Exercise9
Mengni Chen
Nov-05-2024
Now, the question is “does first birth affect women’s life satisfaction?”
No. 1
Question
1. Import 6 waves of women’s data
Answer
#or you use loop to import to avoid repetitive coding, similar to forvalues in stata
library(tidyverse)
library(haven)
library(splitstackshape)
library(ggplot2)
library(janitor)
women1 <- read_dta("wave1_women.dta")
women2 <- read_dta("wave2_women.dta")
women3 <- read_dta("wave3_women.dta")
women4 <- read_dta("wave4_women.dta")
women5 <- read_dta("wave5_women.dta")
women6 <- read_dta("wave6_women.dta")No. 2
Question
2.1: Keep only variables across the 6 waves: id, age, wave, relstat, hlt1, nkidsbio, sat
2.2: clean variables and drop observations when they have missing values
Answer
clean_fun <- function(df) { df %>%
transmute(
id,
age,
wave=as.numeric(wave),
relstat=as_factor(relstat), #make relstat as a factor
relstat=case_when(relstat== "-7 Incomplete data" ~ as.character(NA), #specify when is missing for relstat
TRUE ~ as.character(relstat))%>% as_factor(), #make relstat as a factor again
health=case_when(hlt1<0 ~ as.numeric(NA), #specify when hlt1 is missing
TRUE ~ as.numeric(hlt1)),
childno=case_when(nkidsbio==-7~ as.numeric(NA), #specify when is missing for relstat
TRUE ~ as.numeric(nkidsbio)),
sat=case_when(sat6<0 ~ as.numeric(NA), #specify when sat6 is missing
TRUE ~ as.numeric(sat6)),
)%>% drop_na() }
women1a <- clean_fun(women1)
women2a <- clean_fun(women2)
women3a <- clean_fun(women3)
women4a <- clean_fun(women4)
women5a <- clean_fun(women5)
women6a <- clean_fun(women6)No. 3
Question
3. Keep those who have no kids initially and follow them across 6 waves, and generate a wide-formatted dataset for 6 waves
Answer
women1b <- women1a %>% filter(childno==0)%>% #keep individuals who are childless in the first wave
rename(wave_1=wave, age_1=age, relstat_1=relstat, health_1=health, childno_1=childno, sat_1=sat ) #rename variables
women2b <- women2a %>%
rename(wave_2=wave, age_2=age, relstat_2=relstat, health_2=health, childno_2=childno, sat_2=sat )
women3b <- women3a %>%
rename(wave_3=wave, age_3=age, relstat_3=relstat, health_3=health, childno_3=childno, sat_3=sat )
women4b <- women4a %>%
rename(wave_4=wave, age_4=age, relstat_4=relstat, health_4=health, childno_4=childno, sat_4=sat )
women5b <- women5a %>%
rename(wave_5=wave, age_5=age, relstat_5=relstat, health_5=health, childno_5=childno, sat_5=sat )
women6b <- women6a %>%
rename(wave_6=wave, age_6=age, relstat_6=relstat, health_6=health, childno_6=childno, sat_6=sat )
women_wide <- left_join(women1b, women2b, by = "id") %>% # left join women1b and women2b
left_join(women3b, by = "id") %>% # left join with women3b
left_join(women4b, by = "id") %>% # left join with women4b
left_join(women5b, by = "id") %>% # left join with women5b
left_join(women6b, by = "id") # left join with women6b
#by using left_join I keep those have no kids in the first wave and follow themNo. 4
Question
4. Find out how many women participate in all 6 waves? _____
Answer
women_wide$check <- paste(women_wide$wave_1, women_wide$wave_2, women_wide$wave_3,
women_wide$wave_4, women_wide$wave_5, women_wide$wave_6, sep='-')
tabyl(women_wide,check)## check n percent
## 1-2-3-4-5-6 1479 0.392307692
## 1-2-3-4-5-NA 204 0.054111406
## 1-2-3-4-NA-6 74 0.019628647
## 1-2-3-4-NA-NA 192 0.050928382
## 1-2-3-NA-5-6 72 0.019098143
## 1-2-3-NA-5-NA 27 0.007161804
## 1-2-3-NA-NA-NA 278 0.073740053
## 1-2-NA-4-5-6 58 0.015384615
## 1-2-NA-4-5-NA 14 0.003713528
## 1-2-NA-4-NA-6 2 0.000530504
## 1-2-NA-4-NA-NA 20 0.005305040
## 1-2-NA-NA-NA-NA 401 0.106366048
## 1-NA-3-4-5-6 90 0.023872679
## 1-NA-3-4-5-NA 10 0.002652520
## 1-NA-3-4-NA-6 5 0.001326260
## 1-NA-3-4-NA-NA 20 0.005305040
## 1-NA-3-NA-5-6 9 0.002387268
## 1-NA-3-NA-5-NA 6 0.001591512
## 1-NA-3-NA-NA-NA 35 0.009283820
## 1-NA-NA-4-5-6 1 0.000265252
## 1-NA-NA-4-5-NA 1 0.000265252
## 1-NA-NA-4-NA-NA 1 0.000265252
## 1-NA-NA-NA-NA-NA 771 0.204509284#the answer is that 1479 women participated in all 6 wavesNo. 5
Question
