I have a survey dataset which includes intra-household relationships and I'm trying to write code to identify lone parent households, defined as a household where the parent of a dependent child does not have a cohabiting partner.
Intra-family relationships are coded as:
1 = Spouse, 2= Cohabiting partner, 3 = Son/daughter, 4 = Step son/daughter, 5 = Foster child, 6 = Son-in-law/daughter-in-law, 7 = Parent/guardian, 8 = Step-parent, 9 = Foster parent, 10 = Parent-in-law, 11 = Brother/sister, 12 = Step-brother/sister, 13 = Foster brother/sister, 14 = Brother/sister-in-law, 15 = Grand-child, 16 = Grand-parent, 17 = Other relative, 18 = Other non-relative.
The identifiers of a parent therefore are 7, 8, or 9 in any of a person's relationship columns, however whether their child is dependent (under18) is represented in the child's depchild column. Whether the parent of a depchild has a partner is identified by 1 or 2 in any of the parents relationship columns.
I can't preclude the possibility of multiple families within a given household, e.g. (two lone mothers independently living with two dependent children) therefore the presence of two parents within a household with dependent children does not automatically mean a non-lone-parent household. If there are any lone-parents in a household i.e. a parent of a dependent child who does not have a partner, the household should be tagged as lonepar = 1.
Example Data
household person depchild R01 R02 R03 R04 R05 R06
1 1 1 0 NA 1 7 7 NA NA
2 1 2 0 1 NA 7 7 NA NA
3 1 3 0 3 3 NA 11 NA NA
4 1 4 1 3 3 11 NA NA NA
5 2 1 0 NA 7 16 NA NA NA
6 2 2 0 3 NA 7 NA NA NA
7 2 3 1 15 3 NA NA NA NA
8 3 1 0 NA 18 NA NA NA NA
9 3 2 0 18 NA NA NA NA NA
10 4 1 0 NA NA NA NA NA NA
11 5 1 0 NA 9 NA NA NA NA
12 5 2 1 5 NA 18 NA NA NA
13 5 3 0 2 18 NA NA NA NA
In the above example, dependent children depchild are on rows 4, 7 and 12. The parents of the child on row 4 have spouses, indicated by 1 in R02 and R01 respectively; the household is therefore not a lone-parent household, so should be lonepar = 0. The parent of the depchild on row 7 however (row 6) does not have a spouse 1 or a cohabiting partner 2, the household should therefore be lonepar = 1
Output sought
household person depchild R01 R02 R03 R04 R05 R06 lonepar
1 1 1 0 NA 1 7 7 NA NA 0
2 1 2 0 1 NA 7 7 NA NA 0
3 1 3 0 3 3 NA 11 NA NA 0
4 1 4 1 3 3 11 NA NA NA 0
5 2 1 0 NA 7 16 NA NA NA 1
6 2 2 0 3 NA 7 NA NA NA 1
7 2 3 1 15 3 NA NA NA NA 1
8 3 1 0 NA 18 NA NA NA NA 0
9 3 2 0 18 NA NA NA NA NA 0
10 4 1 0 NA NA NA NA NA NA 0
11 5 1 0 NA 9 NA NA NA NA 0
12 5 2 1 5 NA 18 NA NA NA 0
13 5 3 0 2 18 NA NA NA NA 0
Example Code
df <- data.frame(household = c(1,1,1,1,2,2,2,3,3,4,5,5,5),
person = c(1,2,3,4,1,2,3,1,2,1,1,2,3),
depchild = c(0,0,0,1,0,0,1,0,0,0,0,1,0),
R01 = c(NA, 1, 3, 3, NA, 3, 15, NA, 18, NA, NA, 5,2),
R02 = c(1, NA, 3, 3, 7, NA, 3, 18, NA, NA, 9, NA, 18),
R03 = c(7, 7, NA, 11, 16, 7, rep(NA,5), 18, NA),
R04 = c(7, 7, 11, rep(NA, 10)),
R05 = rep(NA, 13),
R06 = rep(NA, 13))
Rather than concentrating on relationships of the parents, concentrate on the relationships of the dependent children. If a dependent child only has a single relation with a value of 3, 4, or, 5, then that dependent child only has a single parent in the household.
