dmarcelinobr

setwd("~/Documents/Data/BRAZIL/TSE")
todos.arquivos <-""
nomes.bases <- dir(path = getwd(), pattern =".txt")
for(i in 1:length(nomes.bases)){
base <- read.table(nomes.bases[i],header=TRUE, sep=";", 
stringsAsFactors=FALSE)
todos.arquivos <- rbind(todos.arquivos, base)
}
write.table(todos.arquivos, file = "bases_tse.txt",sep=";",
row.names = FALSE, qmethod = "double",

dmarcelinobr

import numpy as np
import matplotlib.pylab as plt
from scipy import stats

# For the last presidential elections
coalition = ['PT','PSDB','PSTU','PSB','PCO','PPS','PCO','PSL', 'PSDC','PRP','PT','PSDB','PDT', 'PSOL','PT','PSDB','PV','PSOL','PSDC','PCB','PRTB','PCO','PSTU']
year =	[2002,2002,2002,2002,2002,2002,2006,2006,2006,2006,2006,2006,2006,2006,2010,2010,2010,2010,2010,2010, 2010,2010,2010]
timeinsec =[312,623,83,133, 83, 257,71,75,75,71,437,613,243,71,639,438,83,62,56,56,56,56,56]
votes = [46.44,23.19,0.47,17.86,0.04,11.97,0,0.06,0.07,0.13, 48.61,41.64,2.64,6.85,46.91,32.61,19.33,0.87, 0.09,0.08,0.06,0.04,0.01]

dmarcelinobr

library(MCMCpack)
## Coping with polls: the more recent, the better they are.
##http://en.wikipedia.org/wiki/Opinion_polling_for_the_Scottish_independence_referendum,_2014

polls = NULL
polls <- data.frame( rbind(
  Survation = c(37, 50, 1010),
  YouGov  =  c(35, 55, 1142),
  TNSBMRB =  c(32, 45, 1003),
  IpsosMORI = c(40, 54, 1006),

dmarcelinobr

#!/Library/Frameworks/Python.framework/Versions/3.3/bin/python3
# -*- coding: utf-8 -*-
from urllib.request import urlopen
from bs4 import BeautifulSoup
import csv
from pymongo import MongoClient
from pandas import DataFrame
def rodaPesquisa():
client = MongoClient()
my_db = client["pesquisas_tse"]

dmarcelinobr

> summary(mcmc)    
forecast class object:
MCMC: 500 draws with adaptation of 500 and a thinning interval of 5.
Predicted at 2014-10-24, using 16 polls from 5 houses.
Evidence starting at 2014-03-10.
Election day 2014-10-26.
Machine runtime: ~ 7 minutes.


Expected results + swing voters:

dmarcelinobr

 ##############  ##################
 # Qual é a probabilidade de A estar a frente de B dada a diferença entre eles e o erro amostral?
 # Abaixo, a diferença é de .505 - .495 = 1%
 
 A = .505
 B = .495
 n = c(2500, 1111, 625, 400)
 
# Resolve a margem de erro por N 
(moe = sqrt((A*(1-A)/n)+(B*(1-B)/n)+(2*((A*B)/n)) ) )

dmarcelinobr

doInstall <- TRUE
require(Amelia)
require(ggplot2)

ANES <- read.csv("ANES.csv")
ANES <- ANES[ANES$year == 2008, -c(1, 11, 17)]  # Limit to just 2008 respondents,
head(ANES)  # remove some non-helpful variables
 
with(ANES, plot(jitter(pid7), jitter(ideo7)))
myModel <- lm(pid7 ~ ideo7 + female + age + south, data = ANES)

dmarcelinobr

forecast class object:
MCMC: 1000 samples with a burnin of 500 and a thinning of 5.
Predicted at 2014-09-20, using 79 polls from 9 houses.
Evidence starting at 2014-02-19.
Election day 2014-10-05.
Machine runtime: ~ 42 minutes.


Expected results + swing voters:  
Expected results: (Pooled)   National samples only 

dmarcelinobr

# Data latest polls
polls = NULL
polls <- data.frame( rbind(
  Opinium = c(43, 47, 1156),
  Survation = c(44, 48, 1000),
  ICM =  c(41, 45, 1175)
))

# set up for decimals
polls[, 1:2] <- polls[, 1:2]/100

dmarcelinobr

R> summary(mcmc2014)
forecast class object:
MCMC: 1000 samples with a burnin of 500 and a thinning of 5.
Predicted at 2014-09-20, using 33 polls from 5 houses.
Evidence starting at 2014-02-19.
Election day 2014-10-05.
Machine runtime: ~ 28 minutes.


Expected results + swing voters: