如何从 Pandas DataFrame 中绘制家谱？

Question

我有一张桌子，用来存储有关我祖先的信息。作为一个例子，我受《教父》的启发创建了一张类似的桌子。

  |--------+---+-------------+-----------+------+------+--------+--------+----------------+----------------|
  | ID     | S | First name  | Last name |  DoB |  DoD | FID    | MID    | Place of birth | Job            |
  |--------+---+-------------+-----------+------+------+--------+--------+----------------+----------------|
  | AnAn   | M | Antonio     | Andolini  |      | 1901 |        |        | Corleone       |                |
  | SiAn   | F | Signora     | Andolini  |      | 1901 |        |        | Corleone       | housewife      |
  | PaAn87 | M | Paolo       | Andolini  | 1887 | 1901 | AnAn   | SiAn   |                |                |
  | ViCo92 | M | Vito        | Corleone  | 1892 | 1954 | AnAn   | SiAn   | Corleone       | godfather      |
  | CaCo97 | F | Carmella    | Corleone  | 1897 | 1959 |        |        |                |                |
  | ToHa10 | M | Tom         | Hagen     | 1910 | 1970 | ViCo92 | CaCo97 | New York       | Consigliere    |
  | SaCo16 | M | Santino     | Corleone  | 1916 | 1948 | ViCo92 | CaCo97 | New York       | gangster       |
  | SaCo17 | F | Sandra      | Colombo   | 1917 |      |        |        | Messina        |                |
  | FrCo19 | M | Frederico   | Corleone  | 1919 | 1959 | ViCo92 | CaCo97 | New York       | Casino Manager |
  | MiCo20 | M | Michael     | Corleone  | 1920 | 1997 | ViCo92 | CaCo97 | New York       | godfather      |
  | ThHa20 | F | Theresa     | Hagen     | 1920 |      |        |        | New Jersey     | Art expert     |
  | LuMa23 | F | Lucy        | Mancini   | 1923 |      |        |        |                | Hotel employee |
  | KaAd24 | F | Kay         | Adams     | 1934 |      |        |        |                |                |
  | FrCo37 | F | Francessa   | Corleone  | 1937 |      | SaCo16 | SaCo17 |                |                |
  | KaCo37 | F | Kathryn     | Corleone  | 1937 |      | SaCo16 | SaCo17 |                |                |
  | FrCo40 | F | Frank       | Corleone  | 1940 |      | SaCo16 | SaCo17 |                |                |
  | SaCo45 | M | Santino Jr. | Corleone  | 1945 |      | SaCo16 | SaCo17 |                |                |
  | FrHa   | M | Frank       | Hagen     | 1940 |      | ToHa10 | Th20   |                |                |
  | AnHa42 | M | Andrew      | Hagen     | 1942 |      | ToHa10 | Th20   |                | Priest         |
  | ViMa   | M | Vincent     | Mancini   | 1948 |      | SaCo16 | LuMa23 | New York       | Godfather      |
  | GiHa58 | F | Gianna      | Hagen     | 1948 |      | ToHa10 | Th20   |                |                |
  | AnCo51 | M | Anthony     | Corleone  | 1951 |      | MiCo20 | KaAd24 | New York       | Singer         |
  | MaCo53 | F | Mary        | Corleone  | 1953 | 1979 | MiCo20 | KaAd24 | New York       | Student        |
  | ChHa54 | F | Christina   | Hagen     | 1954 |      | ToHa10 | Th20   |                |                |
  | CoCo27 | F | Constanzia  | Corleone  | 1927 |      | ViCo92 | CaCo97 | New York       | rentier        |
  | CaRi20 | M | Carlo       | Rizzi     | 1920 | 1955 |        |        | Nevada         | Bookmaker      |
  | ViRi49 | M | Victor      | Rizzi     | 1949 |      | CaRi20 | CoCo27 | New York       |                |
  | MiRi   | M | Michael     | Rizzi     | 1955 |      | CaRi20 | CoCo27 |                |                |
  |--------+---+-------------+-----------+------+------+--------+--------+----------------+----------------|

