查找每一行的第一个交点

Question

我有一个数据框架:

import pandas as pd
data =[[28, ['first'], 'apple edible', 23, 'apple is an edible fruit'],
[28, ['first'], 'apple edible', 34, 'fruit produced by an apple tree'],
[28, ['first'], 'apple edible', 39, 'the apple is a pome edible fruit'],
[21, ['second'], 'green plants', 11, 'plants are green'],
[21, ['second'], 'green plants', 7, 'plant these perennial green flowers']]
df = pd.DataFrame(data, columns=['day', 'group',  'bigram', 'count', 'sentence'])
+---+--------+------------+-----+-----------------------------------+
|day|group   |bigram      |count|sentence                           |
+---+--------+------------+-----+-----------------------------------+
|28 |[first] |apple edible|23   |apple is an edible fruit           |
|28 |[first] |apple edible|34   |fruit produced by an apple tree    |
|28 |[first] |apple edible|39   |the apple is a pome edible fruit   |
|21 |[second]|green plants|11   |plants are green                   |
|21 |[second]|green plants|7    |plant these perennial green flowers|
+---+--------+------------+-----+-----------------------------------+

我需要找到一个句子中重字的交集。此外，首先查找交叉点并标记为True。也就是说，在第一个交叉点之后，其余的交叉点将被标记为False。语序不重要。

所以我想要这样的结果:

+---+--------+------------+-----+--------------------------------+--------+
|day|group   |bigram      |count|sentence                        |        |
+---+--------+------------+-----+--------------------------------+--------+
|28 |[first] |apple edible|23   |apple is an edible fruit        |True    |
|28 |[first] |apple edible|34   |fruit produced by an apple tree |False   |
|28 |[first] |apple edible|39   |the apple is a pome edible fruit|False   |
|21 |[second]|green plants|11   |plant these perennial flowers   |False   |
|21 |[second]|green plants|7    |plants are green                |True    |
+---+--------+------------+-----+--------------------------------+--------+

Answer 1

首先通过将分割值转换为具有issubset的集合来测试所有交集，然后每个bigram只选择第一个Trues:

df['new'] = [set(b.split()).issubset(a.split()) for a,b in zip(df['sentence'],df['bigram'])]
df['new'] = ~df.duplicated(['bigram','new']) & df['new']
print (df)
day     group        bigram  count                             sentence  
0   28   [first]  apple edible     23             apple is an edible fruit   
1   28   [first]  apple edible     34      fruit produced by an apple tree   
2   28   [first]  apple edible     39     the apple is a pome edible fruit   
3   21  [second]  green plants     11                     plants are green   
4   21  [second]  green plants      7  plant these perennial green flowers   
new  
0   True  
1  False  
2  False  
3   True  
4  False  

如果双字母顺序需要交换并且需要第一个交集使用:

df['new'] = ~df.assign(bigram=df['bigram'].apply(lambda x: frozenset(x.split()))).duplicated(['bigram','new']) & df['new']

Answer 2

您可以使用两个步骤，第一个步骤是识别bigram是句子子集的行(使用issubset)，然后只保留第一个True:

# use python sets to identify the matching bigrams
df['intersection'] = [set(a.split()).issubset(b.split())
for a,b in zip(df['bigram'], df['sentence'])]
# select the non-first matches and replace with False
df.loc[~df.index.isin(df.groupby(df['group'].str[0])['intersection'].idxmax()),
'intersection'] = False

输出:

day     group        bigram  count                             sentence  intersection
0   28   [first]  apple edible     23             apple is an edible fruit          True
1   28   [first]  apple edible     34      fruit produced by an apple tree         False
2   28   [first]  apple edible     39     the apple is a pome edible fruit         False
3   21  [second]  green plants     11  plant these perennial green flowers         False
4   21  [second]  green plants      7                     plants are green          True