Essentially, we count up the instances of 3, 4, and 5 in each row for every person in the data frame. Then we group by household. If anyone in that household is a dependent child who only had one instance of a 3, 4, or 5 relationship code, then that household contains a dependent child with only one parent. It is therefore a lone parent household.
library(tidyverse)
df %>%
rowwise() %>%
mutate(n = length(na.omit(match(c(R01, R02, R03, R04, R05, R06), 3:5)))) %>%
group_by(household) %>%
mutate(lonepar = as.numeric(any(depchild == 1 & n == 1))) %>%
select(-n)
#> # A tibble: 12 x 10
#> # Groups: household [5]
#> household person depchild R01 R02 R03 R04 R05 R06 lonepar
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <lgl> <dbl>
#> 1 1 1 0 NA 1 7 7 NA NA 0
#> 2 1 2 0 1 NA 7 7 NA NA 0
#> 3 1 3 0 3 3 NA 11 NA NA 0
#> 4 1 4 1 3 3 11 NA NA NA 0
#> 5 2 1 0 NA 7 16 NA NA NA 1
#> 6 2 2 0 3 NA 7 NA NA NA 1
#> 7 2 3 1 15 3 NA NA NA NA 1
#> 8 3 1 0 NA 18 NA NA NA NA 0
#> 9 3 2 0 18 NA NA NA NA NA 0
#> 10 4 1 0 NA NA NA NA NA NA 0
#> 11 5 1 0 NA 9 NA NA NA NA 1
#> 12 5 2 1 5 NA NA NA NA NA 1
Created on 2022-05-16 by the reprex package (v2.0.1)
Related
This question already has answers here:
Replacing values from a column using a condition in R
(2 answers)
Closed 7 months ago.
I have a data frame that is z-score converted. I want to delete from the data frame (and convert to NA) only those values that are higher or equal to 4, without dropping any row or column. I would appreciate an answer.
Best
You can use the following code:
df <- data.frame(v1 = c(1,3,6,7,3),
v2 = c(2,1,4,6,7),
v3 = c(1,2,3,4,5))
df
#> v1 v2 v3
#> 1 1 2 1
#> 2 3 1 2
#> 3 6 4 3
#> 4 7 6 4
#> 5 3 7 5
is.na(df) <- df >= 4
df
#> v1 v2 v3
#> 1 1 2 1
#> 2 3 1 2
#> 3 NA NA 3
#> 4 NA NA NA
#> 5 3 NA NA
Created on 2022-07-10 by the reprex package (v2.0.1)
you can simply use df[df>=4] <- NA to achieve what you want.
df <- data.frame(replicate(10,sample(0:10,10,rep=TRUE)))
> df
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 2 3 4 5 6 4 3 1 10 6
2 5 7 0 4 3 10 10 3 6 10
3 5 5 0 3 1 3 5 7 2 7
4 7 0 4 1 10 0 5 2 5 0
5 8 8 7 8 4 6 6 10 10 0
6 1 4 1 3 3 8 8 0 4 8
7 6 3 3 6 7 4 10 9 7 2
8 2 1 4 0 7 8 10 1 6 3
9 0 9 6 2 9 6 2 9 0 3
10 8 2 1 0 1 4 0 6 2 8
df[df>=4] <- NA
> df
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 2 3 NA NA NA NA 3 1 NA NA
2 NA NA 0 NA 3 NA NA 3 NA NA
3 NA NA 0 3 1 3 NA NA 2 NA
4 NA 0 NA 1 NA 0 NA 2 NA 0
5 NA NA NA NA NA NA NA NA NA 0
6 1 NA 1 3 3 NA NA 0 NA NA
7 NA 3 3 NA NA NA NA NA NA 2
8 2 1 NA 0 NA NA NA 1 NA 3
9 0 NA NA 2 NA NA 2 NA 0 3
10 NA 2 1 0 1 NA 0 NA 2 NA
Here is one more. Using replace_with_na_all() from naniar package:
Use replace_with_na_all() when you want to replace ALL values that meet a condition across an entire dataset. The syntax here is a little different, and follows the rules for rlang’s expression of simple functions. This means that the function starts with ~, and when referencing a variable, you use .x.
