[英]Filter Pandas DataFrame using value_counts and multiple columns?
I have a dataset of orders and people who have placed those orders.我有一个订单数据集和下订单的人。 Orders have a unique identifier, and buyers have a unique identifier across multiple orders.
订单具有唯一标识符,买家在多个订单中具有唯一标识符。 Here's an example of that dataset:
这是该数据集的示例:
| Order_ID | Order_Date | Buyer_ID |
|----------|------------|----------|
| 123421 | 01/01/19 | a213422 |
| 123421 | 01/01/19 | a213422 |
| 123421 | 01/01/19 | a213422 |
| 346345 | 01/03/19 | a213422 |
| 567868 | 01/05/19 | a346556 |
| 567868 | 01/05/19 | a346556 |
| 234534 | 01/10/19 | a678909 |
I want to be able to filter the dataset to individuals who have only placed one order, even if that order has multiple items:我希望能够将数据集过滤到只下过一个订单的个人,即使该订单有多个项目:
| Order_ID | Order_Date | Buyer_ID |
|----------|------------|----------|
| 567868 | 01/05/19 | a346556 |
| 567868 | 01/05/19 | a346556 |
| 234534 | 01/10/19 | a678909 |
If I try df[df['Buyer_ID'].map(df['Buyer_ID'].value_counts()) == 1]
I get a really weird situation where the resulting dataframe is only rows where there's a 1 to 1 relationship between Order_ID
and Buyer_ID
.如果我尝试
df[df['Buyer_ID'].map(df['Buyer_ID'].value_counts()) == 1]
我会遇到一个非常奇怪的情况,结果数据框只是其中存在 1 到 1 关系的行Order_ID
和Buyer_ID
。 Like this:像这样:
| Order_ID | Order_Date | Buyer_ID |
|----------|------------|----------|
| 346345 | 01/03/19 | a213422 |
| 234534 | 01/10/19 | a678909 |
In the result I want, Buyer_ID
a213422
should not appear at all because that person has more than one Order_ID
.在我想要的结果中,
Buyer_ID
a213422
根本不应该出现,因为那个人有多个Order_ID
。
This leads me to believe that value_counts()
is either not the appropriate way to perform this filter, or I'm doing it wrong.这让我相信
value_counts()
要么不是执行此过滤器的合适方法,要么我做错了。 What would be the appropriate way to perform this filter?执行此过滤器的适当方法是什么?
Method 1: boolean indexing with groupby.transform
方法 1:使用
groupby.transform
布尔索引
df[df.groupby('Buyer_ID')['Order_ID'].transform('nunique').eq(1)]
Method 2: Groupby.filter
方法二:
Groupby.filter
df.groupby('Buyer_ID').filter(lambda x: x['Order_ID'].nunique()==1)
Method 3: boolean indexing
with Series.map
方法 3:使用
Series.map
boolean indexing
df[df['Buyer_ID'].map(df.groupby('Buyer_ID')['Order_ID'].nunique().eq(1))]
Output输出
Order_ID Order_Date Buyer_ID
4 567868 01/05/19 a346556
5 567868 01/05/19 a346556
6 234534 01/10/19 a678909
If you want to remove duplicates use DataFrame.drop_duplicates
at the end:如果要删除重复
DataFrame.drop_duplicates
,请在最后使用DataFrame.drop_duplicates
:
df[df.groupby('Buyer_ID')['Order_ID'].transform('nunique').eq(1)].drop_duplicates()
Order_ID Order_Date Buyer_ID
4 567868 01/05/19 a346556
6 234534 01/10/19 a678909
Here's another way you could do it:这是您可以执行的另一种方法:
import pandas as pd
# | Order_ID | Order_Date | Buyer_ID |
# |----------|------------|----------|
# | 123421 | 01/01/19 | a213422 |
# | 123421 | 01/01/19 | a213422 |
# | 123421 | 01/01/19 | a213422 |
# | 346345 | 01/03/19 | a213422 |
# | 567868 | 01/05/19 | a346556 |
# | 567868 | 01/05/19 | a346556 |
# | 234534 | 01/10/19 | a678909 |
df = pd.DataFrame.from_dict({
"Order_ID": [123421, 123421, 123421, 346345, 567868, 567868, 234534],
"Order_Date": ["01/01/19", "01/01/19", "01/01/19", "01/03/19", "01/05/19", "01/05/19", "01/10/19"],
"Buyer_ID": ["a213422", "a213422", "a213422", "a213422", "a346556", "a346556", "a678909"],
})
buyers_with_one_order = df.groupby(["Buyer_ID"]) \
.agg(num_orders=("Order_ID", pd.Series.nunique)) \
.query("num_orders == 1") \
.reset_index() \
.Buyer_ID
filtered_df = df.merge(buyers_with_one_order).drop_duplicates()
print(filtered_df.to_string(index=False))
# | Order_ID | Order_Date | Buyer_ID |
# |----------|------------|----------|
# | 567868 | 01/05/19 | a346556 |
# | 234534 | 01/10/19 | a678909 |
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.