def Newdf data = dfcopydeep=True dataseq_min=0 dataseq_max=0 dataA = 0 dataB = 0 dataC = 0 dataD = 0 dataE = 0 dataF = 0 dataG = 0 dataH = 0 st = timeperf_counter

可以考虑使用向量化操作，避免逐行迭代操作DataFrame，可以大大提高效率。

首先，可以使用groupby操作，将数据按照Index分组。然后，可以使用apply方法，对每个分组进行操作。在操作过程中，可以使用nsmallest和nlargest方法获取第二小和第二大的值，使用merge方法将这些值与分组合并。最后，可以使用loc方法将计算结果赋值给data的相应列。

以下是可能的优化代码：

def New(df):
    data = df.copy(deep=True)
    data["seq_min"] = data.groupby(['Index'])['Via_Seq'].transform('min')
    data["seq_max"] = data.groupby(['Index'])['Via_Seq'].transform('max')
    group_min = data.groupby(['Index', 'Via_Seq'])['Arrive_TmS'].min().reset_index().rename(columns={'Arrive_TmS': 'A', 'Via_Seq': 'seq_min'})
    group_max = data.groupby(['Index', 'Via_Seq'])['Deptr_TmS'].max().reset_index().rename(columns={'Deptr_TmS': 'H', 'Via_Seq': 'seq_max'})
    group_second_smallest = data.groupby(['Index'])['Via_Seq'].nsmallest(2).reset_index()
    group_second_largest = data.groupby(['Index'])['Via_Seq'].nlargest(2).reset_index()
    group_second_smallest = group_second_smallest.loc[group_second_smallest['level_1'] == 1, ['Index', 'Via_Seq']]
    group_second_largest = group_second_largest.loc[group_second_largest['level_1'] == 1, ['Index', 'Via_Seq']]
    group_second_smallest = group_second_smallest.rename(columns={'Via_Seq': 'second_smallest'})
    group_second_largest = group_second_largest.rename(columns={'Via_Seq': 'second_largest'})
    data = data.merge(group_min, on=['Index', 'seq_min'], how='left')
    data = data.merge(group_max, on=['Index', 'seq_max'], how='left')
    data = data.merge(group_second_smallest, on='Index', how='left')
    data = data.merge(group_second_largest, on='Index', how='left')
    data["C"] = data.loc[data['Via_Seq'] == data['second_smallest'], 'Arrive_TmS']
    data["D"] = data.loc[data['Via_Seq'] == data['second_smallest'], 'Deptr_TmS']
    data["E"] = data.loc[data['Via_Seq'] == data['second_largest'], 'Arrive_TmS']
    data["F"] = data.loc[data['Via_Seq'] == data['second_largest'], 'Deptr_TmS']
    data["G"] = data.loc[data['Via_Seq'] == data['seq_max'], 'Arrive_TmS']
    data['firstFlag'] = data.apply(lambda x: Noequ(x['Via_Seq'], x['seq_min']), axis=1)
    data['lastFlag'] = data.apply(lambda x: Noequ(x['Via_Seq'], x['seq_max']), axis=1)
    return data

这个优化代码的时间复杂度为$O(nlogn)$，比原来的逐行迭代的算法复杂度低得多，应该可以大幅提高程序的效率