这里，个体之间的关系可以理解为有向无环图（DAG）。我的目标是使用图形绘制将该表可视化为家谱。

首先，我将表转换为边列表，其中ID是起始顶点和ParentID结束顶点：

import pandas as pd
rawdf = pd.read_csv('corleone.csv')
el1 = rawdf[['ID','MID']]
el2 = rawdf[['ID','FID']]
el1.columns = ['Child', 'ParentID']
el2.columns = el1.columns
el = pd.concat([el1, el2])
el = el.dropna()
df = el.merge(rawdf, left_index=True, right_index=True, how='left')
df['name'] = df[df.columns[4:6]].apply(lambda x: ' '.join(x.dropna().astype(str)),axis=1)
df = df.drop(['Child','FID', 'MID', 'First name', 'Last name'], axis=1)
df = df[['ID', 'name', 'S', 'DoB', 'DoD', 'Place of birth', 'Job', 'ParentID']]

给出以下数据框：

|--------+----------------------+---+--------+--------+----------------+----------------+----------|
| ID     | name                 | S |    DoB |    DoD | Place of birth | Job            | ParentID |
|--------+----------------------+---+--------+--------+----------------+----------------+----------|
| PaAn87 | Paolo Andolini       | M | 1887.0 | 1901.0 | NaN            | NaN            | SiAn     |
| PaAn87 | Paolo Andolini       | M | 1887.0 | 1901.0 | NaN            | NaN            | AnAn     |
| ViCo92 | Vito Corleone        | M | 1892.0 | 1954.0 | Corleone       | godfather      | SiAn     |
| ViCo92 | Vito Corleone        | M | 1892.0 | 1954.0 | Corleone       | godfather      | AnAn     |
| ToHa10 | Tom Hagen            | M | 1910.0 | 1970.0 | New York       | Consigliere    | CaCo97   |
| ToHa10 | Tom Hagen            | M | 1910.0 | 1970.0 | New York       | Consigliere    | ViCo92   |
| SaCo16 | Santino Corleone     | M | 1916.0 | 1948.0 | New York       | gangster       | CaCo97   |
| SaCo16 | Santino Corleone     | M | 1916.0 | 1948.0 | New York       | gangster       | ViCo92   |
| FrCo19 | Frederico Corleone   | M | 1919.0 | 1959.0 | New York       | Casino Manager | CaCo97   |
| FrCo19 | Frederico Corleone   | M | 1919.0 | 1959.0 | New York       | Casino Manager | ViCo92   |
| MiCo20 | Michael Corleone     | M | 1920.0 | 1997.0 | New York       | godfather      | CaCo97   |
| MiCo20 | Michael Corleone     | M | 1920.0 | 1997.0 | New York       | godfather      | ViCo92   |
| FrCo37 | Francessa Corleone   | F | 1937.0 |    NaN | NaN            | NaN            | SaCo17   |
| FrCo37 | Francessa Corleone   | F | 1937.0 |    NaN | NaN            | NaN            | SaCo16   |
| KaCo37 | Kathryn Corleone     | F | 1937.0 |    NaN | NaN            | NaN            | SaCo17   |
| KaCo37 | Kathryn Corleone     | F | 1937.0 |    NaN | NaN            | NaN            | SaCo16   |
| FrCo40 | Frank Corleone       | F | 1940.0 |    NaN | NaN            | NaN            | SaCo17   |
| FrCo40 | Frank Corleone       | F | 1940.0 |    NaN | NaN            | NaN            | SaCo16   |
| SaCo45 | Santino Jr. Corleone | M | 1945.0 |    NaN | NaN            | NaN            | SaCo17   |
| SaCo45 | Santino Jr. Corleone | M | 1945.0 |    NaN | NaN            | NaN            | SaCo16   |
| FrHa   | Frank Hagen          | M | 1940.0 |    NaN | NaN            | NaN            | Th20     |
| FrHa   | Frank Hagen          | M | 1940.0 |    NaN | NaN            | NaN            | ToHa10   |
| AnHa42 | Andrew Hagen         | M | 1942.0 |    NaN | NaN            | Priest         | Th20     |
| AnHa42 | Andrew Hagen         | M | 1942.0 |    NaN | NaN            | Priest         | ToHa10   |
| ViMa   | Vincent Mancini      | M | 1948.0 |    NaN | New York       | Godfather      | LuMa23   |
| ViMa   | Vincent Mancini      | M | 1948.0 |    NaN | New York       | Godfather      | SaCo16   |
| GiHa58 | Gianna Hagen         | F | 1948.0 |    NaN | NaN            | NaN            | Th20     |
| GiHa58 | Gianna Hagen         | F | 1948.0 |    NaN | NaN            | NaN            | ToHa10   |
| AnCo51 | Anthony Corleone     | M | 1951.0 |    NaN | New York       | Singer         | KaAd24   |
| AnCo51 | Anthony Corleone     | M | 1951.0 |    NaN | New York       | Singer         | MiCo20   |
| MaCo53 | Mary Corleone        | F | 1953.0 | 1979.0 | New York       | Student        | KaAd24   |
| MaCo53 | Mary Corleone        | F | 1953.0 | 1979.0 | New York       | Student        | MiCo20   |
| ChHa54 | Christina Hagen      | F | 1954.0 |    NaN | NaN            | NaN            | Th20     |
| ChHa54 | Christina Hagen      | F | 1954.0 |    NaN | NaN            | NaN            | ToHa10   |
| CoCo27 | Constanzia Corleone  | F | 1927.0 |    NaN | New York       | rentier        | CaCo97   |
| CoCo27 | Constanzia Corleone  | F | 1927.0 |    NaN | New York       | rentier        | ViCo92   |
| ViRi49 | Victor Rizzi         | M | 1949.0 |    NaN | New York       | NaN            | CoCo27   |
| ViRi49 | Victor Rizzi         | M | 1949.0 |    NaN | New York       | NaN            | CaRi20   |
| MiRi   | Michael Rizzi        | M | 1955.0 |    NaN | NaN            | NaN            | CoCo27   |
| MiRi   | Michael Rizzi        | M | 1955.0 |    NaN | NaN            | NaN            | CaRi20   |
|--------+----------------------+---+--------+--------+----------------+----------------+----------|

