How to obtain the nth row of the pascal triangle
<p>Here is my code to find the nth row of pascals triangle</p>

<pre><code>def pascaline(n):
     line = [1]       
     for k in range(max(n,0)):             
         line.append(line[k]*(n-k)/(k+1))             
     return line
</code></pre>

<p>There are two things I would like to ask. First, the outputs integers end with .0 always like in </p>

<pre><code>pascaline(2) = [1, 2.0, 1.0]
</code></pre>

<p>How do I remove those .0 at the end? also, how can I do to start at $n=1$ and not at $0$? For instance, in this case, pascaline(2) should be [1, 1] and not [1, 2, 1]</p>

<p>I know you've got your answer. The problem is you are dealing with floating point numbers, <em>not</em> integers. This is programming, not math. Numbers are represented concretely. I just wanted to compare these two implementations, where the first one let's you save some calculation time by using symmetry. Both are still O(n), though:</p>

<pre><code>def pascal_line(n):
    line = [1]
    mid, even = divmod(n, 2)
    for k in range(1, mid + 1):
        num = int(line[k-1]*(n + 1 - k)/(k))
        line.append(num)
    reverse_it = reversed(line)
    if not even:
        next(reverse_it)
    for n in reverse_it:
        line.append(n)
    return line

def pascal_line2(n):
    line = [1]
    for k in range(1, n + 1):
        num = int(line[k-1]*(n + 1 - k)/(k))
        line.append(num)
    return line
</code></pre>

<p>Now, in action:</p>

<pre><code>&gt;&gt;&gt; for i in range(21):
...   print(pascal_line(i))
... 
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
[1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1]
[1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1]
[1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1]
[1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1]
[1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]
[1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1]
[1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1]
[1, 17, 136, 680, 2380, 6188, 12376, 19448, 24310, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1]
[1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1]
[1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 92378, 75582, 50388, 27132, 11628, 3876, 969, 171, 19, 1]
[1, 20, 190, 1140, 4845, 15504, 38760, 77520, 125970, 167960, 184756, 167960, 125970, 77520, 38760, 15504, 4845, 1140, 190, 20, 1]
</code></pre>

<p>Now, a quick and dirty time-test:</p>

<pre><code>&gt;&gt;&gt; def time_me(f, n):
...   start = time.time()
...   f(n)
...   stop = time.time()
... 
&gt;&gt;&gt; times = [time_me(pascal_line,n) for n in range(10, 1001,10)]
&gt;&gt;&gt; times2 = [time_me(pascal_line2,n) for n in range(10, 1001,10)]
&gt;&gt;&gt; import pandas as pd
&gt;&gt;&gt; import matplotlib.pyplot as plt
&gt;&gt;&gt; n = range(10, 1001,10)
&gt;&gt;&gt; df = pd.DataFrame({'pascal_lines':times, 'pascal_lines2':times2},index=list(n))
&gt;&gt;&gt; df.plot()
&lt;matplotlib.axes._subplots.AxesSubplot object at 0x7f6c1c3d1c18&gt;
&gt;&gt;&gt; plt.savefig('pascal.png')
</code></pre>

<p>Result:</p>

<p><a href="https://i.stack.imgur.com/x2X4n.png" rel="nofollow"><img src="https://i.stack.imgur.com/x2X4n.png" alt="enter image description here"></a></p>

<p>Not sure if it is worth it, since I'm reaching an <code>OverFlow</code> error somewhere soon after n = 1000.</p>

<h2>EDIT</h2>

<p>As others have pointed out, using integer division instead of converting from float to int is more efficient. It also has the added benefit of not throwing: <code>OverflowError: integer division result too large for a float</code> after around n = 1000.</p>

<pre><code>def pascal_line0(n):
    line = [1]
    mid, even = divmod(n, 2)
    for k in range(1, mid + 1):
        num = line[k-1]*(n + 1 - k)//(k)
        line.append(num)
    reverse_it = reversed(line)
    if not even:
        next(reverse_it)
    for n in reverse_it:
        line.append(n)
    return line
</code></pre>

<p><a href="https://i.stack.imgur.com/GGrJP.png" rel="nofollow"><img src="https://i.stack.imgur.com/GGrJP.png" alt="enter image description here"></a></p>

python