Tutorial 2: Exploring your data with ESS
First things first: setting up your environment
In the list following c(, you would be any package you need in between quotation marks.
# List of packages
packages <- c("tidyverse", "fst", "modelsummary") # add any you need here
# Install packages if they aren't installed already
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)
# Load the packages
lapply(packages, library, character.only = TRUE)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
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## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
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## ✖ dplyr::filter() masks stats::filter()
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## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
## `modelsummary` has built-in support to draw text-only (markdown) tables.
## To generate tables in other formats, you must install one or more of
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## install.packages(c(
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## config_modelsummary(factory_default = "markdown")## [[1]]
## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
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## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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## [1] "modelsummary" "fst" "lubridate" "forcats" "stringr"
## [6] "dplyr" "purrr" "readr" "tidyr" "tibble"
## [11] "ggplot2" "tidyverse" "stats" "graphics" "grDevices"
## [16] "utils" "datasets" "methods" "base"Loading Data into R
Let’s load our data, fst style. Remember the data needs to be in the folder where you’re working from. You can enter getwd() to check your working directory.
ess <- read_fst("All-ESS-Data.fst")Important
If you struggle to load the entire ESS dataset on your device (e.g., you get a memory error code), you can individually load the country datasets below after downloading them from the course website. Otherwise, skip this step and jump to “Subsetting”.
Remove hashtags and run the following commands to read each country dataset as specified below.
For both the homework and tutorial, you will need:
#belgium_data <- read.fst("belgium_data.fst")#estonia_data <- read.fst("estonia_data.fst")#france_data <- read.fst("france_data.fst")#norway_data <- read.fst("norway_data.fst")#ireland_data <- read.fst("ireland_data.fst")#portugal_data <- read.fst("portugal_data.fst")#serbia_data <- read.fst("serbia_data.fst")#italy_data <- read.fst("italy_data.fst")Task 1
Provide code and answer.
Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium?
Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)
# Norway_happy is being called, filtered and selected.norway_happy$y <- norway_happy$happy
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10# The contents are being put into a table.# need to remove 77, 88, 99 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 99 to NA
norway_happy$y[norway_happy$y %in% 77:99] <- NA
# checking again
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005# Now, let's determine the median of 'y'.
median_y <- median(norway_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8# Using tidyverse syntax principles, we can get both the mean and median at once.
norway_happy %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 7.975005 8belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$y <- belgium_happy$happy
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 50 27 104 194 234 830 999 3503 6521 3402 1565 3 16 3# need to remove 77, 88, 99 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 99 to NA
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA
# checking again
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334# Now, let's determine the median of 'y'.
median_y <- median(belgium_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8# Using tidyverse syntax principles, we can get both the mean and median at once.
belgium_happy %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 7.737334 8Norway has a 7.975 happiness rating out of 10, while Belgium has a 7.838 rating. You can see that Norway has a higher mean than Belgium.
Task 2
Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.
Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.
ireland_alcbnge <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_alcbnge$y <- ireland_alcbnge$alcbnge
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6# Recode values 6 through 8 to NA
ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NA# Converting to categories to get mode as a category instead of a number
df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category), ### here you would put the categories in order you want them to appear or else it will appear alphabetically
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239According to the table, Irish people most commonly drink alcohol on a weekly basis.
Task 3
Provide code and answer.
Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)
summary(portugal_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 3.000
## Median : 5.000
## Mean : 6.426
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :14604# Calling portugal_plnftr, filtering and selecting it so it can be summarized.serbia_plnftr <- ess %>%
filter(cntry == "RS") %>%
select(plnftr)
summary(serbia_plnftr)## plnftr
## Min. : 0.000
## 1st Qu.: 0.000
## Median : 4.000
## Mean : 4.983
## 3rd Qu.: 8.000
## Max. :88.000
## NA's :1505 # Calling serbia_plnftr, filtering and selecting it so it can be summarized.Portugal’s first quartile has a low rating for how much they plan the future, but it is still higher than Serbia’s non-existent rating. Howver, both countries’ third quartile rate themselves the same. Despite this, the discrepancies in their first quartiles explain why Portugal has a higher median, mean and NA’s.
Task 4
Provide code and answer.
Prompt and question: using the variables stfdem and gndr, answer the following: on average, who is more dissastified with democracy in Italy, men or women? Explain while referring to the output generated.
Info on variable here: https://ess.sikt.no/en/variable/query/stfdem/page/1
italy_data <- ess %>%
filter(cntry == "IT")
# Convert gender and stfdem
italy_data <- italy_data %>%
mutate(
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
TRUE ~ as.character(gndr)
),
stfdem = ifelse(stfdem %in% c(77, 88), NA, stfdem), # Convert stfdem values
)
means_by_gender <- italy_data %>%
group_by(gndr) %>% # here you are "grouping by" your second variable
summarize(stfdem = mean(stfdem, na.rm = TRUE)) # here you are summarizing your variable of interest
print(means_by_gender)## # A tibble: 3 × 2
## gndr stfdem
## <chr> <dbl>
## 1 9 3.25
## 2 Female 4.69
## 3 Male 4.78The table shows that females are more disatisfied with Italian democracy because their satisfaction score is lower (4.69 compared to 4.78).
Task 5
Provide code and answer.
Prompt: Interpret the boxplot graph of stfedu and stfhlth that we generated already: according to ESS data, would we say that the median French person is more satisfied with the education system or health services? Explain.
Change the boxplot graph: provide the code to change some of the key labels: (1) Change the title to: Boxplot of satisfaction with the state of education vs. health services; (2) Remove the x-axis label; (3) Change the y-axis label to: Satisfaction (0-10).
Hint: copy the boxplot code above and just replace or cut what is asked.
france_data <- ess %>%
filter(cntry == "FR")
france_data %>%
# Setting values to NA
mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
# Reshaping the data
select(stfedu, stfhlth) %>%
gather(variable, value, c(stfedu, stfhlth)) %>%
# Creating the boxplot
ggplot(aes(x = variable, y = value)) +
geom_boxplot() +
labs(y = "Y-axis", x = "X-axis", title = "Boxplot of stfedu vs. stfhlth") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
# The boxplot shows that the median French person is more satisfied with
health services than the education system, since the line is shown to be
higher (around 7, while the line for education system is 5).
france_data <- ess %>%
filter(cntry == "FR")
france_data %>%
# Setting values to NA
mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
# Reshaping the data
select(stfedu, stfhlth) %>%
gather(variable, value, c(stfedu, stfhlth)) %>%
# Creating the boxplot
ggplot(aes(x = variable, y = value)) +
geom_boxplot() +
labs(y = "Y-axis", x = "", title = "Boxplot of satisfaction with the state of education vs. health services") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).