提问人：Anonymous 提问时间：5/8/2023 最后编辑：Anonymous 更新时间：5/10/2023 访问量：89
通过匹配多个列和条件对数据帧进行排序
Sorting the dataframe by matching multiple columns and conditions
问：
输入数据：
| source_element | source_text | source_entity | target_element | target_text | target_entity |
|---------------|-------------------|----------------|---------------|-------------|----------------|
| vvvvvv | ['body'] | name | aaaaa | ceo | code |
| vvvvvv | ['body'] | name | bbbbb | cap | code |
| vvvvvv | ['body', 'fat'] | name | ccccc | fat | name |
| hhhhhh | ['man', 'hat'] | name | kkkkk | hat | name |
| kkkkk | ['hat'] | name | lllll | 700 | price |
| kkkkk | ['hat'] | name | fffff | mud | code |
| eeeee | ['strong'] | name | zzzzz | 300 | price |
| ssssss | ['head'] | name | ddddd | zip | code |
| ssssss | ['head', 'strong']| name | eeeee | strong | name |
| ssssss | ['head', 'part'] | name | uuuuu | part | name |
| uuuuu | ['part'] | name | xxxxx | 190 | price |
输出：
| source_text | source_entity | target_text | target_entity |
|-----------------|----------------|----------------|-------------------|
| ['body', 'fat']| name | 'ceo' | code |
| ['body', 'fat']| name | 'cap' | code |
| ['man', 'hat'] | name | '700' | price |
| ['man', 'hat'] | name | 'mud' | code |
| ['head', 'strong'] | name | 'zip' | code |
| ['head', 'strong'] | name | '300' | price |
| ['head', 'part'] | name | 'zip' | code |
| ['head', 'part'] | name | '190' | price |
我需要满足以下条件的输出：
1.如果source_entity和target_entity是“name”，则该特定行的source_element或target_element与整个DataFrame中任何行的source_element匹配，将target_text和text_entity替换为匹配的数据帧行，并删除source_entity行，target_entity为“name”。
2.保留其他不受上述条件影响的数据帧行
答：
1赞
inquirer
5/8/2023
#1
import pandas as pd
cond = (df['source_entity'] == 'name') & (df['target_entity'] == 'name')
bbb = df[~df.index.isin(df[cond].index)]
def fff(x):
a = bbb[bbb['source_element'] == df.loc[x, 'source_element']]
b = bbb[bbb['source_element'] == df.loc[x, 'target_element']]
if len(a) > 0:
df.loc[a.index, 'source_text'] = df.loc[x, 'source_text']
return x
elif len(b) > 0:
df.loc[b.index, 'source_text'] = df.loc[x, 'source_text']
return x
aaa = [fff(i) for i in df[cond].index]
ind = df.index.isin(aaa)
print(df[~ind][['source_text', 'source_entity', 'target_text', 'target_entity']])
输出
source_text source_entity target_text target_entity
0 'body fat' name 'ceo' code
1 'body fat' name 'cap' code
4 'man hat' name '700' price
5 'head strong' name 'zip' code
如果需要重置索引：
print(df[~ind][['source_text', 'source_entity', 'target_text', 'target_entity']].reset_index(drop=True))
更新
这里使用了一种完全不同的方法，其中生成（创建）字符串。
“source_text”列数据将转换为字符串。df['source_text'].str.join(',')
和上次一样，我们得到的字符串没有双“名称”。bbb
但是，在函数中，获取了 'source_element'， 'target_element' 的匹配索引，如果找到它们，我们得到 .一切都写在列表中。在行中生成。source_text列是动态创建的。生成的列表由 连接。数据帧已创建。fff'source_text'aaaqqqassign(source_text=i[1])np.vstack
import numpy as np
import pandas as pd
df['source_text'] = df['source_text'].str.join(',')
cond = (df['source_entity'] == 'name') & (df['target_entity'] == 'name')
bbb = df[~df.index.isin(df[cond].index)]
def fff(x):
a = bbb[bbb['source_element'] == df.loc[x, 'source_element']]
b = bbb[bbb['source_element'] == df.loc[x, 'target_element']]
if len(a) > 0 or len(b) > 0:
tex = df.loc[x, 'source_text']
return [list(a.index) + list(b.index), tex]
aaa = [fff(i) for i in df[cond].index]
qqq = np.array(
[df.loc[i[0], ['source_entity', 'target_text', 'target_entity']].copy().assign(source_text=i[1]).values for i in aaa])
qqq = np.vstack(qqq)
df1 = pd.DataFrame(qqq, columns=['source_entity', 'target_text', 'target_entity', 'source_text'])
df1['source_text'] = df1['source_text'].str.split(',')
print(df1[['source_text', 'source_entity', 'target_text', 'target_entity']])
输出
source_text source_entity target_text target_entity
0 body,fat name ceo code
1 body,fat name cap code
2 man,hat name 700 price
3 man,hat name mud code
4 head,strong name zip code
5 head,strong name 300 price
6 head,part name zip code
7 head,part name 190 price
If you need to return the lists back:
df1['source_text'] = df1['source_text'].str.split(',')
评论
0赞
Anonymous
5/9/2023
Can You modify the code If I include extar input data: [('kkkkk' , 'hat', name, 'fffff', 'mud', code), ('eeeee', 'strong', name, 'zzzzz', '300', price), ('ssssss', 'head', 'part', name, 'uuuuu', 'part,' name),('uuuuu', 'part', name, 'xxxxx', '190', price)] Extra output: [('man', 'hat', name, 'mud', code), ('head', 'strong', name, '300', price), ('head', 'part', name, 'zip' , code), ('head', 'part', name , '190', price)]
0赞
inquirer
5/10/2023
@Anonymous post a table and what you want to get with explanations. By the fact that you have given it is not clear what the table and the expected result consist of.
0赞
inquirer
5/10/2023
To quickly understand: what are the differences from the previous version? And the data in the string is exactly in the list: ['body', 'fat'] or is it a string: 'body', 'fat'?
1赞
Anonymous
5/10/2023
It is exactly like list of strings :['body', 'fat'] and one more difference is I added these two rows in input data, ('ssssss', ['head', 'part'], 'name', 'uuuuu', 'part', 'name'), ('uuuuu', ['part'], 'name', 'xxxxx', '190', 'price');. The output differs from previous version a bit. But the conditions remain same!
1赞
Anonymous
5/10/2023
for idx, row in df[cond].iterrows(): a = df.loc[(df['source_element'] == row['source_element']) & (df['source_entity'] == 'name')] b = df.loc[(df['source_element'] == row['target_element']) & (df['source_entity'] == 'name')] source_text = row['source_text'] if len(a) > 0: df.loc[a.index, 'source_text'] = [source_text] * len(a) if len(b) > 0: df.loc[b.index, 'source_text'] = [source_text] * len(b) I modified the above code like this. But I am missing this row in output: ['head', 'strong'] name 'zip' code
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