# hist(v,e)

hist(v, e) -> e, counts

Compute the histogram of `v` using a vector/range `e` as the edges for the bins. The result will be a vector of length `length(e) - 1`, such that the element at location `i` satisfies `sum(e[i] .< v .<= e[i+1])`. Note: Julia does not ignore `NaN` values in the computation.

## Examples

1. Compute histogram with default number of bins:

``````julia> v = [1, 2, 1, 3, 2, 4, 1, 2, 3];
julia> e, counts = hist(v)
([1, 2, 3, 4], [3, 3, 2])``````

This example computes the histogram of the vector `v` using the default number of bins. It returns the range `e` which represents the edges of the bins, and `counts` which contains the count of elements in each bin.

2. Compute histogram with a specific number of bins:

``````julia> v = [1.5, 2.7, 3.8, 4.9, 5.5];
julia> e, counts = hist(v, 3)
([1.5, 3.0, 4.5, 6.0], [1, 1, 3])``````

This example computes the histogram of the vector `v` using 3 bins. It returns the range `e` with the bin edges and `counts` with the count of elements in each bin.

3. Handle NaN values in the computation:
``````julia> v = [1, 2, NaN, 3, 4, NaN, 5];
julia> e, counts = hist(v)
([1.0, 2.0, 3.0, 4.0, 5.0], [1, 1, 1, 1, 1])``````

The `hist` function does not ignore `NaN` values in the computation. It treats them as separate elements in the histogram.

Common mistake example:

``````julia> v = [1, 2, 3, 4, 5];
julia> e, counts = hist(v, -2)
ERROR: ArgumentError: number of bins must be non-negative``````

In this example, a negative number of bins is provided, which is not allowed. Ensure that the number of bins is a non-negative integer when using the `hist` function.