Parse FabioLolix/LinuxTimeline to get the chain of ancestry for Linux distros and find the longest

linuxtree.py

#!/usr/bin/env python3

import requests
try:
    from requests_cache import CachedSession as MySession
except:
    MySession = requests.Session


def unstring(s):
    if s.startswith('"') and s.endswith('"'): return s[1:-1]
    return s

def decorate(distros, name, depth=0, chain=[]):
    distros[name]["depth"] = depth
    distros[name]["chain"] = chain[:]
    children = [x for x in distros if distros[x]["p"] == name]
    for c in children:
        cchain = chain[:]
        cchain.append(name)
        decorate(distros, c, depth + 1, cchain)

distros = {}

session = MySession()
resp = session.get("https://raw.githubusercontent.com/FabioLolix/LinuxTimeline/master/ldt.csv")
c = resp.content.decode("utf-8")

for line in c.split("\n"):
    parts = [unstring(x) for x in line.split(",")]
    distro = None
    if parts[0] in ["N"]: #, "#N", "//N"]:
        distro = parts[1].strip()
        parent = parts[3].strip()
    if distro:
        distros[distro] = {"p": parent}

unparents = [x for x in distros if not distros[x]["p"]]
for u in sorted(unparents):
    decorate(distros, u, 0, [])

bydepth = {}
for name, details in distros.items():
    d = details["depth"]
    if d not in bydepth:
        bydepth[d] = []
    bydepth[d].append(name)

for n in sorted(list(bydepth.keys()))[-3:]:
    print("")
    print("Chains of length", n + 1)
    print("==================")
    s = []
    for d in bydepth[n]:
        s.append(' -> '.join((distros[d]["chain"] + [d])))
    print("\n".join(sorted(s)))