https://cran.r-project.org/web/packages/naniar/vignettes/replace-with-na.html
library(naniar)
library(dplyr)
df %>%
replace_with_na_all(condition = ~.x > 4)
v1 v2 v3
<dbl> <dbl> <dbl>
1 1 2 1
2 3 1 2
3 NA 4 3
4 NA NA 4
5 3 NA NA
Though the solution by #Quinten is very concise, just add an approach in tidyverse
library(dplyr)
set.seed(123)
df <- data.frame(
x = sample(1:10, 7),
y = sample(1:10, 7)
)
df %>%
mutate(
across(.fns = ~ if_else(.x >= 4, NA_integer_, .x))
)
#> x y
#> 1 3 NA
#> 2 NA NA
#> 3 2 1
#> 4 NA 2
#> 5 NA 3
#> 6 NA NA
#> 7 1 NA
Created on 2022-07-10 by the reprex package (v2.0.1)
In base R, we can use replace():
df <- replace(df, df > 4, NA_real_)
Output
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 NA NA 3 NA 1 3 1 1 NA NA
2 1 NA 2 NA NA 3 NA NA 2 0
3 NA 1 NA 2 2 1 NA NA 4 1
4 NA NA 0 NA NA NA 0 2 4 NA
5 NA 1 NA 3 0 NA 4 NA 2 3
6 0 3 NA 0 NA NA 1 1 NA 2
7 3 NA NA NA 2 2 NA 2 NA 4
8 NA 1 0 2 NA NA 2 NA NA NA
9 NA 3 NA 2 4 NA NA 0 1 3
10 1 3 NA 3 NA NA 3 4 NA NA
Or use replace in dplyr:
library(dplyr)
df %>%
mutate(across(everything(), ~ replace(.x, .x > 4, NA_real_)))
Data
set.seed(321)
df <- data.frame(replicate(10, sample(0:10, 10, rep = TRUE)))
If the columns are numeric, an option is also to use ^ on a logical matrix (df >= 4) to return NA for TRUE values and 1 for FALSE, then multiply with original data so that those elements corresponding to NA returns NA and the ones with 1 returns the original element
NA^(df >= 4) * df
I have a dataset that looks like this:
set.seed(999)
col1 = sample.int(10, 10)
col2 = sample.int(10, 10)
col3 = sample.int(10, 10)
col4 = sample.int(10, 10)
col5 = sample.int(10, 10)
col_data = data.frame(col1, col2, col3, col4, col5)
col1 col2 col3 col4 col5
1 4 8 3 9 8
2 7 5 9 7 10
3 1 7 7 8 2
4 6 6 5 5 4
5 8 10 8 3 7
6 2 3 1 2 6
7 5 9 2 1 1
8 10 2 4 4 3
9 9 1 10 6 9
10 3 4 6 10 5
I would like to create new columns in this dataset that :
Find out the position (i.e. column number) for the first "9" in each row
Find out the position (i.e. column number) for the first "7" in each row
Find out the position (i.e. column number) for the first "1" in each row
Find out the position (i.e. column number) for the first "10" in each row
Find out the position (i.e. column number) for the first "4" in each row
I thought this might be easier to do if the data was a matrix, and then convert it back to a data frame:
col_d = as.matrix(col_data)
first_4 = apply(col_d == 9, 1, which.max)
first_7 = apply(col_d == 7, 1, which.max)
first_1 = apply(col_d == 1, 1, which.max)
first_10 = apply(col_d == 10, 1, which.max)
first_4 = apply(col_d == 4, 1, which.max)
final = cbind(col_data, first_4, first_7, first_1, first_10, first_4)
But this does not appear to be working:
col1 col2 col3 col4 col5 first_4 first_7 first_1 first_10 first_4
1 4 8 3 9 8 1 1 1 1 1
2 7 5 9 7 10 1 1 1 5 1
3 1 7 7 8 2 1 2 1 1 1
4 6 6 5 5 4 5 1 1 1 5
5 8 10 8 3 7 1 5 1 2 1
6 2 3 1 2 6 1 1 3 1 1
7 5 9 2 1 1 1 1 4 1 1
8 10 2 4 4 3 3 1 1 1 3
9 9 1 10 6 9 1 1 2 3 1
10 3 4 6 10 5 2 1 1 4 2
For example: In the first row, there is no 10 - but the value of "first_10" is 1
Is there a way to resolve this error?
