Question

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', 'B', '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', 'B', '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', 'B', '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！

Answer 1

一种方法是将列表转换为集合，并在list of string上使用listcomp，然后将time与小于当前time的列表进行比较

s = df['list of string'].map(set)
t = pd.to_datetime(df.time)

df['result'] = [sum(len(x & y) != 0 for y in s[t.iloc[i] > t]) 
                                        for i,x in enumerate(s)]

Out[283]:
                  time list of string  result
0  2019-06-18 09:05:00      [A, B, C]       0
1  2019-06-19 09:05:00         [A, C]       1
2  2019-06-19 09:05:00            [D]       0
3  2019-06-20 09:05:00            [C]       2
4  2019-06-20 09:05:00      [A, B, C]       2

通过查看两个列表是否具有共同价值来计算熊猫的累积计数

1 个答案: