python - 将 pandas 数据框列转换为具有源和目标的 networkx 图

Question

我在 pandas 中有一个 DataFrame，其中包含有关人员及时位置的信息。它大约有 300+ 百万行。

这是一个示例，其中每个 Name 都分配给一个唯一index的 by并按group.byand 排序：NameYear

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)
df['Author_Grouped_Index'] = df.groupby(['Name']).ngroup()
df.sort_values(['Name', 'Year'], ascending=[False, True])

输出：

+-------+-------+------+---------------+----------------------+
| Index | Name  | Year | Address       | Name_Grouped_Index   |
+-------+-------+------+---------------+----------------------+
| 5     | Steve | 2018 | Canada        | 1                    |
+-------+-------+------+---------------+----------------------+
| 15    | Steve | 2018 | NewYork       | 1                    |
+-------+-------+------+---------------+----------------------+
| 16    | Steve | 2018 | California    | 1                    |
+-------+-------+------+---------------+----------------------+
| 6     | Steve | 2019 | Canada        | 1                    |
+-------+-------+------+---------------+----------------------+
| 7     | Steve | 2019 | Canada        | 1                    |
+-------+-------+------+---------------+----------------------+
| 8     | Steve | 2020 | California    | 1                    |
+-------+-------+------+---------------+----------------------+
| 9     | Steve | 2020 | Canada        | 1                    |
+-------+-------+------+---------------+----------------------+
| 13    | Steve | 2021 | California    | 1                    |
+-------+-------+------+---------------+----------------------+
| 14    | Steve | 2022 | California    | 1                    |
+-------+-------+------+---------------+----------------------+
| 17    | Steve | 2022 | NewYork       | 1                    |
+-------+-------+------+---------------+----------------------+
| 0     | John  | 2018 | Beverly hills | 0                    |
+-------+-------+------+---------------+----------------------+
| 1     | John  | 2018 | Beverly hills | 0                    |
+-------+-------+------+---------------+----------------------+
| 2     | John  | 2019 | Beverly hills | 0                    |
+-------+-------+------+---------------+----------------------+
| 3     | John  | 2019 | Orange county | 0                    |
+-------+-------+------+---------------+----------------------+
| 4     | John  | 2019 | New York      | 0                    |
+-------+-------+------+---------------+----------------------+
| 10    | John  | 2020 | Canada        | 0                    |
+-------+-------+------+---------------+----------------------+
| 11    | John  | 2021 | Canada        | 0                    |
+-------+-------+------+---------------+----------------------+
| 12    | John  | 2021 | Beverly hills | 0                    |
+-------+-------+------+---------------+----------------------+

我想获取网络图矩阵（邻接矩阵），以查看地址之间的总变化。换句话说，例如，有多少人在 2018 年从“加拿大”搬到“加利福尼亚”。

理想输出：

1)地址列的直接图。从技术上讲，将地址列转换为“源”和“目标”两列，其中“目标”值是下一行的“源”。最好计算另一列“重量”中的配对，而不是重复配对。

+------------+------------+------+--------+
| Source     | Target     | Year | Weight |
+------------+------------+------+--------+
| Canada     | NewYork    | 2018 |        |
+------------+------------+------+--------+
| NewYork    | California | 2018 |        |
+------------+------------+------+--------+
| California | Canada     | 2019 |        |
+------------+------------+------+--------+
| Canada     | Canada     | 2019 |        |
+------------+------------+------+--------+
| Canada     | California | 2020 |        |
+------------+------------+------+--------+
| California | Canada     | 2020 |        |
+------------+------------+------+--------+
| Canada     | California | 2021 |        |
+------------+------------+------+--------+
| California | California | 2022 |        |
+------------+------------+------+--------+
| California | NewYork    | 2022 |        |
+------------+------------+------+--------+

或者

2）一个矩阵来说明地址之间的总变化。

+---------------+--------+---------+------------+---------------+---------------+
| From \ To     | Canada | NewYork | California | Beverly hills | Orange county |
+---------------+--------+---------+------------+---------------+---------------+
| Canada        | 2      | 2       | 2          | 2             | 0             |
+---------------+--------+---------+------------+---------------+---------------+
| NewYork       | 1      | 0       | 1          | 0             | 0             |
+---------------+--------+---------+------------+---------------+---------------+
| California    | 2      | 1       | 1          | 0             | 0             |
+---------------+--------+---------+------------+---------------+---------------+
| Beverly hills | 0      | 0       | 0          | 2             | 1             |
+---------------+--------+---------+------------+---------------+---------------+
| Orange county | 0      | 1       | 0          | 0             | 0             |
+---------------+--------+---------+------------+---------------+---------------+

标签： pythondataframegraphnetworkxadjacency-matrix

Answer 1

这不是最漂亮的代码，但至少您可以按照每个步骤进行操作。我选择了第二个选项，因为您可以轻松地从此连接矩阵制作图表。您在制作 networkx 图表方面需要帮助吗？矩阵的行和列是：['Beverly hills', 'Orange County', 'New York', 'Canada', 'California', 'NewYork'] 您对每个人的 newyork 拼写不同，所以它出现了两次。

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)
df['Author_Grouped_Index'] = df.groupby(['Name']).ngroup()
df.sort_values(['Name', 'Year'], ascending=[False, True])

print (df)
dictionary_ = {} # where each person went
places = [] # all of the places
for index, row in df.iterrows():
    if row['Author_Grouped_Index'] not in dictionary_:
        dictionary_[row['Author_Grouped_Index']] = []
        dictionary_[row['Author_Grouped_Index']].append(row["Address"])
    else:
        dictionary_[row['Author_Grouped_Index']].append(row["Address"])
    if row["Address"] not in places:
        places.append(row["Address"])


print (dictionary_)

new_dictionary = {} #number of times each place visited
for key, value in dictionary_.items():
    for x in range(len(value)-1):
        move = value[x] + "-" + value[x+1]
        if not move in new_dictionary:
            new_dictionary[move] = 1
        else:
            new_dictionary[move] += 1

print (new_dictionary)
print (places)
import numpy as np
array = np.zeros((len(places),len(places)), dtype=int)
for x, place in enumerate(places):
    for y, place_2 in enumerate(places):

        move_2 = (place + "-" + place_2)
        try:
            array[x,y] = (new_dictionary[move_2])
        except:
            array[x,y] = 0

print (array)