Thank you!
How about
apply(col_data == 7, 1, function(x) {ifelse(sum(x)==0, NA, which.max(x))})
[1] NA 1 2 NA 5 NA NA NA NA NA
apply(col_data == 10, 1, function(x) {ifelse(sum(x)==0, NA, which.max(x))})
[1] NA 5 NA NA 2 NA NA 1 3 4
You may change NA whatever you want, that it means there is no that number(i.e 7 or 10)
get second one
apply(col_data == 7, 1, function(x) {ifelse(sum(x)==0, NA, which(x)[2])})
get last one
apply(col_data == 7, 1, function(x) {ifelse(sum(x)==0, NA, dplyr::last(which(x)))})
Use max.col:
nr <- c(9, 7, 1, 10, 4)
nr <- setNames(nr, paste0("first_", nr))
cbind(col_data, sapply(nr, function(x) {
. <- col_data == x
tt <- max.col(., "first")
is.na(tt) <- tt == 1 & !.[,1]
tt
}))
# col1 col2 col3 col4 col5 first_9 first_7 first_1 first_10 first_4
#1 4 8 3 9 8 4 NA NA NA 1
#2 7 5 9 7 10 3 1 NA 5 NA
#3 1 7 7 8 2 NA 2 1 NA NA
#4 6 6 5 5 4 NA NA NA NA 5
#5 8 10 8 3 7 NA 5 NA 2 NA
#6 2 3 1 2 6 NA NA 3 NA NA
#7 5 9 2 1 1 2 NA 4 NA NA
#8 10 2 4 4 3 NA NA NA 1 3
#9 9 1 10 6 9 1 NA 2 3 NA
#10 3 4 6 10 5 NA NA NA 4 2
For the last:
nr <- c(9, 7, 1, 10, 4)
nr <- setNames(nr, paste0("last_", nr))
cbind(col_data, sapply(nr, function(x) {
. <- col_data == x
tt <- max.col(., "last")
is.na(tt) <- rowSums(.) == 0
tt
}))
# col1 col2 col3 col4 col5 last_9 last_7 last_1 last_10 last_4
#1 4 8 3 9 8 4 NA NA NA 1
#2 7 5 9 7 10 3 4 NA 5 NA
#3 1 7 7 8 2 NA 3 1 NA NA
#4 6 6 5 5 4 NA NA NA NA 5
#5 8 10 8 3 7 NA 5 NA 2 NA
#6 2 3 1 2 6 NA NA 3 NA NA
#7 5 9 2 1 1 2 NA 5 NA NA
#8 10 2 4 4 3 NA NA NA 1 4
#9 9 1 10 6 9 5 NA 2 3 NA
#10 3 4 6 10 5 NA NA NA 4 2
And for the second match:
nr <- c(9, 7, 1, 10, 4)
nr <- setNames(nr, paste0("2nd_", nr))
cbind(col_data, sapply(nr, function(x) {
. <- which(col_data == x, TRUE)
. <- tapply(.[,2], .[,1], `[`, 2)
replace(rep(NA_integer_, nrow(col_data)), as.integer(names(.)), .)
