# How Much Money Should Machines Earn?

Every inch of sky’s got a star
Every inch of skin’s got a scar

I think that a very good way to start with R is doing an interactive visualization of some open data because you will train many important skills of a data scientist: loading, cleaning, transforming and combinig data and performing a suitable visualization. Doing it interactive will give you an idea of the power of R as well, because you will also realise that you are able to handle indirectly other programing languages such as JavaScript.

That’s precisely what I’ve done today. I combined two interesting datasets:

• The probability of computerisation of 702 detailed occupations, obtained by Carl Benedikt Frey and Michael A. Osborne from the University of Oxford, using a Gaussian process classifier and published in this paper in 2013.
• Statistics of jobs from (employments, median annual wages and typical education needed for entry) from the US Bureau of Labor, available here.

Apart from using `dplyr` to manipulate data and `highcharter` to do the visualization, I used `tabulizer` package to extract the table of probabilities of computerisation from the `pdf`: it makes this task extremely easy.

This is the resulting plot:

If you want to examine it in depth, here you have a full size version.

These are some of my insights (its corresponding figures are obtained directly from the dataset):

• There is a moderate negative correlation between wages and probability of computerisation.
• Around 45% of US employments are threatened by machines (have a computerisation probability higher than 80%): half of them do not require formal education to entry.
• In fact, 78% of jobs which do not require formal education to entry are threatened by machines: 0% which require a master’s degree are.
• Teachers are absolutely irreplaceable (0% are threatened by machines) but they earn a 2.2% less then the average wage (unfortunately, I’m afraid this phenomenon occurs in many other countries as well).
• Don’t study for librarian or archivist: it seems a bad way to invest your time
• Mathematicians will survive to machines

The code of this experiment is available here.

# Bayesian Blood

The fourth, the fifth, the minor fall and the major lift (Hallelujah, Leonard Cohen)

Next problem is extracted from MacKay’s Information Theory, Inference and Learning Algorithms:

Two people have left traces of their own blood at the scene of a crime. A suspect, Oliver, is tested and found to have type ‘O’ blood. The blood groups of the two traces are found to be of type ‘O’ (a common type in the local population, having frequency 60%) and of type ‘AB’ (a rare type, with frequency 1%). Do these data give evidence in favor of the proposition that Oliver was one of the people who left blood at the scene?

To answer the question, let’s first remember the probability form of Bayes theorem:

$p(H|D)=\dfrac{p(H)p(D|H)}{p(D)}$

where:

• p(H) is the probability of the hypothesis H before we see the data, called the prior
• p(H|D) is the probablity of the hyothesis after we see the data, called the posterior
• p(D|H) is the probability of the data under the hypothesis, called the likelihood
• p(D)is the probability of the data under any hypothesis, called the normalizing constant

If we have two hypothesis, A and B, we can write the ratio of posterior probabilities like this:

$\dfrac{p(A|D)}{p(B|D)}=\dfrac{p(A)p(D|A)}{p(B)p(D|B)}$

If p(A)=1-p(B) (what means that A and B are mutually exclusive and collective exhaustive), then we can rewrite the ratio of the priors and the ratio of the posteriors as odds. Writing o(A) for odds in favor of A, we get the odds form of Bayes theorem:

$o(A|D)=o(A)\dfrac{p(D|A)}{p(D|B)}$

Dividing through by o(A) we have:

$\dfrac{o(A|D)}{o(A)}=\dfrac{p(D|A)}{p(D|B)}$

The term on the left is the ratio of the posteriors and prior odds. The term on the right is the likelihood ratio, also called the Bayes factor. If it is greater than 1, that means that the data were more likely under A than under B. And since the odds ratio is also greater than 1, that means that the odds are greater, in light of the data, than they were before. If the Bayes factor is less than 1, that means the data were less likely under A than under B, so th odds in favor of A go down.

Let’s go back to our initial problem. If Oliver left his blood at the crime scene, the probability of the data is just the probability that a random member of the population has type ‘AB’ blood, which is 1%. If Oliver did not leave blood at the scene, what is the the chance of finding two people, one with type ‘O’ and one with type ‘AB’? There are two ways it might happen: the first person we choose might have type ‘O’ and the second ‘AB’, or the other way around. So the probability in this case is 2(0.6)(0.01)=1.2%. Dividing probabilities of both scenarios we obtain a Bayes factor of 0.83, and we conclude that the blood data is evidence against Oliver’s guilt.

Once I read this example, I decided to replicate it using real data of blood type distribution by country from here. After cleaning data, I have this nice data set to work with:

For each country, I get the most common blood type (the one which the suspect has) and the least common and replicate the previous calculations. For example, in the case of Spain, the most common type is ‘O+’ with 36% and the least one is ‘AB-‘ with 0.5%. The Bayes factor is 0.005/(2(0.36)(0.005))=1.39 so data support the hypothesis of guilt in this case. Next chart shows Bayes factor accross countries:

• Sometimes data consistent with a hypothesis are not necessarily in favor of the hypothesis
• How different is the distribution of blood types between countries!
• If you are a estonian ‘A+’ murderer, choose carefully your accomplice

