Python 在没有循环的情况下遍历数组的列

我的数据框如下所示：

           asks                                                        bids                                   lastUpdated   
0   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
1   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
2   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
3   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464

midprice
0.091965
0.091965
0.091965
0.091965

我需要的是创建一列

midprice

，每行等于

asks[0][0]+bids[0][0]/2

你知道如何在没有循环的情况下做到这一点吗？像lambda函数？ 类似这样：

df.assign（中间价=lambda x:[x['bids'][0]+x['asks'][0]）*0.5）

输出应为如下所示的列：

           asks                                                        bids                                   lastUpdated   
0   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
1   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
2   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464     
3   [[0.09245, 1654], [0.09265000000000001, 506], ...   [[0.09148, 121], [0.09141, 183], [0.09134, 550...   2020-04-19 00:02:24.464

midprice
0.091965
0.091965
0.091965
0.091965

坦斯克

df = (df
      .assign(topBid = df.bids.apply(lambda bid: bid[0][0] if bid else np.nan))
      .assign(topAsk = df.asks.apply(lambda ask: ask[0][0] if ask else np.nan))
      .assign(midprice = lambda x: (x.topBid + x.topAsk) * 0.5))