Using for loops, while, tidyverse, or packages to create a dataset with matching characteristics from a previous one (sampling)
I'm working with panel data. We assessed children in 2019 and 2020. Therefore, I have two datasets (2019 and 2020) and I want to create a third dataset matching the data from the second dataset (2020) that matches the characteristics of the first dataset (2019).This third dataset will have fewer participants, but they'll share the same characteristics of their "peers" from 2019. So, the proportion of boys and girls will be about the same of 2019, the mother's age will be about the same, etc.
Examaple:
Code:
df_2019 = structure(list(asqse_quest = c(24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24), year_completed_cat = structure(c(2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), levels = c("18", "19", "20", "21", "22", "23", "24"), class = "factor"),
sex_male = c(1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0,
1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1,
0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0), momage = c(36,
39, 22, 20, 29, 40, 31, 37, 29, 38, 24, 35, 32, 30, 32, 31,
29, 21, 28, 29, 40, 21, 38, 29, 28, 33, 25, 25, 30, 29, 25,
27, 28, 31, 24, 28, 35, 29, 17, 35, 32, 29, 27, 24, 29, 25,
28, 24, 21, 26), momed = c(4, 4, 2, 2, 4, 3, 2, 3, 2, 4,
3, 4, 4, 4, 4, 4, 3, 4, 3, 4, 4, 2, 2, 4, 4, 4, 4, 4, 4,
4, 2, 4, 3, 3, 3, 3, 4, 4, 2, 4, 4, 3, 2, 2, 3, 4, 4, 3,
2, 4), income = c(4, 4, 2, 3, 4, 1, 2, 5, 4, 4, 5, 4, 4,
4, 4, 4, 4, 2, 3, 3, 4, 2, 3, 4, 4, 4, 5, 4, 3, 3, 4, 4,
3, 4, 1, 4, 2, 4, 3, 4, 4, 3, 4, 3, 4, 4, 4, 3, 4, 4)), class = "data.frame", row.names = c(NA,
-50L))
df_2020 = structure(list(asqse_quest = c(24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24,
24, 24, 24, 24, 24, 24, 24, 24, 24, 24), year_completed_cat = structure(c(3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L), levels = c("18", "19", "20", "21", "22", "23", "24"), class = "factor"),
sex_male = c(1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1,
0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1), momage = c(23,
26, 33, 34, 29, 26, 23, 29, 40, 36, 33, 18, 31, 31, 31, 32,
34, 35, 29, 37, 19, 30, 33, 25, 32, 35, 37, 27, 23, 29, 28,
26, 30, 27, 38, 28, 29, 39, 26, 25, 29, 39, 35, 32, 20, 38,
31, 27, 28, 23), momed = c(2, 4, 4, 3, 4, 3, 2, 2, 3, 4,
1, 2, 2, 4, 4, 4, 4, 2, 4, 4, 2, 4, 4, 4, 2, 4, 4, 2, 4,
2, 1, 4, 3, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 4, 2, 4, 4, 4,
4, 1), income = c(2, 4, 4, 4, 4, 5, 3, 2, 2, 4, 1, 3, 4,
5, 1, 4, 3, 1, 4, 5, 5, 4, 4, 4, 3, 4, 4, 2, 4, 5, 1, 4,
4, 1, 4, 4, 4, 4, 3, 4, 4, 4, 5, 4, 2, 4, 4, 4, 4, 4)), class = "data.frame", row.names = c(NA,
-50L))
Created on 2024-07-12 with reprex v2.1.0
You can try the MatchIt package, which has a function that performs propensity score matching.
We first combine the two datasets with bind_rows, assigning an id to distinguish the two datasets:
library(dplyr)
data <- bind_rows(df_2019, df_2020, .id="year") |>
mutate(year=+(year==1)) # 1=2019 (treated), 0=2020 (controls)
Rows corresponding to year==1 are your treated (from 2019 data) and year==0 corresponds to your controls (from 2020 data).
To find controls that match as closely to the treated as possible, we can use the matchit function. There are a number of arguments, and for brevity, we'll just use the defaults.
library(MatchIt)
We'll first try to match exactly on years completed, sex, and mother's age and see if we have any luck.
match_obj <- matchit(year ~ asqse_quest+year_completed_cat+sex_male+momage+momed+income,
data = data,
exact= ~ year_completed_cat+sex_male+momage,
replace = FALSE)
#Error in `matchit()`:
#! No matches were found.
No surprise because the two datasets don't match at all on years completed. Let's make our matching condition less strict
match_obj <- matchit(year ~ asqse_quest+year_completed_cat+sex_male+momage+momed+income,
data = data,
exact= ~ sex_male+momage,
replace = FALSE)
There's no error this time, but we get a warning
#Warning message:
#Fewer control units than treated units in some `exact` strata; not all treated units will get a match.
That's ok. Now summarise the results.
summary(match_obj)
...
Sample Sizes:
Control Treated
All 50 50
Matched 25 25
Unmatched 25 25
Discarded 0 0
The output says that we found 25 controls from the original 50. Other useful information is given but I've omitted here for brevity. Now use match.data to get the matches, together with the original treated.
matched_data <- match.data(match_obj)
Now we simply filter out the treated and we are left with the matching controls:
df_2020_new <- filter(matched_data, year==0)
head(df_2020_new)
asqse_quest year_completed_cat sex_male momage momed income
1 24 20 1 23 2 2
2 24 20 1 26 4 4
3 24 20 1 33 4 4
4 24 20 1 34 3 4
5 24 20 0 29 4 4
6 24 20 1 26 3 5
7 24 20 0 23 2 3
8 24 20 1 29 2 2
9 24 20 0 40 3 2
10 24 20 1 36 4 4
Check the help page for matchit to see how you can modify the methods for matching. There's too much detail to give here, but this is the basic idea.
- grid.arrange's relative height argument didn't works well in RMarkdown
- How apply a function to every element of a matrix?
- Polar plot in R using ggplot2
- Find total length of time in multiple values of a dataset that overlaps with a given period of time
- Faster than R's `simplify2array`?
- What is the most simple way to label a data.table frequency chart in R?
- How to show dates in a secondary x-axis while the primary x-axis is numerical?
- online compiler for r-markdown document
- Unecessarily large column widths in HTML tables published by R Quarto when the table header contains special characters
- Bootstrapping a vector of results, by group in R
- Speeding Up Monte Carlo Simulations for Weather-Dependent Claims
- Is there an R class which allows to list 2 elements but shows only the first of them by default when called?
- Convert day of week number to weekday name in R
- How do you use "<<-" (scoping assignment) in R?
- Using reactivePoll to monitor and process file changes
- How to find out which graph is most similar to another graph?
- ggplot2: rectangular background behind axis text in circular bar plot
- How to simulate button click using rvest
- CMIP6, automating by month
- Covert dummy variables to single categorical in R?
- Animated reducing bar chart with gganimate
- Create an object class matrix/array from data frame
- How to add a vector of names to `sort_by()` in base R
- Among many subfolders, find CSVs starting with a given string and within ZIPs or not, and merge them while adding their names in a new column with R
- How to add a greek letter to an axis label to a 3D plot in plotly?
- bigrquery + dplyr error when using functions and sourcing from separate R script
- Shiny App Error: /v1/applications/ 400 - Validation Error Execution halted
- Rupee Unicode replacement not working in R
- Mutate multiple variable to create multiple new variables
- How to use the Stata equivalent of a foreach loop in R with mutate()