This is the code of the experiment:

```library(rvest)
library(dplyr)
library(stringr)
library(DT)
library(highcharter)

# Webscapring of the table with the distribution of blood types
url <- "http://www.rhesusnegative.net/themission/bloodtypefrequencies/"
blood <- url %>%
html_node(xpath='/html/body/center/table') %>%
html_table(fill=TRUE)

# Some data cleansing
blood %>% slice(-c(66:68)) -> blood

blood[,-c(1:2)] %>%
sapply(gsub, pattern=",", replacement=".") %>%
as.data.frame %>%
sapply(gsub, pattern=".79.2", replacement=".79") %>%
as.data.frame-> blood[,-c(1:2)]

blood %>%
sapply(gsub, pattern="%|,", replacement="") %>%
as.data.frame -> blood

blood[,-1] = apply(blood[,-1], 2, function(x) as.numeric(as.character(x)))

blood[,-c(1:2)] %>% mutate_all(funs( . / 100)) -> blood[,-c(1:2)]

# And finally, we have a nice data set
datatable(blood,
rownames = FALSE,
options = list(
searching = FALSE,
pageLength = 10)) %>%
formatPercentage(3:10, 2)

# Calculate the Bayes factor
blood %>%
mutate(factor=apply(blood[,-c(1,2)], 1, function(x) {min(x)/(2*min(x)*max(x))})) %>%
arrange(factor)-> blood

# Data Visualization
highchart() %>%
hc_chart(type = "column") %>%
hc_title(text = "Bayesian Blood") %>%
hc_subtitle(text = "An experiment about the Bayes Factor") %>%
hc_xAxis(categories = blood\$Country,
labels = list(rotation=-90, style = list(fontSize = "12px")))  %>%
hc_yAxis(plotBands = list(list(from = 0, to = 1, color = "rgba(255,215,0, 0.8)"))) %>%
color = "rgba(255, 0, 0, 0.5)",
name = "Bayes Factor")%>%
hc_yAxis(min=0.5) %>%
hc_tooltip(pointFormat = "{point.y:.2f}") %>%
hc_legend(enabled = FALSE) %>%
hc_exporting(enabled = TRUE) %>%
hc_chart(zoomType = "xy")
```

# Visualising The Evolution Of Migration Flows With rCharts

Heaven we hope is just up the road (Atlas, Coldplay)

Following with the analysis of migration flows, I have done next two visualizations. These charts are called bump charts and are very suitable to represent rankings. This is what I have done:

• Obtaining top 20 countries of the world according to % of migrants respect its population
• To do this, I divide total number of migrants between 1960 and 2009 by the mean population in the same period.
• I do the same to obtain top 20 countries of the world according to % of immigrants.
• In both cases, I only consider countries with population greater than 2 million.
• For these countries, I calculate % of migrants in each decade (60’s, 70’s, 80’s, 90’s and 00’s), dividing total number of migrants by mean population each decade
• I do the same in the case of immigrants.
• Instead of representing directly % of migrants and immigrants, I represent the ranking of countries according these indicators by decade

This is the bump chart of migrants:

And this is the one of immigrants:

• There is a permanent exodus in Puerto Rico: all decades (except 70’s) is located in the top 1 of countries with most migrants respect its population
• Ireland is also living a diaspora although in the 00’s decade has lost some positions
• Albania, Georgia and Bosnia and Herzegovina are gaining positions. Is East Europe gradually becoming uncomfortable?
• Jamaica is also moving up in this sad competition.
• On the other hand, Hong Kong and Israel are persistently leaders as receivers
• Saudi Arabia has presented an impressive growth receiving immigrants since 70’s
• United States does not appear in the immigrants ranking
• Singapore is gaining positions: in the 00’s decade is the third receiver country
• Also in the 00s, Switzerland is the first European country in the ranking, holding the fifth position

I like using rCharts as well as using Enigma data sets, as I have done previously. This is the code:

```library(data.table)
library(rCharts)
library(dplyr)
population %>%
filter(indicator_name=="Population, total") %>%
as.data.frame %>%
summarise(population=mean(value)) %>%
plyr::rename(., c("country_name"="country")) -> population2
populflows %>% filter(!is.na(total_migrants)) %>%
group_by(migration_year, destination_country) %>%
summarise(inmigrants = sum(total_migrants))  %>%
populflows %>% filter(!is.na(total_migrants)) %>%
group_by(migration_year, country_of_origin) %>%
summarise(migrants = sum(total_migrants)) %>%
# Join of data sets
migrants %>%
merge(inmigrants, by = c("country", "decade")) %>%
merge(population2, by = c("country", "decade")) %>%
mutate(p_migrants=migrants/population, p_inmigrants=inmigrants/population) -> populflows2
# Global Indicators
populflows2 %>%
group_by(country) %>%
summarise(migrants=sum(migrants), inmigrants=sum(inmigrants), population=mean(population)) %>%
mutate(p_migrants=migrants/population, p_inmigrants=inmigrants/population)  %>%
filter(population > 2000000)  %>%
mutate(rank_migrants = dense_rank(desc(p_migrants)), rank_inmigrants = dense_rank(desc(p_inmigrants))) -> global
# Migrants dataset
global %>%
filter(rank_migrants<=20) %>%
select(country) %>%
merge(populflows2, by = "country") %>%
plyr::ddply("decade", transform, rank = dense_rank(p_migrants)) -> migrants_rank
# Migrants dataset
global %>%
filter(rank_inmigrants<=20) %>%
select(country) %>%
merge(populflows2, by = "country") %>%
plyr::ddply("decade", transform, rank = dense_rank(p_inmigrants)) -> inmigrants_rank
# Function for plotting
plotBumpChart <- function(df){
bump_chart = Rickshaw\$new()
mycolors = ggthemes::tableau_color_pal("tableau20")(20)
bump_chart\$layer(rank ~ decade2, group = 'country_code', data = df, type = 'line', interpolation = 'none', colors = mycolors)
bump_chart\$set(slider = TRUE, highlight = TRUE, legend=TRUE)
bump_chart\$yAxis(tickFormat = "#!  function(y) { if (y == 0) { return '' } else { return String((21-y)) } } !#")
bump_chart\$hoverDetail(yFormatter = "#! function(y){return (21-y)} !#")
return(bump_chart)
}
plotBumpChart(migrants_rank)
plotBumpChart(inmigrants_rank)
```