Question

我试图根据它们在两列之间的顺序关系进行分组。

d = {'df1':[10,20, 30, 60, 70, 40, 30, 70], 'df2':[20, 30, 40, 80, 70, 50, 90, 100]}

df = pd.DataFrame(data = d)
df

   df1  df2
0   10  20
1   20  30
2   30  40
3   60  80
4   80  70
5   40  50
6   30  90
7   70  100

我期望结果如下：

为了更加清楚：-df1和df2具有基于其顺序的关系。例如，10与20有直接关系，10与30至20有间接关系。10与40至20和30有间接关系。再举一个例子，让我们以80与70和70有直接关系。与100到70之间存在间接关系。这适用于其余的列值。

  df1  |    df2
  -----|-------------------
0   10 | 20, 30, 40, 50, 90
1   20 | 30, 40, 50, 90
2   30 | 40, 50, 90
3   60 | 80, 70, 100
4   80 | 70, 100
5   40 | 50
6   70 | 100

我正在尝试使用以下脚本，但无法成功。

(df.groupby('df1')
   .agg({ 'df2' : ','.join})
   .reset_index()
   .reindex(columns=df.columns))

有人可以帮助应对这一挑战吗？如果在堆栈溢出中有任何类似的解决方案，请告诉我。

编辑： 第一个答案与上面的示例完美配合，但是当我尝试使用想要的数据时，它无法正常工作。我的真实数据如下所示。

    df1 df2
0   10  20
1   10  30
2   10  80
3   10  90
4   10  120
5   10  140
6   10  170
7   20  180
8   30  40
9   30  165
10  30  175
11  40  20
12  40  50
13  50  60
14  60  70
15  70  180
16  80  180
17  90  100
18  100 110
19  110 180
20  120 130
21  130 180
22  140 150
23  150 160
24  160 165
25  165 180
26  165 200
27  170 175
28  175 180
29  175 200
30  180 190
31  190 200
32  200 210
33  210 220
34  220 230
35  230 240
36  240 -

Answer 1

一种可能的解决方案：

import pandas as pd
from itertools import chain

l1 = [10, 20, 30, 60, 80, 40, 30, 70]
l2 = [20, 30, 40, 80, 70, 50, 90, 100]

d = dict()
for i, j in zip(l1, l2):
    if i == j:
        continue
    d.setdefault(i, []).append(j)

for k in d:
    d[k].extend(chain.from_iterable(d.get(v, []) for v in d[k]))

df = pd.DataFrame({'df1': list(d.keys()), 'df2': [', '.join(str(v) for v in d[k]) for k in d]})
print(df)

打印：

   df1                 df2
0   10  20, 30, 40, 90, 50
1   20      30, 40, 90, 50
2   30          40, 90, 50
3   60         80, 70, 100
4   80             70, 100
5   40                  50
6   70                 100

编辑：基于新输入数据的其他解决方案。现在，我正在检查路径中可能存在的圆圈：

import pandas as pd

data = '''
0   10  20
1   10  30
2   10  80
3   10  90
4   10  120
5   10  140
6   10  170
7   20  180
8   30  40
9   30  165
10  30  175
11  40  20
12  40  50
13  50  60
14  60  70
15  70  180
16  80  180
17  90  100
18  100 110
19  110 180
20  120 130
21  130 180
22  140 150
23  150 160
24  160 165
25  165 180
26  165 200
27  170 175
28  175 180
29  175 200
30  180 190
31  190 200
32  200 210
33  210 220
34  220 230
35  230 240
36  240 -
'''

df1, df2 = [], []
for line in data.splitlines()[:-1]: # <--- get rid of last `-` character
    line = line.strip().split()
    if not line:
        continue

    df1.append(int(line[1]))
    df2.append(int(line[2]))

from pprint import pprint

d = dict()
for i, j in zip(df1, df2):
    if i == j:
        continue
    d.setdefault(i, []).append(j)

for k in d:
    seen = set()
    for v in d[k]:
        for val in d.get(v, []):
            if val not in seen:
                seen.add(val)
                d[k].append(val)


df = pd.DataFrame({'df1': list(d.keys()), 'df2': [', '.join(str(v) for v in d[k]) for k in d]})
print(df)