}))
# col1 col2 col3 col4 col5 2nd_9 2nd_7 2nd_1 2nd_10 2nd_4
#1 4 8 3 9 8 NA NA NA NA NA
#2 7 5 9 7 10 NA 4 NA NA NA
#3 1 7 7 8 2 NA 3 NA NA NA
#4 6 6 5 5 4 NA NA NA NA NA
#5 8 10 8 3 7 NA NA NA NA NA
#6 2 3 1 2 6 NA NA NA NA NA
#7 5 9 2 1 1 NA NA 5 NA NA
#8 10 2 4 4 3 NA NA NA NA 4
#9 9 1 10 6 9 5 NA NA NA NA
#10 3 4 6 10 5 NA NA NA NA NA
Or using apply on one column.
#First
apply(col_data == 9, 1, function(x) if(any(x)) which.max(x) else NA)
# [1] 4 3 NA NA NA NA 2 NA 1 NA
#Last
apply(col_data == 9, 1, function(x) if(any(x)) tail(which(x), 1) else NA)
# [1] 4 3 NA NA NA NA 2 NA 5 NA
#Second
apply(col_data == 9, 1, function(x) if(any(x)) which(x)[2] else NA)
# [1] NA NA NA NA NA NA NA NA 5 NA
I have a dataframe like this:
household person R01 R02 R03 R04 R05
1 1 1 NA 1 7 7 NA
2 1 2 1 NA 7 7 NA
3 1 3 3 3 NA 11 NA
4 1 4 3 3 11 NA NA
5 2 1 NA 7 16 NA NA
6 2 2 3 NA 7 NA NA
7 2 3 15 3 NA NA NA
and I'm trying add new columns which are the grouped transposed versions of columns R01 to R05, like this:
household person R01 R02 R03 R04 R05 R01x R02x R03x R04x R05x
1 1 1 NA 1 7 7 NA NA 1 3 3 NA
2 1 2 1 NA 7 7 NA 1 NA 3 3 NA
3 1 3 3 3 NA 11 NA 7 7 NA 11 NA
4 1 4 3 3 11 NA NA 7 7 11 NA NA
5 2 1 NA 7 16 NA NA NA 3 15 NA NA
6 2 2 3 NA 7 NA NA 7 NA 3 NA NA
7 2 3 15 3 NA NA NA 16 7 NA NA NA
I have tried various attempts using t() and reshaping using gather() and spread() but I don't think they are designed to do this as I'm moving the data around rather than just reshaping it.
Example Code
df <- data.frame(household = c(rep(1,4),rep(2,3)),
person = c(1:4,1:3),
R01 = c(NA,1,3,3,NA,3,15),
R02 = c(1,NA,3,3,7,NA,3),
R03 = c(7,7,NA,11,16,7,NA),
R04 = c(7,7,11,rep(NA,4)),
R05 = rep(NA,7))
Referring to my previous answer, you can transpose the matrx within group_modify():
library(dplyr)
df %>%
group_by(household) %>%
group_modify(~ {
mat <- t(.x[-1][1:nrow(.x)])
colnames(mat) <- paste0(rownames(mat), "x")
cbind(.x, mat)
}) %>%
ungroup()
# # A tibble: 7 × 11
# household person R01 R02 R03 R04 R05 R01x R02x R03x R04x
# <dbl> <int> <dbl> <dbl> <dbl> <dbl> <lgl> <dbl> <dbl> <dbl> <dbl>
# 1 1 1 NA 1 7 7 NA NA 1 3 3
# 2 1 2 1 NA 7 7 NA 1 NA 3 3
# 3 1 3 3 3 NA 11 NA 7 7 NA 11
# 4 1 4 3 3 11 NA NA 7 7 11 NA
# 5 2 1 NA 7 16 NA NA NA 3 15 NA
# 6 2 2 3 NA 7 NA NA 7 NA 3 NA
# 7 2 3 15 3 NA NA NA 16 7 NA NA
Partly using a previous answer, here's a way to do it.
