Li Liu_Josselyn_Homework_1
Li Liu Josselyn
2024-01-18
# 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
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors## [[1]]
## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
## [7] "tidyr" "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [13] "grDevices" "utils" "datasets" "methods" "base"
##
## [[2]]
## [1] "fst" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
##
## [[3]]
## [1] "modelsummary" "fst" "lubridate" "forcats" "stringr"
## [6] "dplyr" "purrr" "readr" "tidyr" "tibble"
## [11] "ggplot2" "tidyverse" "stats" "graphics" "grDevices"
## [16] "utils" "datasets" "methods" "base"ess <- read_fst("All-ESS-Data.fst")unique(ess$cntry) ## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"Loading data
belgium_data <- read.fst("belgium_data.fst")france_data <- read.fst("france_data.fst")norway_data <- read.fst("norway_data.fst")ireland_data <- read.fst("ireland_data.fst")portugal_data <- read.fst("portugal_data.fst")serbia_data <- read.fst("serbia_data.fst")italy_data <- read.fst("italy_data.fst")##Task1
Provide code and answer.
Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium?
Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.
Here we filter to a specific value (Norway) within the cntry column and select variable of interest
norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)norway_happy$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#Recode values 77 through 99 to NA
norway_happy$y[norway_happy$y %in% 77:99] <- NA
#checking again
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$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# Recode values 77 through 99 to NA
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA
# checking again
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565Here we calculate the average for the variable ‘happy’ for Norway
mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005Here is the average for the variable ‘happy’ for Belgium
mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334The mean happiness for Norway (7.98) can be compared to the mean hapiness for Belgium (7.74), Norway reports higher levels of hapiness.
##Task2
Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
More info here: https://ess-search.nsd.no/en/variable/0c65116e-7481-4ca6-b1d9-f237db99a694.
Hint: need to convert numeric value entries to categories as specified in the variable information link. We did similar steps for Estonia and the climate change attitude variable.
Here we convert numeric values to categories
ireland_alcbnge <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_alcbnge$y <- ireland_alcbnge$alcbnge
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NAHere we calculate the mean and median for Ireland
mean_y <- mean(ireland_alcbnge$y, na.rm = TRUE)
median_y <- median(ireland_alcbnge$y, na.rm = TRUE)
cat("Mean of 'Ireland':", mean_y, "\n")## Mean of 'Ireland': 3.066977cat("Median of 'Ireland':", median_y, "\n")## Median of 'Ireland': 3Here we calculate the mode for Ireland categories
df <- ireland_alcbnge %>%
mutate(
ireland_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_
),
ireland_category = fct_relevel(factor(ireland_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$ireland_category)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239Here we check the most commom category
get_mode <- function(v) {
tbl <- table(v)
mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
return(mode_vals)
}
mode_values <- get_mode(df$y_category)## Warning in max(tbl): no non-missing arguments to max; returning -Infcat("Mode of binge drinking category:", paste(mode_values, collapse = ", "), "\n")## Mode of binge drinking category:The most common category selected, for Irish respondents, for frequency of binge drinking is weekly.
##Task3
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.
Here we filter to a specific value (Portugal) within the cntry column and select variable of interest
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)Here we filter to a specific value (Serbia) within the cntry column and select variable of interest
serbia_plnftr <- ess %>%
filter(cntry == "RS") %>%
select(plnftr)Here we use summary function to get data from Portugal and Serbia
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 :14604summary(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 :1505When comparing the two countries Portugal and Serbia with the variable plnftr I noticed in Portugal more people just take each day as it comes, but in contrasts Serbia has more people that plan for the future.
##Task4
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
)
italy_data$gndr[italy_data$gndr%in%9]<- NAHere we compute the mean and average for men
# Compute mean for male
mean_male_lrscale <- italy_data %>%
filter(gndr == "Male") %>%
summarize(mean_lrscale_men = mean(lrscale, na.rm = TRUE))
print(mean_male_lrscale)## mean_lrscale_men
## 1 24.32775Here we compute the mean and average by gender
means_by_gender <- italy_data %>%
group_by(gndr) %>%
summarize(lrscale = mean(lrscale, na.rm = TRUE))
print(means_by_gender)## # A tibble: 3 × 2
## gndr lrscale
## <chr> <dbl>
## 1 Female 28.3
## 2 Male 24.3
## 3 <NA> 22.2On average, women is more dissastified with democracy in Italy.
##Task5
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 %>%
# 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 = "Sastifaction(0-10)", title = "Boxplot of sastifaction with the state of education vs. health services") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
The median French person is more satisfied with the health services
because the sastifaction level is higher than the sastifaction level of
education.