如果我有这样的桌子

|---------------------|------------------|
|      time           | list of string   |
|---------------------|------------------|
| 2019-06-18 09:05:00 |   ['A','B','C']|
|---------------------|------------------|
| 2019-06-19 09:05:00 |   ['A','C']     |
|---------------------|------------------|
| 2019-06-19 09:05:00 |   ['B','C']     |
|---------------------|------------------|
| 2019-06-20 09:05:00 |   ['C']          |
|---------------------|------------------|
| 2019-06-20 09:05:00 |   ['A','C']|
|---------------------|------------------|

对于每一行，我想知道当前时间戳之前的几行与当前字符串列表至少有一个公共值。

慢速代码如下：

results = [] for i in range(len(df)):
    current_t = df['time'].iloc[i]
    current_string = df['list_of_string'].iloc[i]
    df_before_t = df[df['time']<current_t]
    cumm_count = 0
    for row in df_before_t['list_of_string']:
        if (set(current_string) & set(row)):
            cumm_count += 1
    results.append(cumm_count)

所以结果表将是：

|---------------------|------------------|---------------------|
|      time           | list of string   |   result            |
|---------------------|------------------|---------------------|
| 2019-06-18 09:05:00 |   ['A','C']|           0         |
|---------------------|------------------|---------------------|
| 2019-06-19 09:05:00 |   ['A','C']     |           1         |
|---------------------|------------------|---------------------|
| 2019-06-19 09:05:00 |   ['D']          |           0         |
|---------------------|------------------|---------------------|
| 2019-06-20 09:05:00 |   ['C']          |           2         |
|---------------------|------------------|---------------------|
| 2019-06-20 09:05:00 |   ['A','C']|           2         |
|---------------------|------------------|---------------------|

我当前拥有的数据集相对较大，我希望获得帮助以快速处理此数据。非常感谢！