Generating evenly distributed multiples/samples within a range
<p><strong>Specific instance of Problem</strong><br>
I have an int range from 1-100. I want to generate n total numbers within this range that are <strong>as evenly distributed as possible</strong> and include the first and last values.</p>

<p><strong>Example</strong>  </p>

<pre><code>start = 1, end = 100, n = 5   
Output: [1, 25, 50, 75, 100]

start = 1, end = 100, n = 4   
Output: [1, 33, 66, 100]

start = 1, end = 100, n = 2   
Output: [1, 100]
</code></pre>

<p><strong>What I currently have</strong><br>
I actually have a working approach but I keep feeling I am over thinking this and missing something more simple? Is this the most efficient approach or could this be improved?</p>

<pre class="lang-python prettyprint-override"><code>def steps(start, end, n):
    n = min(end, max(n, 2) - 1)
    mult = end / float(n)
    yield start
    for scale in xrange(1, n+1):
        val = int(mult * scale)
        if val != start:
            yield val
</code></pre>

<p>Note, I am ensuring that this function will always return at least the lower and upper limit values of the range. So, I force <code>n &gt;= 2</code></p>

<p>Just for search reference, I am using this to sample image frames from a rendered sequence, where you would usually want the first, middle, last. But I wanted to be able to scale a bit better to handle really long image sequences and get better coverage. </p>

<p><strong>Solved: From the selected answer</strong></p>

<p>I ended up using this slightly modified version of @vartec's answer, to be a generator, and also cap the <code>n</code> value for safety:</p>

<pre><code>def steps(start,end,n):
    n = min(end, max(n, 2))
    step = (end-start)/float(n-1)
    return (int(round(start+x*step)) for x in xrange(n))
</code></pre>
