P20 (*)リストからK番目の要素を削除する。

def remove_at(l, num):
    return [v for i,v in enumerate(l) if (i+1) != num] if isinstance(l, list) else None

if __name__ == "__main__":
    print remove_at("a", 3) == None
    print remove_at(["a"], -1) == ["a"]
    print remove_at(["a"], 0) == ["a"]
    print remove_at(["a"], 1) == []
    print remove_at(["a", "b"], -2) == ["a", "b"]
    print remove_at(["a", "b"], -1) == ["a", "b"]
    print remove_at(["a", "b"], 0) == ["a", "b"]
    print remove_at(["a", "b"], 1) == ["b"]
    print remove_at(["a", "b"], 2) == ["a"]
    print remove_at(["a", "b", "c", "d"], 2) == ["a", "c", "d"]