我在大熊猫中有DataFrame，其中包含有关人的位置的及时信息。它大约有300+百万行。
示例：
import pandas as pd
inp = [{'Name': 'John', 'Year':2018, 'Address':'Beverly hills'}, {'Name': 'John', 'Year':2018, 'Address':'Beverly hills'}, {'Name': 'John', 'Year':2019, 'Address':'Beverly hills'}, {'Name': 'John', 'Year':2019, 'Address':'Orange county'}, {'Name': 'John', 'Year':2019, 'Address':'New York'}, {'Name': 'Steve', 'Year':2018, 'Address':'Canada'}, {'Name': 'Steve', 'Year':2019, 'Address':'Canada'}, {'Name': 'Steve', 'Year':2019, 'Address':'Canada'}, {'Name': 'Steve', 'Year':2020, 'Address':'California'}, {'Name': 'Steve', 'Year':2020, 'Address':'Canada'}, {'Name': 'John', 'Year':2020, 'Address':'Canada'}, {'Name': 'John', 'Year':2021, 'Address':'Canada'}, {'Name': 'John', 'Year':2021, 'Address':'Beverly hills'}, {'Name': 'Steve', 'Year':2021, 'Address':'California'}, {'Name': 'Steve', 'Year':2022, 'Address':'California'}, {'Name': 'Steve', 'Year':2018, 'Address':'NewYork'}, {'Name': 'Steve', 'Year':2018, 'Address':'California'}, {'Name': 'Steve', 'Year':2022, 'Address':'NewYork'}]
df = pd.DataFrame(inp)
print (df)
输出：
Address Name Year
0 Beverly hills John 2018
1 Beverly hills John 2018
2 Beverly hills John 2019
3 Orange county John 2019
4 New York John 2019
5 Canada Steve 2018
6 Canada Steve 2019
7 Canada Steve 2019
8 California Steve 2020
9 Canada Steve 2020
10 Canada John 2020
11 Canada John 2021
12 Beverly hills John 2021
13 California Steve 2021
14 California Steve 2022
15 NewYork Steve 2018
16 California Steve 2018
17 NewYork Steve 2022
我想计算特定年份中地址之间的变化的total。换句话说，2018年有多少人从“加拿大”移居到“加利福尼亚”。
理想的输出：
1]每年的矩阵如下。示例：2019年（包括2018年至2019年）中的所有地址更改。
+---------------+---------------+---------------+----------+------------+
| From\ To | Beverly hills | Orange county | New York | California |
+---------------+---------------+---------------+----------+------------+
| Beverly hills | 0 | 1 | 0 | 0 |
+---------------+---------------+---------------+----------+------------+
| Orange county | 0 | 0 | 1 | 0 |
+---------------+---------------+---------------+----------+------------+
| New York | 0 | 2 | 0 | 0 |
+---------------+---------------+---------------+----------+------------+
| California | 0 | 0 | 0 | 0 |
+---------------+---------------+---------------+----------+------------+
[2）全年地址更改。
+---------------+---------------+------+------+------+
| Address 1 | Address 2 | 2018 | 2019 | 2020 |
+---------------+---------------+------+------+------+
| Beverly hills | Orange county | 0 | 1 | 0 |
+---------------+---------------+------+------+------+
| New York | Canada | 0 | 0 | 1 |
+---------------+---------------+------+------+------+
| Canada | New York | 1 | 0 | 0 |
+---------------+---------------+------+------+------+
| California | Canada | 0 | 1 | 2 |
+---------------+---------------+------+------+------+
到目前为止我的解决方案：感谢@QuangHoang，我可以使用以下代码捕获“年份”的更改和“地址”的更改：
groups = df.groupby('Name')
for col in ['Year', 'Address']:
df[f'cng-{col}'] = groups[col].shift().fillna(df[col]).ne(df[col]).astype(int)
groups[col].shift()在每个名称中将对应的列移动1。 fillna(df[col]将每个（移位的）组的第一行填充为原始文件，表示没有更改。最后，ne(df[col])将移位后的值与原始值进行比较以进行更改。
收益率：
+----+---------------+-------+------+----------+-------------+
| ID | Address | Name | Year | cng-Year | cng-Address |
+----+---------------+-------+------+----------+-------------+
| 0 | Beverly hills | John | 2018 | 0 | 0 |
+----+---------------+-------+------+----------+-------------+
| 1 | Beverly hills | John | 2018 | 0 | 0 |
+----+---------------+-------+------+----------+-------------+
| 2 | Beverly hills | John | 2019 | 1 | 0 |
+----+---------------+-------+------+----------+-------------+
| 3 | Orange county | John | 2019 | 0 | 1 |
+----+---------------+-------+------+----------+-------------+
| 4 | New York | John | 2019 | 0 | 1 |
+----+---------------+-------+------+----------+-------------+
| 10 | Canada | John | 2020 | 1 | 1 |
+----+---------------+-------+------+----------+-------------+
| 11 | Canada | John | 2021 | 1 | 0 |
+----+---------------+-------+------+----------+-------------+
| 12 | Beverly hills | John | 2021 | 0 | 1 |
+----+---------------+-------+------+----------+-------------+
| 5 | Canada | Steve | 2018 | 0 | 0 |
+----+---------------+-------+------+----------+-------------+
| 15 | NewYork | Steve | 2018 | 1 | 1 |
+----+---------------+-------+------+----------+-------------+
| 16 | California | Steve | 2018 | 0 | 1 |
+----+---------------+-------+------+----------+-------------+
| 6 | Canada | Steve | 2019 | 1 | 0 |
+----+---------------+-------+------+----------+-------------+
| 7 | Canada | Steve | 2019 | 0 | 0 |
+----+---------------+-------+------+----------+-------------+
| 8 | California | Steve | 2020 | 1 | 1 |
+----+---------------+-------+------+----------+-------------+
| 9 | Canada | Steve | 2020 | 0 | 1 |
+----+---------------+-------+------+----------+-------------+
| 13 | California | Steve | 2021 | 1 | 1 |
+----+---------------+-------+------+----------+-------------+
| 14 | California | Steve | 2022 | 1 | 0 |
+----+---------------+-------+------+----------+-------------+
| 17 | NewYork | Steve | 2022 | 1 | 1 |
+----+---------------+-------+------+----------+-------------+
如果我理解问题的话..
df.drop_duplicates().groupby(['Name','Year']).size().reset_index(name="changes")
具有此输出
Name Year changes
0 John 2018 1
1 John 2019 3
2 John 2020 1
3 John 2021 2
4 Steve 2018 3
5 Steve 2019 1
6 Steve 2020 2
7 Steve 2021 1
8 Steve 2022 2