在 Ruby 中合并和交错两个数组

It's not exactly elegant, but it works for arrays of any size:

>> a.map.with_index { |x, i| [x, i == a.size - 2 ? s.last : s.first] }.flatten[0..-2]
#=> ["Cat", "and", "Dog", "&", "Mouse"]

You can do that with:

a.zip(s).flatten.compact

This won't give a result array in the order Chris asked for, but if the order of the resulting array doesn't matter, you can just use a |= b. If you don't want to mutate a, you can write a | b and assign the result to a variable.

See the set union documentation for the Array class at http://www.ruby-doc.org/core/classes/Array.html#M000275.

This answer assumes that you don't want duplicate array elements. If you want to allow duplicate elements in your final array, a += b should do the trick. Again, if you don't want to mutate a, use a + b and assign the result to a variable.

In response to some of the comments on this page, these two solutions will work with arrays of any size.

s.inject(a, :<<)


s   #=> ["and", "&"]
a   #=> ["Cat", "Dog", "Mouse", "and", "&"]

It doesn't give you the order you asked for, but it's a nice way of merging two arrays by appending to the one.

arr = [0, 1]
arr + [2, 3, 4]


//outputs [0, 1, 2, 3, 4]

Here's a solution that allows interleaving multiple arrays of different sizes (general solution):

arr = [["Cat", "Dog", "Mouse", "boo", "zoo"],
["and", "&"],
["hello", "there", "you"]]


first, *rest = *arr; first.zip(*rest).flatten.compact
=> ["Cat", "and", "hello", "Dog", "&", "there", "Mouse", "you", "boo", "zoo"]

How about a more general solution that works even if the first array isn't the longest and accepts any number of arrays?

a = [
["and", "&"],
["Cat", "Dog", "Mouse"]
]


b = a.max_by(&:length)
a -= [b]
b.zip(*a).flatten.compact


=> ["Cat", "and", "Dog", "&", "Mouse"]

If you don't want duplicate, why not just use the union operator :

new_array = a | s

One way to do the interleave and also guarantee which one is the biggest array for the zip method, is to fill up one of the arrays with nil until the other array size. This way, you also guarantee which element of which array will be on first position:

preferred_arr = ["Cat", "Dog", "Mouse"]
other_arr = ["and","&","are","great","friends"]


preferred_arr << nil while preferred_arr.length < other_arr.length
preferred_arr.zip(other_arr).flatten.compact
#=> ["Cat", "and", "Dog", "&", "Mouse", "are", "great", "friends"]

To handle the situation where both a & s are not of the same size:

a.zip(s).flatten.compact | s

• .compact will remove nil when a is larger than s
• | s will add the remaining items from s when a is smaller than s

Interleave 2D array of any size

arr = [["Cat", "Dog", "Mouse"],
["and", "&"],
["hello", "there", "you", "boo", "zoo"]]


max_count = arr.map(&:count).max
max_count.times.map{|i| arr.map{|a| a[i]}}.flatten.compact


#=> ["Cat", "and", "hello", "Dog", "&", "there", "Mouse", "you", "boo", "zoo"]

A very clear way to merge multiple arrays is to unpack them into one array. This works in practically the same way for many languages, so I'd prefer this method due to its simplicity and developer familiarity with it.

a = ["Cat", "Dog", "Mouse"]
s = ["and", "&"]


[*a, *s]
#=> ["Cat", "Dog", "Mouse", "and", "&"]

def merge_and_interleave(arr_a, arr_b)
final_arr = []


until arr_a.empty? && arr_b.empty?
final_arr << arr_a.shift unless arr_a.empty?
final_arr << arr_b.shift unless arr_b.empty?
end


final_arr
end