Python Pandas-提取唯一的列组合，并在另一个表中对它们进行计数

任务1:

我有这样的桌子：

+----------+------------+----------+------------+----------+------------+-------+
| a_name_0 | id_qname_0 | a_name_1 | id_qname_1 | a_name_2 | id_qname_2 | count |
+----------+------------+----------+------------+----------+------------+-------+
| country  | 1          | NAN      | NAN        | NAN      | NAN        | 100   |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 2          | city     | 8          | NAN      | NAN        | 20    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 2          | city     | 9          | NAN      | NAN        | 80    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 3          | age      | 4          | sex      | 6          | 40    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 3          | age      | 5          | sex      | 7          | 60    |
+----------+------------+----------+------------+----------+------------+-------+

{'a_name_0':'country','id_qname_0':1}
{'a_name_0':'region','id_qname_0':2, 'a_name_1':'city','id_qname_1':8}
{'a_name_0':'region','id_qname_0':2, 'a_name_1':'city','id_qname_1':9}

我需要将每一行串联起来，删除NaN，并在大小可变的字典中转换序列，例如，前两个dict将如下所示：

+----------+------------+----------+------------+----------+------------+-------+
| a_name_0 | id_qname_0 | a_name_1 | id_qname_1 | a_name_2 | id_qname_2 | count |
+----------+------------+----------+------------+----------+------------+-------+
| country  | 1          | NAN      | NAN        | NAN      | NAN        | 100   |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 2          | city     | 8          | NAN      | NAN        | 20    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 2          | city     | 9          | NAN      | NAN        | 80    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 3          | age      | 4          | sex      | 6          | 40    |
+----------+------------+----------+------------+----------+------------+-------+
| region   | 3          | age      | 5          | sex      | 7          | 60    |
+----------+------------+----------+------------+----------+------------+-------+

{'a_name_0':'country','id_qname_0':1}
{'a_name_0':'region','id_qname_0':2, 'a_name_1':'city','id_qname_1':8}
{'a_name_0':'region','id_qname_0':2, 'a_name_1':'city','id_qname_1':9}

之后的每一本词典都应该存储在一个列表中

任务2。

使用下表，我必须计算上一步dict中列的外观：

+----------+------------+----------+------------+----------+
| id       | country    | city     | age        | sex      | 
+----------+------------+----------+------------+----------+
| 1        | 1          | NAN      | NAN        | NAN      | 
+----------+------------+----------+------------+----------+
| 2        | 1          | 8        | NAN        | NAN      | 
+----------+------------+----------+------------+----------+

如果有更快的映射解决方案，请提供建议，因为我将要做的事情可能会非常混乱。

---

答案对我没有帮助，因为我需要迭代器来提取参数以及计算它们的外观

您可以通过使用

orient='r'

（

记录

）删除

count

列并将所有行转换为目录列表，然后在字典理解中过滤掉缺少值的目录：

L = [{k:v for k, v in x.items() if pd.notna(v)} for x in df.drop('count', 1).to_dict('r')]
print (L)
[{'a_name_0': 'country', 'id_qname_0': 1},
 {'a_name_0': 'region', 'id_qname_0': 2, 'a_name_1': 'city', 'id_qname_1': 8.0}, 
 {'a_name_0': 'region', 'id_qname_0': 2, 'a_name_1': 'city', 'id_qname_1': 9.0}, 
 {'a_name_0': 'region', 'id_qname_0': 3, 'a_name_1': 'age', 
 'id_qname_1': 4.0, 'a_name_2': 'sex', 'id_qname_2': 6.0},
 {'a_name_0': 'region', 'id_qname_0': 3, 'a_name_1': 'age',
 'id_qname_1': 5.0, 'a_name_2': 'sex', 'id_qname_2': 7.0}]

无法100%确定第二个数据帧：

L1 = [dict(zip(list(x.values())[::2], list(x.values())[1::2])) for x in L]
df = pd.DataFrame(L1)
print (df)
   country  region  city  age  sex
0      1.0     NaN   NaN  NaN  NaN
1      NaN     2.0   8.0  NaN  NaN
2      NaN     2.0   9.0  NaN  NaN
3      NaN     3.0   NaN  4.0  6.0
4      NaN     3.0   NaN  5.0  7.0

---

您可以通过使用

orient='r'

（

records

）删除

count

列并将所有行转换为目录列表，然后在字典理解中过滤掉缺少值的目录：

L = [{k:v for k, v in x.items() if pd.notna(v)} for x in df.drop('count', 1).to_dict('r')]
print (L)
[{'a_name_0': 'country', 'id_qname_0': 1},
 {'a_name_0': 'region', 'id_qname_0': 2, 'a_name_1': 'city', 'id_qname_1': 8.0}, 
 {'a_name_0': 'region', 'id_qname_0': 2, 'a_name_1': 'city', 'id_qname_1': 9.0}, 
 {'a_name_0': 'region', 'id_qname_0': 3, 'a_name_1': 'age', 
 'id_qname_1': 4.0, 'a_name_2': 'sex', 'id_qname_2': 6.0},
 {'a_name_0': 'region', 'id_qname_0': 3, 'a_name_1': 'age',
 'id_qname_1': 5.0, 'a_name_2': 'sex', 'id_qname_2': 7.0}]

无法100%确定第二个数据帧：

L1 = [dict(zip(list(x.values())[::2], list(x.values())[1::2])) for x in L]
df = pd.DataFrame(L1)
print (df)
   country  region  city  age  sex
0      1.0     NaN   NaN  NaN  NaN
1      NaN     2.0   8.0  NaN  NaN
2      NaN     2.0   9.0  NaN  NaN
3      NaN     3.0   NaN  4.0  6.0
4      NaN     3.0   NaN  5.0  7.0

---

很好，比我的解决方案df.drop（'count'，axis=1）中的index行的

更好。iterrows（）：list\uu.append（row.dropna（）.to_json（orient='index'））

@datanoveler list理解是一个好办法：但我会采用任何有效的解决方案，因为这不是大数据案例。谢谢你，我的朋友乔泽夫！你又做了一次；）很好，比我的解决方案df.drop（'count'，axis=1）中的index行的

更好。iterrows（）：list\uu.append（row.dropna（）.to_json（orient='index'））

@datanoveler list理解是一个好办法：但我会采用任何有效的解决方案，因为这不是大数据案例。谢谢你，我的朋友乔泽夫！你又做了一次；）如何提取第二个

数据帧的
id
？@jezrael'id'与此无关。我只需要扫描dict中的列是否在行中并对它们进行计数。如何提取第二个数据帧的
id
？@jezrael'id'与此无关。我只需要扫描dict中的列是否在行中，并对它们进行计数。