Austin Atendido - 1 year ago 79

Ruby Question

I'm trying to figure out how to find a count of the most frequent element in an array of integers. I can think of a few methods that might be helpful but when I get to writing an expression inside the block I get complete lost on how to compare an element with the next and previous element. Any ideas? All help is really really appreciated!!!

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Answer Source

An easy was is to determine all the unique values, convert each to its count in the array, then determine the largest count.

```
def max_count(arr)
arr.uniq.map { |n| arr.count(n) }.max
end
```

For example:

```
arr = [1,2,4,3,2,6,3,4,2]
max_count(arr)
#=> 3
```

There are three steps:

```
a = arr.uniq
#=> [1, 2, 4, 3, 6]
b = a.map { |n| arr.count(n) }
#=> [1, 3, 2, 2, 1]
b.max
#=> 3
```

A somewhat more efficient way (because the elements of `arr`

are enumerated only once) is to use a counting hash:

```
def max_count(arr)
arr.each_with_object(Hash.new(0)) { |n,h| h[n] += 1 }.values.max
end
max_count(arr)
#=> 3
```

We have:

```
a = arr.each_with_object(Hash.new(0)) { |n,h| h[n] += 1 }
#=> {1=>1, 2=>3, 4=>2, 3=>2, 6=>1}
b = a.values
#=> [1, 3, 2, 2, 1]
b.max
#=> 3
```

See Hash::new for an explanation of `Hash.new(0)`

. Briefly, if `h = Hash.new(0)`

and `h`

does not have a key `k`

, `h[k]`

will return the default value, which here is zero. `h[k] += 1`

expands to `h[k] = h[k] + 1`

, so if `h`

does not have a key `k`

, this becomes `h[k] = 0 + 1`

. On the other hand, if, say, `h[k] => 2`

, then `h[k] = h[k] + 1 #=> h[k] = 3 + 1`

.

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