Split the dataframe according to their group
Get their number of columns with at least one non-NA (important to do the transposition)
Reduce their size using the length size created in step 2, and do the transposition.
Swap (again) the colnames and rownames which were swapped (first) in the transposition.
Bind the columns with the original dataframe.
l <- split(df[startsWith(colnames(df), "R")], df$household)
len <- lapply(l, \(l) ncol(l) - (sum(sapply(l, \(x) any(!is.na(x))))))
l <- mapply(\(x, y) t(x[1:(length(x) - y)]), l, len, SIMPLIFY = F)
l <- lapply(l, function(x){
r <- paste0(rownames(x), "x")
c <- colnames(x)
rownames(x) <- c
colnames(x) <- r
data.frame(x)
})
cbind(df, bind_rows(l))
output
household person R01 R02 R03 R04 R05 R01x R02x R03x R04x
1 1 1 NA 1 7 7 NA NA 1 3 3
2 1 2 1 NA 7 7 NA 1 NA 3 3
3 1 3 3 3 NA 11 NA 7 7 NA 11
4 1 4 3 3 11 NA NA 7 7 11 NA
5 2 1 NA 7 16 NA NA NA 3 15 NA
6 2 2 3 NA 7 NA NA 7 NA 3 NA
7 2 3 15 3 NA NA NA 16 7 NA NA
df %>%
left_join(pivot_longer(.,starts_with('R'), names_to = 'name',
names_pattern = "(\\d+)", values_drop_na = TRUE,
names_transform = list(name = as.integer)) %>%
pivot_wider(c(household,name), names_from = person,
names_glue = "R0{person}x"),
by = c('household', person = 'name'))
household person R01 R02 R03 R04 R05 R01x R02x R03x R04x
1 1 1 NA 1 7 7 NA NA 1 3 3
2 1 2 1 NA 7 7 NA 1 NA 3 3
3 1 3 3 3 NA 11 NA 7 7 NA 11
4 1 4 3 3 11 NA NA 7 7 11 NA
5 2 1 NA 7 16 NA NA NA 3 15 NA
6 2 2 3 NA 7 NA NA 7 NA 3 NA
7 2 3 15 3 NA NA NA 16 7 NA NA
Another solution:
df %>%
left_join(
reshape2::recast(.,household+variable~person,id.var = c('household', 'person'))%>%
group_by(household) %>%
mutate(person = seq_along(variable), variable = NULL))
household person R01 R02 R03 R04 R05 1 2 3 4
1 1 1 NA 1 7 7 NA NA 1 3 3
2 1 2 1 NA 7 7 NA 1 NA 3 3
3 1 3 3 3 NA 11 NA 7 7 NA 11
4 1 4 3 3 11 NA NA 7 7 11 NA
5 2 1 NA 7 16 NA NA NA 3 15 NA
6 2 2 3 NA 7 NA NA 7 NA 3 NA
7 2 3 15 3 NA NA NA 16 7 NA NA
Here's a way to do it.
