分配给数组并替换出现的nil值

Greetings!

When assigning a value to an array as in the following, how could I replace the nils by 0?

如下所示为数组赋值时,如何将nils替换为0?

array = [1,2,3]
array[10] = 2
array # => [1, 2, 3, nil, nil, nil, nil, nil, nil, nil, 2]

If not possible when assigning, how would I do it the best way afterwards? I thought of array.map { |e| e.nil? ? 0 : e }, but well…

如果在分配时不可能,我将如何以最佳方式进行分配?我想到了array.map {| e | e.nil? ? 0:e},但是......

Thanks!

6 个解决方案

#1

There is no built-in function to replace nil in an array, so yes, map is the way to go. If a shorter version would make you happier, you could do:

没有内置函数来替换数组中的nil,所以是的,map是要走的路。如果较短的版本会让你更快乐,你可以这样做:

array.map {|e| e ? e : 0}

#2

To change the array after assignment:

要在分配后更改数组:

array.map! { |x| x || 0 }

Note that this also converts false to 0.

请注意,这也会将false转换为0。

If you want to use zeros during assignment, it's a little messy:

如果你想在任务期间使用零,那就有点乱了:

i = 10
a = [1, 2, 3]
a += ([0] * (i - a.size)) << 2
# => [1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 2]

#3

To change the array in place

要更改阵列

array.map!{|x|x ?x:0}

If the array can contain false you'll need to use this instead

如果数组可以包含false,则需要使用它

array.map!{|x|x.nil? ? 0:x}

#4

nil.to_i is 0, if all the numbers are integers then below should work. I think It is also the shortest answer here.

nil.to_i为0,如果所有数字都是整数,则下面应该有效。我认为这也是最简短的答案。

array.map!(&:to_i)

#5

a.select { |i| i }

This answer is too short so I am adding a few more words.

这个答案太短了,所以我补充说了几句话。

#6

Another approach would be to define your own function for adding a value to the array.

另一种方法是定义自己的函数,以便为数组添加值。

class Array
  def addpad(index,newval)
    concat(Array.new(index-size,0)) if index > size
    self[index] = newval
  end
end

a = [1,2,3]
a.addpad(10,2)
a => [1,2,3,0,0,0,0,0,0,0,2]

#1