5. Find out how many childless women have their first child over the 6 waves? _____
Step1: first transform the data from wide to long
Step2: define the transition from childless to first child. Note that first childbearing could be a single birth or a twin
Step3: do tabulations to find out the number of first-childbearing event
Answer for Step 1
women_long<- merged.stack(women_wide, #dataset for transfrom
var.stubs = c("age", "wave", "relstat", "health","childno", "sat"),
#var.stubs is to specify the prefixes of the variable groups
sep = "_") %>%
#sep is to specify the character that separates the "variable name" from the "times" in the source
drop_na(wave)
#drop the observations which did not join the waveAnswer for Step 2
women_long <- women_long %>%
group_by(id) %>%
mutate(
firstbirth=case_when( childno!=lag(childno, 1) & lag(childno, 1)==0 & childno>0 ~ 1,
TRUE ~ 0),
#when the person has 0 children at t-1 while has at least 1 child at t, define it first childbirth
twin=case_when( childno!=lag(childno, 1) & lag(childno, 1)==0 & childno==1 ~ "single birth", #single birth
childno!=lag(childno, 1) & lag(childno, 1)==0 & childno==2 ~ "likely twin", #twin birth
TRUE ~ "No birth") %>% as_factor()
#when the person has 0 children at t-1 while has 1 child at t, define it a single birth
#when the person has 0 children at t-1 while has 2 children at t, define it a likely twin birth
) Answer for Step 3
tabyl(women_long,firstbirth)## firstbirth n percent
## 0 14674 0.97670394
## 1 350 0.02329606tabyl(women_long,check,twin )## check No birth likely twin single birth
## 1-2-3-4-5-6 8656 9 209
## 1-2-3-4-5-NA 1001 0 19
## 1-2-3-4-NA-6 361 0 9
## 1-2-3-4-NA-NA 757 1 10
## 1-2-3-NA-5-6 352 1 7
## 1-2-3-NA-5-NA 103 1 4
## 1-2-3-NA-NA-NA 815 0 19
## 1-2-NA-4-5-6 280 2 8
## 1-2-NA-4-5-NA 55 0 1
## 1-2-NA-4-NA-6 8 0 0
## 1-2-NA-4-NA-NA 58 0 2
## 1-2-NA-NA-NA-NA 788 0 14
## 1-NA-3-4-5-6 427 2 21
## 1-NA-3-4-5-NA 39 0 1
## 1-NA-3-4-NA-6 20 0 0
## 1-NA-3-4-NA-NA 57 0 3
## 1-NA-3-NA-5-6 33 1 2
## 1-NA-3-NA-5-NA 17 0 1
## 1-NA-3-NA-NA-NA 67 0 3
## 1-NA-NA-4-5-6 4 0 0
## 1-NA-NA-4-5-NA 3 0 0
## 1-NA-NA-4-NA-NA 2 0 0
## 1-NA-NA-NA-NA-NA 771 0 0#350 women become mothers over the 6 waves; 333 women have given a single birth;17 women have given a birth of twinsNo. 6
Question
6. Randomly select 10 individuals who have their first child over the 6 waves, and plot the life satisfaction
Step1: please find out at which wave the first childbearing happened
Step2: please randomly select 10 individuals who have first child over the 6 waves
Step3: please plot the life satisfaction of these 10 individuals and also highlight the time of first childbirth in the graph
Answer for Step1
women_long <- women_long %>%
group_by(id) %>%
mutate(
birthwave=case_when(firstbirth==1 ~ wave)
#define the time when firstkid is 1, meaning that the person experience first childbearing event at this wave
)Answer for Step2
sample<- sample (women_long$id[women_long$firstbirth==1], size=10) #randomly select 10 individuals#or you can just
birth_id <- women_long %>%
filter(firstbirth==1) %>%
select(id)#restrict to individuals who have first birth
sample_id <- sample(birth_id$id, size=10) #randomly select 10
sample <- filter(women_long, id%in%sample_id) #find the 10 individuals in the women_long data set by matching idAnswer for Step3
ggplot(data=sample )+ #use ggplot to see changes of sat over time
geom_point(mapping=aes(x=wave,y=sat))+
geom_vline(mapping=aes(xintercept = birthwave ))+
facet_wrap(~ id, ncol=5) #this is new, to plot sat by id