How to draw a family tree from a Pandas DataFrame?

Question

I have a table where I store information about my ancestors. As an example, I created a similar table inspired by The Godfather.

  |--------+---+-------------+-----------+------+------+--------+--------+----------------+----------------|
  | 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 |                |                |
  |--------+---+-------------+-----------+------+------+--------+--------+----------------+----------------|

Here, the relationship between individuals can be understood as a directed acyclic graph (DAG). My goal is to use graph drawing to visualize this table as a family tree.

Firstly I transform the table into an edge list where ID is the start vertex and ParentID the end vertex:

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']]

Which gives the following DataFrame:

|--------+----------------------+---+--------+--------+----------------+----------------+----------|
| 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   |
|--------+----------------------+---+--------+--------+----------------+----------------+----------|

Then, I use graphviz to generate a 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()

Which looks like this:

The issue that I face is that there are many aspects that I'd like to modify, such as:

having one color for males and another one for females
having names instead of IDs
having arrows looking like family tree arrows
being able to add additional information in each box, such as DoB, DoD, etc.

I don't know if it is possible to do that with graphviz (cannot find how in the documentation), and if not I'd be interested in ideas on how I could implement that.

user_5709240 · Answer

I improved the drawing but it still not reach my expectations. So here is the code with some comments on modifications.

Blank cells blank instead of NaN:
- keep_default_na=False
Replacing each blank in ParentID by a specific string:
- el.replace('', np.nan, regex=True, inplace = True)
- t = pd.DataFrame({'tmp':['no_entry'+str(i) for i in range(el.shape[0])]})
- el['ParentID'].fillna(t['tmp'], inplace=True)

import pandas as pd
import numpy as np
rawdf = pd.read_csv('corleone.csv', keep_default_na=False)
el1 = rawdf[['ID','MID']]
el2 = rawdf[['ID','FID']]
el1.columns = ['Child', 'ParentID']
el2.columns = el1.columns
el = pd.concat([el1, el2])
el.replace('', np.nan, regex=True, inplace = True)
t = pd.DataFrame({'tmp':['no_entry'+str(i) for i in range(el.shape[0])]})
el['ParentID'].fillna(t['tmp'], inplace=True)
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']]

Grouping edges with same starting and ending node, and having square edges
- graph_attr={"concentrate": "true", "splines":"ortho"})
Having nodes displaying name, job, DoB, Place of birth, DoD
- label=...
Defining node color according to sex
- _attributes={'color':'lightpink' if row['S']=='F' else 'lightblue'if row['S']=='M' else 'lightgray'}

from graphviz import Digraph
f = Digraph('neato', format='jpg', encoding='utf8', filename='corleone', node_attr={'style': 'filled'},  graph_attr={"concentrate": "true", "splines":"ortho"})
f.attr('node', shape='box')
for index, row in df.iterrows():
    f.node(row['ID'],
           label=
             row['name']
              + '
' + 
             row['Job'] 
             + '
'+ 
             row['DoB'] 
             + '
' + 
             row['Place of birth']
             + '
†' + 
             row['DoD'],
           _attributes={'color':'lightpink' if row['S']=='F' else 'lightblue'if row['S']=='M' else 'lightgray'})
for index, row in df.iterrows():
    f.edge(str(row["ParentID"]), str(row["ID"]), label='')  
f.view()

The result looks like this:

Which is much better. Still, two main flaws remain:

edges between parents and children are all split when they should look like this
I was not able to remove unnecessary newlines and death symbols

user_4238408 · Answer

Here's what I mean:

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()

I was able to get something like this. You can modify it more:

How to draw a family tree from a Pandas DataFrame?

2 Answers