How to download a list of all KEGG orthologs with respect to KEGG module (or pathway) in Python?
1
1
Entering edit mode
7 weeks ago
O.rka ▴ 290

Is there a way to download a table of all KEGG orthologs that go into specific modules and pathways? If possible, a tool in Python or commandline? I want to run GSEA to do gene set enrichment of KEGG modules or pathways. I already ran KOFAMSCAN but now I need the actual database and I'm having a difficult time finding this information.

kegg database python • 414 views
ADD COMMENT
0
Entering edit mode
ADD REPLY
0
Entering edit mode
7 weeks ago
O.rka ▴ 290

First download the json: https://www.kegg.jp/kegg-bin/get_htext?ko00001.keg

or do some fancy urllib and StringIO to pull directly from web:

A bit messy but here's my Python code. Forgive the lack of documentation but it's pretty straightforward. I coded it up on a plane ride:

import pandas as pd
from collections import * 

database = list()
for _, v in pd.read_json("/Users/jespinoz/Downloads/ko00001.json").iterrows():
    d = v["children"]
    cat_1 = d["name"]
    for child_1 in d["children"]:
        cat_2 = child_1["name"] # Module?
        for child_2 in child_1["children"]:
            cat_3 = child_2["name"]
            if "children" in child_2:
                for child_3 in child_2["children"]:
                    cat_4 = child_3["name"]
                    fields = [cat_1, cat_2, cat_3, cat_4]
                    database.append(fields)
df_kegg = pd.DataFrame(database, columns=["Level_A", "Level_B", "Level_C", "Level_D"])


def parse_ko_identifiers(x):
    x = x.upper()
    kos = list()
    elements = x.split(" ")
    for word in elements:
        if word:

            conditions = [
                word[0] == "K",
                word[1:].isnumeric(),
                len(word) == 6,
            ]
            if all(conditions):
                kos.append(word)
    return set(kos)
df_kegg["Level_D-KOs"] = df_kegg["Level_D"].map(parse_ko_identifiers)

database_expanded = dict()
for i, row in df_kegg.iterrows():
    for id_ko in row["Level_D-KOs"]:
        database_expanded[id_ko] = row
df_kegg_expanded = pd.DataFrame(database_expanded).T
df_kegg_expanded.index.name = "KO"
df_kegg_expanded.columns = df_kegg_expanded.columns.map(lambda x: (x.split("-")[0], x))

for id_cat in ["Level_A", "Level_B", "Level_C"]:
    df_kegg_expanded[(id_cat, "ID")] = df_kegg_expanded[(id_cat, id_cat)].map(lambda x: x.split(" ")[0])
    df_kegg_expanded[(id_cat, "Name")] = df_kegg_expanded[(id_cat, id_cat)].map(lambda x: " ".join(x.split(" ")[1:]))

def f(x):
    if "; " in x:
        return x.split("; ")[1]
    else:
        return x
df_kegg_expanded[("Level_D", "Name")] = df_kegg_expanded[("Level_D", "Level_D")].map(f)
df_kegg_expanded.columns = df_kegg_expanded.columns.map(lambda x: (x[0], "Full") if x[0] == x[1] else x)
df_kegg_expanded = df_kegg_expanded.sort_index(axis=1)

Looks like this: enter image description here

ADD COMMENT

Login before adding your answer.

Traffic: 1235 users visited in the last hour
Help About
FAQ
Access RSS
API
Stats

Use of this site constitutes acceptance of our User Agreement and Privacy Policy.

Powered by the version 2.3.6