library(dplyr)
transposed_df <- df %>%
group_split(household) %>%
lapply(\(x){
select(x, -1:-2) %>%
t() %>%
head(nrow(x)) %>%
as_tibble() %>%
setNames(paste0(names(x)[-1:-2], 'x'))
}) %>%
bind_rows()
df %>%
bind_cols(transposed_df)
#> household person R01 R02 R03 R04 R05 R01x R02x R03x R04x
#> 1 1 1 NA 1 7 7 NA NA 1 3 3
#> 2 1 2 1 NA 7 7 NA 1 NA 3 3
#> 3 1 3 3 3 NA 11 NA 7 7 NA 11
#> 4 1 4 3 3 11 NA NA 7 7 11 NA
#> 5 2 1 NA 7 16 NA NA NA 3 15 NA
#> 6 2 2 3 NA 7 NA NA 7 NA 3 NA
#> 7 2 3 15 3 NA NA NA 16 7 NA NA
Issue
I have a dataframe of familial relationships coded with integers, where R01 is the relationship of person N to person 1, R02 their relationship to person 2 etc.
However, only the lower.tri of each family matrix is coded, so I am trying to write a function to match the correct relationship in the upper.tri.
Relationships
The relationships are coded in integers as follows:
1 = Spouse, 2 = Cohabiting partner, 3 = Son/daughter, 4 = Step son/daughter, 5 = Foster child, 6 = Son-in-law/daughter-in-law, 7 = Parent/guardian, 8 = Step-parent, 9 = Foster parent, 10 = Parent-in-law, 11 = Brother/sister, 12 = Step-brother/sister, 13 = Foster brother/sister, 14 = Brother/sister-in-law, 15 = Grand-child, 16 = Grand-parent, 17 = Other relative, 18 = Other non-relative.
thus the relationships are:
rel = c("1" = 1, "2" = 2, "3" = 7, "4" = 8, "5" = 9, "6" = 10, "7" = 3, "8" = 4, "9" = 5, "10" = 6, "11" = 11, "12" = 12, "13" = 13, "14" = 14, "15" = 16, "16" = 15, "17" = 17, "18" = 18)
Example Data
household person R01 R02 R03 R04 R05 R06
1 1 1 NA NA NA NA NA NA
2 1 2 1 NA NA NA NA NA
3 1 3 3 3 NA NA NA NA
4 1 4 3 3 11 NA NA NA
5 2 1 NA NA NA NA NA NA
6 2 2 3 NA NA NA NA NA
7 2 3 15 3 NA NA NA NA
8 3 1 NA NA NA NA NA NA
9 3 2 18 NA NA NA NA NA
10 4 1 NA NA NA NA NA NA
11 5 1 NA NA NA NA NA NA
12 5 2 5 NA NA NA NA NA
Required Output
household person R01 R02 R03 R04 R05 R06
1 1 1 NA 1 7 7 NA NA
2 1 2 1 NA 7 7 NA NA
3 1 3 3 3 NA 11 NA NA
4 1 4 3 3 11 NA NA NA
5 2 1 NA 1 16 NA NA NA
6 2 2 3 NA 1 NA NA NA
7 2 3 15 3 NA NA NA NA
8 3 1 NA 18 NA NA NA NA
9 3 2 18 NA NA NA NA NA
10 4 1 NA NA NA NA NA NA
11 5 1 NA 9 NA NA NA NA
12 5 2 5 NA NA NA NA NA
Example Code
df <- data.frame(household = c(1,1,1,1,2,2,2,3,3,4,5,5),
person = c(1,2,3,4,1,2,3,1,2,1,1,2),
R01 = c(NA, 1, 3, 3, NA, 3, 15, NA, 18, NA, NA, 5),
R02 = c(NA, NA, 3, 3, NA, NA, 3, rep(NA, 5)),
R03 = c(rep(NA,3), 11, rep(NA, 8)),
R04 = rep(NA, 12),
R05 = rep(NA, 12),
R06 = rep(NA, 12))
I know it's possible to write a function to do the matrix match and then apply it to each household with dplyr, however I'm not great at functions yet so I'm running into issues in a few areas.
