lukmdo

default, prometheus-kube-state-metrics, NetworkPolicy (networking.k8s.io) has changed:
  # Source: prometheus/templates/kube-state-metrics-networkpolicy.yaml
  apiVersion: networking.k8s.io/v1
  kind: NetworkPolicy
  metadata:
    name: prometheus-kube-state-metrics
    labels:
      component: "kube-state-metrics"
      app: prometheus
      release: prometheus

lukmdo

--- master.txt	2018-12-02 19:09:21.000000000 +0000
+++ unified_labels.txt	2018-12-02 19:09:40.000000000 +0000
@@ -4,11 +4,11 @@
 kind: ConfigMap
 metadata:
   labels:
-    app: prometheus
-    chart: prometheus-7.4.5
     component: "alertmanager"
-    heritage: Tiller

lukmdo

"""
Tools for drawing Python object reference graphs with graphviz.

You can find documentation online at https://mg.pov.lt/objgraph/

Copyright (c) 2008-2017 Marius Gedminas <[email protected]> and contributors

Released under the MIT licence.
"""
# Permission is hereby granted, free of charge, to any person obtaining a

lukmdo

"""
For non inlplace dict setdefault.
"""
base_dict = {'a': 'a-ORIG', 'b': 'b-ORIG', 'd': 'd-ORIG'}
defaults = {'a': 'A', 'b': None, 'c': 0}
overrides = defaults

# Update by override. Like: 
# - base_dict.update(overrides) -> new dict BUT `dict.update` is inplace
# - base_dict + overrides -> new dict BUT unsupported operand

lukmdo

from __future__ import print_function
import inspect


def fn():
    return inspect.stack()[1][3]


class Example(object):

lukmdo

ps -ef

telnet

stats
stats slabs

stats items

stats cachedump SLAB_ID NUM_HOW_MANY

lukmdo

Github API response dumps

Public Request Resoponse Headers

HTTP/1.1 200 OK
Status: 200 OK
Server: GitHub.com
Access-Control-Allow-Credentials: true
Access-Control-Allow-Origin: *
Access-Control-Expose-Headers: ETag, Link, X-GitHub-OTP, X-RateLimit-Limit, X-RateLimit-Remaining, X-RateLimit-Reset, X-OAuth-Scopes, X-Accepted-OAuth-Scopes, X-Poll-Interval

lukmdo

def solution(items):
    if not items:
        return

    count = 0
    best = None

    for n in items:
        if count == 0:
            best = n

lukmdo

def solution(items):
    if not items:
        return

    best = None
    best_count = 0

    current_count = 0
    last = items[0]
    for current in items:

lukmdo

def test_solution():
    assert solution([]) == None
    assert solution([1]) == 1
    assert solution([1, 2]) == None
    assert solution([1, 2, 3]) == None
    assert solution([1, 2, 1]) == 1
    assert solution([1, 2, 1, 2]) == None


if __name__ == "__main__":