打印：

    df1                                                df2
0    10  20, 30, 80, 90, 120, 140, 170, 180, 40, 165, 1...
1    20                  180, 190, 200, 210, 220, 230, 240
2    30  40, 165, 175, 20, 50, 180, 200, 190, 210, 220,...
3    40  20, 50, 180, 190, 200, 210, 220, 230, 240, 60, 70
4    50          60, 70, 180, 190, 200, 210, 220, 230, 240
5    60              70, 180, 190, 200, 210, 220, 230, 240
6    70                  180, 190, 200, 210, 220, 230, 240
7    80                  180, 190, 200, 210, 220, 230, 240
8    90        100, 110, 180, 190, 200, 210, 220, 230, 240
9   100             110, 180, 190, 200, 210, 220, 230, 240
10  110                  180, 190, 200, 210, 220, 230, 240
11  120             130, 180, 190, 200, 210, 220, 230, 240
12  130                  180, 190, 200, 210, 220, 230, 240
13  140   150, 160, 165, 180, 200, 190, 210, 220, 230, 240
14  150        160, 165, 180, 200, 190, 210, 220, 230, 240
15  160             165, 180, 200, 190, 210, 220, 230, 240
16  165             180, 200, 190, 210, 200, 220, 230, 240
17  170             175, 180, 200, 190, 210, 220, 230, 240
18  175             180, 200, 190, 210, 200, 220, 230, 240
19  180                       190, 200, 210, 220, 230, 240
20  190                            200, 210, 220, 230, 240
21  200                                 210, 220, 230, 240
22  210                                      220, 230, 240
23  220                                           230, 240
24  230                                                240

或pprint(d, width=250)：

{10: [20, 30, 80, 90, 120, 140, 170, 180, 40, 165, 175, 100, 130, 150, 190, 20, 50, 200, 110, 160, 60, 210, 70, 220, 230, 240],
 20: [180, 190, 200, 210, 220, 230, 240],
 30: [40, 165, 175, 20, 50, 180, 200, 190, 210, 220, 230, 240, 60, 70],
 40: [20, 50, 180, 190, 200, 210, 220, 230, 240, 60, 70],
 50: [60, 70, 180, 190, 200, 210, 220, 230, 240],
 60: [70, 180, 190, 200, 210, 220, 230, 240],
 70: [180, 190, 200, 210, 220, 230, 240],
 80: [180, 190, 200, 210, 220, 230, 240],
 90: [100, 110, 180, 190, 200, 210, 220, 230, 240],
 100: [110, 180, 190, 200, 210, 220, 230, 240],
 110: [180, 190, 200, 210, 220, 230, 240],
 120: [130, 180, 190, 200, 210, 220, 230, 240],
 130: [180, 190, 200, 210, 220, 230, 240],
 140: [150, 160, 165, 180, 200, 190, 210, 220, 230, 240],
 150: [160, 165, 180, 200, 190, 210, 220, 230, 240],
 160: [165, 180, 200, 190, 210, 220, 230, 240],
 165: [180, 200, 190, 210, 200, 220, 230, 240],
 170: [175, 180, 200, 190, 210, 220, 230, 240],
 175: [180, 200, 190, 210, 200, 220, 230, 240],
 180: [190, 200, 210, 220, 230, 240],
 190: [200, 210, 220, 230, 240],
 200: [210, 220, 230, 240],
 210: [220, 230, 240],
 220: [230, 240],
 230: [240]}

编辑2：如果df是带有“ df1”和“ df2”列的输入数据框：

from pprint import pprint

d = dict()
for i, j in zip(df.df1, df.df2):
    if i == j:
        continue
    if j == '-':   # <-- this will remove the `-` character in df2
        continue
    d.setdefault(i, []).append(j)

for k in d:
    seen = set()
    for v in d[k]:
        for val in d.get(v, []):
            if val not in seen:
                seen.add(val)
                d[k].append(val)


df = pd.DataFrame({'df1': list(d.keys()), 'df2': [', '.join(str(v) for v in d[k]) for k in d]})
print(df)

Answer 2

您好，感谢您的澄清，我有一个具有递归功能的解决方案，您可以尝试。对于大型数据帧，可能效率不高，但效果很好。该函数返回一个列表，但是您可以根据需要编辑结果系列以将该列表加入字符串中。

def get_related(df1, related):
    # get directly related values
    next_vals = df.loc[df['df1'] == df1, 'df2'].values.tolist()
    # remove links to self (will cause recursion issues)
    next_vals = list(set(next_vals) - set([df1]))
    # add to running list
    related = related + next_vals
    # continue to next level
    if any(next_val in df['df1'].unique() for next_val in next_vals):
        for next_val in next_vals:
            related = related + get_related(next_val, related)
    # get unique list
    return list(set(related))

df['df1'].apply(lambda x: get_related(x, []))

Answer 3

这应该可以解决问题：

def recursive_walk(df, node):
    parents=df.loc[(df['df1']==node) & (df['df2']!=node), 'df2'].tolist()
    if(len(parents)==0):
        yield node
    else:
        for parent in parents:
            yield parent
            lst=[el for el in recursive_walk(df, parent)]
            for el in lst:
                yield el

df['tree']=df.apply(lambda x: list(set([el for el in recursive_walk(df, x['df2'])]+[x['df2']])), axis=1)

输出：

   df1  df2                  tree
0   10   20  [40, 50, 20, 90, 30]
1   20   30      [40, 50, 90, 30]
2   30   40              [40, 50]
3   60   80                  [80]
4   70   70             [100, 70]
5   40   50                  [50]
6   30   90                  [90]
7   70  100                 [100]

（*）我还检查了扩展的数据帧-相当快，我不会共享输出，因为我的IDE正在截断它；）

如何根据数据帧列的顺序关系对其分组

3 个答案: