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    <p>You could start with something like:</p> <pre><code>set(lst1).issubset(lst2) </code></pre> <p>to see if lst1 is contained within lst2 ignoring order. If it passes the test that one list is contained within the other, then you could do something like:</p> <pre><code>ii = lst2.index(lst1[0]) if lst2[ii:ii+len(lst1)] == lst1: return True else: return False </code></pre> <p>Originally I stated that the first check was irrelevant given the second, although you'd have to properly handle the <code>ValueError</code> if the first element of lst1 is not in lst2. </p> <p><strong>Edit:</strong> Just as a side note, I compared a version of my code vs yan's and mine is significantly faster under almost all use cases, especially if len(lst1) is larger (up to 200x speedup vs yan's implementation). Give it a try with the <code>timeit</code> module.</p> <pre><code>def check(lst1,lst2): try: ii = lst2.index(lst1[0]) except ValueError: return False if lst2[ii:ii+len(lst1)] == lst1: return True else: return False </code></pre> <p>As an explanation of how it works <code>ii = lst2.index(lst1[0])</code> finds the index in <code>lst2</code> that matches the first element of <code>lst1</code>. If that item is missing from <code>lst2</code> it catches the <code>ValueError</code> and returns <code>False</code>. If that element does exist, <code>lst2[ii:ii+len(lst1)] == lst1</code> compares all of <code>lst1</code> vs the sublist of <code>lst2</code> starting from the matched element and taking the next <code>len(lst)</code> elements. </p>
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    1. POpython list and sublist
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    1. CONo explicit looping, anyway.
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      1. USMatt Curtis
    2. COYour solution intrigues me, especially this line: if lst2[ii:ii+len(lst1)] == lst1:. Do you mind explaining it to me?
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    3. COadded actual implementation and explanation
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