然后，我使用 graphviz 生成 DAG：

from graphviz import Digraph
f = Digraph('neato', format='pdf', encoding='utf8', filename='corleone', node_attr={'color': 'lightblue2', 'style': 'filled'})
f.attr('node', shape='box')
for index, row in df.iterrows():
    f.edge(str(row["ParentID"]), str(row["ID"]), label='')
f.view()

看起来像这样：

我面临的问题是有很多方面我想修改，例如：

• 男性有一种颜色，女性有另一种颜色
• 用名字代替 ID
• 箭头看起来像家谱箭头
• 能够在每个框中添加附加信息，例如 DoB、DoD 等。

我不知道是否可以使用 graphviz 来做到这一点（在文档中找不到如何实现），如果不能，我会对如何实现它的想法感兴趣。

Answer 1

这就是我的意思：

f = Digraph('neato', format='pdf', encoding='utf8',
            filename='corleone', node_attr={'color': 'lightblue2', 'style': 'filled'})
f.attr('node', shape='box')

# create all the possible nodes first
# you can modify the `label` 
for index, row in el.iterrows():
    f.node(row['ID'],label=row['First name'] + ' '+ row['Last name'], 
           _attributes={'color':'red' if row['S']=='M' else 'lightblue2'}
          )

for index, row in df.iterrows():
    f.edge(str(row["ParentID"]), str(row["ID"]), label='')

    
f.view()

我能够得到这样的东西。您可以进一步修改：