smarnach

This command can be used to retrieve pre-filtered logs from AWS:

aws logs filter-log-events \
    --log-group-name /aws/eks/<cluster_name>/cluster \
    --log-stream-name-prefix kube-apiserver-audit \
    --start-time <time_since_epoch_in_ms> \
    --filter-pattern '"v1beta1"' \
    --region <region>

smarnach

fn annihilate(polymer: &[u8], remove: u8) -> Vec<u8> {
    let mut result = Vec::new();
    for &c in polymer {
        if c == remove || c ^ 32 == remove {
            continue;
        }
        if result.is_empty() || result.last().unwrap() ^ c != 32 {
            result.push(c);
        } else {
            result.pop();

smarnach

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# Run with
#     sudo consul watch -type=service -service=redis ./consul-iptables.py

import json
import subprocess
import sys

def main():

smarnach

---

# Example sandbox configuration
# for single server community
# installs

- name: Bootstrap instance(s)
  hosts: all
  gather_facts: no
  become: True

smarnach

A comment on https://nedbatchelder.com/text/names.html

Myth: Python assigns mutable and immutable values differently.

While it is true that Python's assignment semantics are the same for both mutable and immutable types, the results look different for augmented assignments. E.g. the code

a = [1, 2, 3]

b = a

smarnach

As of: 2015-08-19

This is how to set up a combined devstack/insights virtualbox.

Run these commands on your host computer, in a new "combined-devstack" folder:

mkdir combined-devstack
cd combined-devstack
curl -L https://raw.githubusercontent.com/edx/configuration/master/vagrant/release/analyticstack/Vagrantfile > Vagrantfile

smarnach

>>> import ctypes
>>> buffer = (ctypes.c_char * 5)(*"abcde")
>>> ctypes.cast(buffer, ctypes.c_char_p)
c_char_p(3073075392)

smarnach

class A(object):
    def __init__(self, i):
        self.i = i
    def __radd__(self, other):
        return self.i

a = [A(42)]

# Test with built-in sum()
print(sum(a))

smarnach

In [1]: import numpy

In [2]: a = numpy.array([0.5, 1.3, 2.4, 10.3, 10.8, 10.2, 7.6, 3.2, 2.9])

In [3]: numpy.unwrap(2.0 * a) / 2.0
Out[3]: 
array([ 0.5       ,  1.3       ,  2.4       ,  0.87522204,  1.37522204,
        0.77522204,  1.31681469,  0.05840735, -0.24159265])

smarnach

import timeit
import numpy
import scipy.ndimage.morphology

def f0(a, kernel):
    b = numpy.convolve(a, kernel, mode="same") > 1
    b[1:] |= b[:-1]
    b[:-1] |= b[1:]
    return b