CamDavidsonPilon

from lifelines.utils import concordance_index
#assuming your data is in a dataframe

turnover_frequencies = df['turnover']
turnover_censored = df['turnover_censored'].astype(bool)
redshift = df['redshift']

ci = concordance_index(turnover_frequencies, redshift, turnover_censored)

print 2.0 * ci - 1.0

CamDavidsonPilon

CamDavidsonPilon

import pymc as pymc
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
 
def linear_setup ( ind_cols, dep_col):
    '''
        Inputs:list of strings for the independent variables,
        single string for the dependent variable
        Output: PyMC Model

CamDavidsonPilon

CamDavidsonPilon

# aftering import one_bucket_theta.py, https://gist.github.com/CamDavidsonPilon/8750e37242c4942c1984

from datetime import datetime

S = sc.parallelize([
    {'start': datetime(2000,10,11), 'end': datetime(2001,01,01)},
    {'start': datetime(2001,01,01), 'end': datetime(2002,01,01)},
    {'start': datetime(2002,01,01), 'end': datetime(2003,01,01)},
    {'start': datetime(2003,01,01), 'end': datetime(2004,01,01)},
    {'start': datetime(2004,01,01), 'end': datetime(2005,01,01)},

CamDavidsonPilon

from random import randint
from math import sqrt


def theta_join(S,T, join_condition = lambda s,t: s==t):
    """
    This implements the 1-bucket-theta algorithm found in 
    http://www.ccs.neu.edu/home/mirek/papers/2011-SIGMOD-ParallelJoins.pdf
    
    S,T: rdds of (key, value)

CamDavidsonPilon

{
 "metadata": {
  "name": "",
  "signature": "sha256:6c05f68f3874a728b39c0c5b111d6f90a2c8408facda8e5807cccf7fa086e4c6"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [

CamDavidsonPilon

CamDavidsonPilon

,0.01,0.04,0.06,0.09,0.11,0.14,0.16,0.19,0.22,0.24,0.27,0.29,0.32,0.35,0.37,0.40,0.42,0.45,0.47,0.50
750,5.3862512173306086e-05,0.00018431921564253194,0.00031355844405735097,0.00042940089912076783,0.0005413071841306608,0.0006332685115330671,0.0007476803364486353,0.0008116122112611769,0.0009076983192572439,0.0009633715254949598,0.0010418319478846378,0.0011075363347131727,0.0011778591911268729,0.00121129482248262,0.001259617421627161,0.0012865336098251596,0.0012832222488798127,0.001354041119620631,0.0013650443271181398,0.001315568626527553
1500,2.650184911783411e-05,9.209479830082467e-05,0.0001528774978470515,0.00020787484098875884,0.0002699076554215256,0.0003258727426287213,0.00035788667314687285,0.0004111510809490972,0.0004359117655034314,0.00047771994212569596,0.0005288493018160943,0.0005515370514559378,0.000580798803038502,0.000606253199804767,0.0006222858253985169,0.0006462444695299379,0.0006482967265578801,0.0006743000904418961,0.0006642287272113608,0.0006635825250303563
2250,1.754325587021407e-05,6.23366

CamDavidsonPilon

import numpy as np
from numpy.random import binomial, beta
import pandas as pd

N = np.r_[ 750*np.arange(1,30) ] 
p_assignment = 0.5

def sample_beta_posterior(N, C):
    return beta(1+C,N-C+1)