Question

我有一个df：

df2 = pd.DataFrame({
    'ID': ['James', 'James', 'James',
           'Max', 'Max', 'Max', 'Max', 'Max',
           'Park', 'Park', 'Park',
           'Tom', 'Tom', 'Tom', 'Tom'],
    'From_num': [78, 420, 'Started', 298, 78, 36, 298, 'Started', 28, 311, 'Started', 60, 520, 99, 'Started'],
    'To_num': [96, 78, 420, 36, 78, 78, 36, 298, 112, 28, 311, 150, 520, 78, 99],
    'Date': ['2020-05-12', '2020-02-02', '2019-06-18',
             '2019-08-26', '2019-06-20', '2019-01-30', '2018-10-23',
             '2018-08-29', '2020-05-21', '2019-11-22',
             '2019-04-12', '2019-10-16', '2019-08-26', '2018-12-11', '2018-10-09']})

它看起来像这样：

       ID From_num  To_num        Date
0   James       78      96  2020-05-12
1   James      420      78  2020-02-02
2   James  Started     420  2019-06-18
3     Max      298      36  2019-08-26
4     Max       78      78  2019-06-20
5     Max       36      78  2019-01-30
6     Max      298      36  2018-10-23
7     Max  Started     298  2018-08-29
8    Park       28     112  2020-05-21
9    Park      311      28  2019-11-22
10   Park  Started     311  2019-04-12
11    Tom       60     150  2019-10-16
12    Tom      520     520  2019-08-26
13    Tom       99      78  2018-12-11
14    Tom  Started      99  2018-10-09

我希望为每个ID（人名）创建一个新数据框，其中任一列在组中包含数字78（无论78在From_num或To_num中出现，还是在这两者中都出现），并删除这两列都不包含的人78，在这种情况下为“公园”。我已经写了这样的代码：

find_nn = df2.groupby('ID').apply(lambda x: x[['From_num', 'To_num']].isin([78]).any())
find_nn.columns = ['from_bool', 'to_bool']
find_nn['bool_result'] = find_nn['from_bool'] | find_nn['to_bool']
bool_nn = find_nn['bool_result'].reset_index()
df2_new = pd.merge(left=df2, right=bool_nn, on='ID', copy=False)
df2_new = df2_new[df2_new['bool_result'] == True]

它正在工作，但是非常冗余且缓慢，因为在我的实际情况下，数据集更加复杂。如果您有任何更好的主意，请提供帮助。非常感谢！！期望像这样：

       ID From_num  To_num        Date
0   James       78      96  2020-05-12
1   James      420      78  2020-02-02
2   James  Started     420  2019-06-18
3     Max      298      36  2019-08-26
4     Max       78      78  2019-06-20
5     Max       36      78  2019-01-30
6     Max      298      36  2018-10-23
7     Max  Started     298  2018-08-29
11    Tom       60     150  2019-10-16
12    Tom      520     520  2019-08-26
13    Tom       99      78  2018-12-11
14    Tom  Started      99  2018-10-09

Answer 1

让我们尝试filter

df1 = df2.groupby('ID').filter(lambda x : x[['From_num','To_num']].eq(78).any().any())
       ID From_num  To_num        Date
0   James       78      96  2020-05-12
1   James      420      78  2020-02-02
2   James  Started     420  2019-06-18
3     Max      298      36  2019-08-26
4     Max       78      78  2019-06-20
5     Max       36      78  2019-01-30
6     Max      298      36  2018-10-23
7     Max  Started     298  2018-08-29
11    Tom       60     150  2019-10-16
12    Tom      520     520  2019-08-26
13    Tom       99      78  2018-12-11
14    Tom  Started      99  2018-10-09

为了速度

m=df2[['From_num','To_num']].eq(78).any(axis=1).groupby(df2.ID).transform('any')
df1=df2[m]

Answer 2

这是获取相同数据的更简单方法。您可以将2个过滤器应用于df2。第一行说，过滤df2，其中From_num或To_num = 78，然后获取这些行的ID。在下一行，我们用这些ID过滤df2。

ids = df2[(df2.From_num == 78) | (df2.To_num == 78)]['ID'].unique()
df2_new = df2[df2['ID'].isin(ids)]

Answer 3

这对你来说是个好人：

df2[df2['ID'].isin((df2.set_index(['ID','Date']).stack() == 78).any(level=0).loc[lambda x:x].index)]

输出：

       ID From_num  To_num        Date
0   James       78      96  2020-05-12
1   James      420      78  2020-02-02
2   James  Started     420  2019-06-18
3     Max      298      36  2019-08-26
4     Max       78      78  2019-06-20
5     Max       36      78  2019-01-30
6     Max      298      36  2018-10-23
7     Max  Started     298  2018-08-29
11    Tom       60     150  2019-10-16
12    Tom      520     520  2019-08-26
13    Tom       99      78  2018-12-11
14    Tom  Started      99  2018-10-09

熊猫：从另一个包含组内特定值的df创建新的df

3 个答案: