如何通过将另一列中的两个值分组为两个值来添加两列值 [英] How to add two columns of values from grouping two by two values from another column
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问题描述
由于要分组文本值(在另一列中具有相同的值)，因此我想创建一个新的pandas数据框.例如，我得到了以下数据框:
I would like to create a new pandas data-frame as a result of grouping text values which has the same value in other column. So for instance, I got the following dataframe:
example_dct = {
"text": {
"0": "this is my text 1",
"1": "this is my text 2",
"2": "this is my text 3",
"3": "this is my text 4",
"4": "this is my text 5"
},
"article_id": {
"0": "#0001_01_xml",
"1": "#0001_01_xml",
"2": "#0001_02_xml",
"3": "#0001_03_xml",
"4": "#0001_03_xml"
}
}
df_example = pd.DataFrame.from_dict(example_dct)
print(df_example)
text article_id
0 this is my text 1 #0001_01_xml
1 this is my text 2 #0001_01_xml
2 this is my text 3 #0001_02_xml
3 this is my text 4 #0001_03_xml
4 this is my text 5 #0001_03_xml
我想通过以下方式创建两个新列:
I would like to create two new columns in the following way:
text_1 text_2 article_id
0 'this is my text 1' 'this is my text 2' #0001_01_xml
1 'this is my text 4' 'this is my text 5' #0001_03_xml
如果有> 2个具有相同id值的文本值，例如:
In the case that there is >2 text values with the same id value, example:
example_dct = {
"text": {
"0": "this is my text 1",
"1": "this is my text 2",
"2": "this is my text 3",
"3": "this is my text 4",
"4": "this is my text 5",
"5": "this is my text 6",
},
"article_id": {
"0": "#0001_01_xml",
"1": "#0001_01_xml",
"2": "#0001_02_xml",
"3": "#0001_03_xml",
"4": "#0001_03_xml",
"5": "#0001_03_xml",
}
}
然后输出数据帧应该是将1乘1的文本串联的结果:
Then the output dataframe should be the result of concatenating 1 by 1 texts:
text_1 text_2 article_id
0 'this is my text 1' 'this is my text 2' #0001_01_xml
1 'this is my text 4' 'this is my text 5' #0001_03_xml
2 'this is my text 4' 'this is my text 6' #0001_03_xml
3 'this is my text 5' 'this is my text 6' #0001_03_xml
此外，我想创建另一个与此数据集相似的数据集，但仅使用那些没有通用article_id的列(因此group by的倒数).
示例:
Furthermore, I would like to create another dataset similar to this one, but just with those columns that does not have common article_id (so the inverse of group by).
Example:
text_1 text_2 article_id_1 article_id_2
0 'this is my text 1' 'this is my text 3' #0001_01_xml. "#0001_02_xml"
1 'this is my text 1' 'this is my text 4' #0001_01_xml" #0001_03_xml"
2 'this is my text 1' 'this is my text 5' #0001_01_xml. "#0001_03_xml"
3 'this is my text 1' 'this is my text 6' #0001_01_xml "#0001_03_xml"
4 'this is my text 2' 'this is my text 3' #0001_02_xml "#0001_03_xml"
5 'this is my text 2' 'this is my text 4' #0001_02_xml "#0001_03_xml"
6 'this is my text 2' 'this is my text 5' #0001_02_xml "#0001_03_xml"
7 'this is my text 2' 'this is my text 6' #0001_02_xml "#0001_03_xml"
..
..
..
..
..
有什么主意我该怎么做?
Any ideas how can I make this approach?
推荐答案
在扁平化列表推导中，首次使用每组2个值的组合时，默认情况下会省略1个值的组:
For first use combinations of 2 values per groups in flattened list compreehnsion, there are groups with 1 values omitted by default:
example_dct = {
"text": {
"0": "this is my text 1",
"1": "this is my text 2",
"2": "this is my text 3",
"3": "this is my text 4",
"4": "this is my text 5",
"5": "this is my text 6",
},
"article_id": {
"0": "#0001_01_xml",
"1": "#0001_01_xml",
"2": "#0001_02_xml",
"3": "#0001_03_xml",
"4": "#0001_03_xml",
"5": "#0001_03_xml",
}
}
df = pd.DataFrame.from_dict(example_dct)
from itertools import combinations
L = [y + (name,) for name, x in df.groupby('article_id')['text'] for y in combinations(x, 2)]
df1 = pd.DataFrame(L, columns=['text_1','text_2', 'article_id'])
print(df1)
text_1 text_2 article_id
0 this is my text 1 this is my text 2 #0001_01_xml
1 this is my text 4 this is my text 5 #0001_03_xml
2 this is my text 4 this is my text 6 #0001_03_xml
3 this is my text 5 this is my text 6 #0001_03_xml
因此，如果将值0001_02_xml更改为0001_03_xml，则会得到:
So if changed values 0001_02_xml to 0001_03_xml get:
example_dct = {
"text": {
"0": "this is my text 1",
"1": "this is my text 2",
"2": "this is my text 3",
"3": "this is my text 4",
"4": "this is my text 5",
"5": "this is my text 6",
},
"article_id": {
"0": "#0001_01_xml",
"1": "#0001_01_xml",
"2": "#0001_03_xml",
"3": "#0001_03_xml",
"4": "#0001_03_xml",
"5": "#0001_03_xml",
}
}
df = pd.DataFrame.from_dict(example_dct)
from itertools import combinations
L = [y + (name,) for name, x in df.groupby('article_id')['text'] for y in combinations(x, 2)]
df1 = pd.DataFrame(L, columns=['text_1','text_2', 'article_id'])
print(df1)
text_1 text_2 article_id
0 this is my text 1 this is my text 2 #0001_01_xml
1 this is my text 3 this is my text 4 #0001_03_xml
2 this is my text 3 this is my text 5 #0001_03_xml
3 this is my text 3 this is my text 6 #0001_03_xml
4 this is my text 4 this is my text 5 #0001_03_xml
5 this is my text 4 this is my text 6 #0001_03_xml
6 this is my text 5 this is my text 6 #0001_03_xml
第二次使用:
df2 = (df.assign(a=1).merge(df.assign(a=1), on='a', suffixes=('_1','_2'))
.merge(df1, indicator=True, how='left')
.query('_merge == "left_only" & article_id_1 != article_id_2')
[['text_1','text_2', 'article_id_1','article_id_2']]
)
print (df2)
text_1 text_2 article_id_1 article_id_2
2 this is my text 1 this is my text 3 #0001_01_xml #0001_02_xml
3 this is my text 1 this is my text 4 #0001_01_xml #0001_03_xml
4 this is my text 1 this is my text 5 #0001_01_xml #0001_03_xml
5 this is my text 1 this is my text 6 #0001_01_xml #0001_03_xml
8 this is my text 2 this is my text 3 #0001_01_xml #0001_02_xml
9 this is my text 2 this is my text 4 #0001_01_xml #0001_03_xml
10 this is my text 2 this is my text 5 #0001_01_xml #0001_03_xml
11 this is my text 2 this is my text 6 #0001_01_xml #0001_03_xml
12 this is my text 3 this is my text 1 #0001_02_xml #0001_01_xml
13 this is my text 3 this is my text 2 #0001_02_xml #0001_01_xml
15 this is my text 3 this is my text 4 #0001_02_xml #0001_03_xml
16 this is my text 3 this is my text 5 #0001_02_xml #0001_03_xml
17 this is my text 3 this is my text 6 #0001_02_xml #0001_03_xml
18 this is my text 4 this is my text 1 #0001_03_xml #0001_01_xml
19 this is my text 4 this is my text 2 #0001_03_xml #0001_01_xml
20 this is my text 4 this is my text 3 #0001_03_xml #0001_02_xml
24 this is my text 5 this is my text 1 #0001_03_xml #0001_01_xml
25 this is my text 5 this is my text 2 #0001_03_xml #0001_01_xml
26 this is my text 5 this is my text 3 #0001_03_xml #0001_02_xml
30 this is my text 6 this is my text 1 #0001_03_xml #0001_01_xml
31 this is my text 6 this is my text 2 #0001_03_xml #0001_01_xml
32 this is my text 6 this is my text 3 #0001_03_xml #0001_02_xml
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