/Array/i/zip.html

Perl 2016/07/11

関数の自作例

sub zip {
    my ($lst1, $lst2) = @_;
    my @src1 = @$lst1;
    my @src2 = @$lst2;
    my @result = ();
    while (@src1 && @src2) {
        my $elem1 = shift(@src1);
        my $elem2 = shift(@src2);
        push(@result, [$elem1, $elem2]);
    }
    \@result;
}

my $lst1 = [1, 2, 3];
my $lst2 = [10, 20, 30];

my $zipped = zip($lst1, $lst2);

print Dumper $zipped;
# =>
# $VAR1 = [
#           [
#             1,
#             10
#           ],
#           [
#             2,
#             20
#           ],
#           [
#             3,
#             30
#           ]
#     ];