Nagpal_Angela_Homework1
2024-01-16
packages <- c("tidyverse", "fst", "modelsummary")
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)
lapply(packages, library, character.only = TRUE)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.4
## ✔ 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
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## ✖ dplyr::filter() masks stats::filter()
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## [16] "utils" "datasets" "methods" "base"ess <- read_fst("All-ESS-Data.fst")rm(list=ls()); gc() ## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1098003 58.7 2196105 117.3 NA 1502405 80.3
## Vcells 1888757 14.5 1072251184 8180.7 16384 1304873991 9955.4packages <- c("tidyverse", "fst", "modelsummary")
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)
lapply(packages, library, character.only = TRUE)## [[1]]
## [1] "fstcore" "modelsummary" "fst" "lubridate" "forcats"
## [6] "stringr" "dplyr" "purrr" "readr" "tidyr"
## [11] "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [16] "grDevices" "utils" "datasets" "methods" "base"
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## [1] "fstcore" "modelsummary" "fst" "lubridate" "forcats"
## [6] "stringr" "dplyr" "purrr" "readr" "tidyr"
## [11] "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [16] "grDevices" "utils" "datasets" "methods" "base"
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## [1] "fstcore" "modelsummary" "fst" "lubridate" "forcats"
## [6] "stringr" "dplyr" "purrr" "readr" "tidyr"
## [11] "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [16] "grDevices" "utils" "datasets" "methods" "base"ess <- read_fst("All-ESS-Data.fst")Task 1: 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?
belgium_happy <- ess %>%
filter(cntry == "BE") %>%
select(happy)belgium_happy$y <- belgium_happy$happy
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 50 27 104 194 234 830 999 3503 6521 3402 1565 3 16 3mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.838519Norway_happy <- ess %>%
filter(cntry == "NO") %>%
select(happy)Norway_happy$y <- Norway_happy$happy
table(Norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10mean_y <- mean(Norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 8.076377#mean of belgium happy is 7.84. and mean of norway happy is 8.08, therefore Norway reports a higher level of happiness.
ireland_data <- read.fst("ireland_data.fst")Task 2: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
ireland_ccdrinking <- ess %>%
filter(cntry == "IL") %>%
select(alcbnge)ireland_ccdrinking$y <- ireland_ccdrinking$alcbnge
table(ireland_ccdrinking$y)##
## 1 2 3 4 5 7 8 9
## 15 153 149 264 362 5 69 1545ireland_ccdrinking$y[ireland_ccdrinking$y %in% 6:8] <- NA
table(ireland_ccdrinking$y)##
## 1 2 3 4 5 9
## 15 153 149 264 362 1545mean_y <- mean(ireland_ccdrinking$y, na.rm = TRUE)
median_y <- median(ireland_ccdrinking$y, na.rm = TRUE)
cat("Mean of 'y':", mean_y, "\n")## Mean of 'y': 7.049437cat("Median of 'y':", median_y, "\n")## Median of 'y': 9df <- ireland_ccdrinking %>%
mutate(
y_category = case_when(
y == 1 ~ "Don't Binge Drink",
y == 2 ~ "Rarely Binge Drink",
y == 3 ~ "Somewhat Binge Drink",
y == 4 ~ "Frequently Binge Drink",
y == 5 ~ "Extreme Binge Drink",
TRUE ~ NA_character_ ),
y_category = fct_relevel(factor(y_category),
"Don't Binge Drink",
"Rarely Binge Drink",
"Somewhat Binge Drink",
"Frequently Binge Drink",
"Extreme Binge Drink")
)table(df$y_category)##
## Don't Binge Drink Rarely Binge Drink Somewhat Binge Drink
## 15 153 149
## Frequently Binge Drink Extreme Binge Drink
## 264 362get_mode <- function(v) {
tbl <- table(v)
mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
return(mode_vals)
}
mode_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")## Mode of y category: Extreme Binge Drink#The most common stratergy for irish respondents is “extreme binge drink”
task 3: 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.
result <- ess %>%
filter(cntry %in% c("PT", "SE")) %>%
mutate(plnftr = recode(plnftr, `77` = NA_real_, `88` = NA_real_, `99` = NA_real_)) %>%
group_by(cntry) %>%
summarize(mean_plnftr = mean(plnftr, na.rm = TRUE))
print(result)## # A tibble: 2 × 2
## cntry mean_plnftr
## <chr> <dbl>
## 1 PT 5.42
## 2 SE 5.02#as Portuals mean is higher than serbia, they tend to take everyday as it comes more, in comparison to serbia.
Task 4: 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.
italy_data <- ess %>%
filter(cntry == "IT")
italy_data <- italy_data %>%
mutate(
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
TRUE ~ as.character(gndr)
),
lrscale = ifelse(lrscale %in% c(77, 88), NA, lrscale) # Convert lrscale values
)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 5.202087mean_female_lrscale <- italy_data %>%
filter(gndr == "Female") %>%
summarize(mean_lrscale_female = mean(lrscale, na.rm = TRUE))
print(mean_female_lrscale)## mean_lrscale_female
## 1 5.086377means_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 9 4.6
## 2 Female 5.09
## 3 Male 5.20so therefore, Men are more dissatisfied with democracy in Italy, in comparison to women.
Task 5: 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 %>%
mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
select(stfedu, stfhlth) %>%
gather(variable, value, c(stfedu, stfhlth)) %>%
ggplot(aes(x = variable, y = value)) +
geom_boxplot() +
labs(y = "Satisfaction", title = "Boxplot of satisfaction with the state of education vs. health services") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
# from the boxplot it looks like french citizens are more satisfied wiht
the health system in comparison to the education system.