You can make the relationship matrix symmetric in each household, and at the same time recode the elements according to rel.
library(dplyr)
df %>%
group_by(household) %>%
group_modify(~ {
mat <- as.matrix(.x[-1][1:nrow(.x)])
mat[upper.tri(mat)] <- recode(t(mat)[upper.tri(mat)], !!!rel)
cbind(.x[1], mat)
}) %>%
ungroup()
# A tibble: 12 × 6
household person R01 R02 R03 R04
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 NA 1 7 7
2 1 2 1 NA 7 7
3 1 3 3 3 NA 11
4 1 4 3 3 11 NA
5 2 1 NA 7 16 NA
6 2 2 3 NA 7 NA
7 2 3 15 3 NA NA
8 3 1 NA 18 NA NA
9 3 2 18 NA NA NA
10 4 1 NA NA NA NA
11 5 1 NA 9 NA NA
12 5 2 5 NA NA NA
Here's a way to do it mostly using base R.
First, create f, a function that replace the upper triangle of a matrix with the matching value from the rel vector and the lower triangle of the same matrix.
Then, split your data according to the household, compute the lengths of each group so that the resulting matrix has the right number of columns, and then apply the function to each groups. Finally, bind_rows and cbind with the original data set.
f <- function(m) {
m[upper.tri(m)] <- match(t(m)[upper.tri(m)], rel)
m
}
l <- split(df[3:6], df$household)
len <- lapply(l, \(l) ncol(l) - (sum(sapply(l, \(x) any(!is.na(x)))) + 1))
l <- mapply(\(x, y) x[1:(length(x) - y)], l, len, SIMPLIFY = F)
cbind(df[1:2],
dplyr::bind_rows(lapply(l, f)))
output
household person R01 R02 R03 R04
1 1 1 NA 1 7 7
2 1 2 1 NA 7 7
3 1 3 3 3 NA 11
4 1 4 3 3 11 NA
5 2 1 NA 7 16 NA
6 2 2 3 NA 7 NA
7 2 3 15 3 NA NA
8 3 1 NA 18 NA NA
9 3 2 18 NA NA NA
10 4 1 NA NA NA NA
11 5 1 NA 9 NA NA
12 5 2 5 NA NA NA
This question already has answers here:
How to reshape data from long to wide format
(14 answers)
Closed 3 years ago.
I have a dataframe with 3 column: participant ID, questionID and a column containing wether or not they gave the correct (1) response or not (0).
It looks like this:
> head(df)
# A tibble: 6 x 3
ID questionID correct
<dbl> <int> <dbl>
1 1 1 1
2 2 2 0
3 3 3 1
4 4 4 0
5 5 5 0
6 6 6 1
And can be recreated using:
set.seed(0)
df <- tibble(ID = seq(1, 100, 1),
questionID = rep(seq(1, 10,), 10),
correct = base::sample(c(0, 1), size = 100, replace = TRUE))
Now I would like each question to have their own column, with the ultimate goal of fitting a 2PL model to it. The data should for that purpose look like 1 row per participant, and 11 columns (ID and 10 question Columns).
How do I achieve this?
You can use pivot_wider from the tidyr package:
df %>%
pivot_wider(names_from = questionID,
values_from = correct,
names_prefix = "questionID_")
# A tibble: 100 x 11
ID questionID_1 questionID_2 questionID_3 questionID_4
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 NA NA NA
2 2 NA 0 NA NA
3 3 NA NA 0 NA
4 4 NA NA NA 1
5 5 NA NA NA NA
6 6 NA NA NA NA
7 7 NA NA NA NA
8 8 NA NA NA NA
9 9 NA NA NA NA
10 10 NA NA NA NA
# ... with 90 more rows, and 6 more variables: questionID_5 <dbl>,
# questionID_6 <dbl>, questionID_7 <dbl>, questionID_8 <dbl>,
# questionID_9 <dbl>, questionID_10 <dbl>
Using data.table you can use dcast
df <- data.frame(ID=c(1,2,3,4,5,6), questionID= c(1,22,13,4,35,8),correct=c(1,0,1,0,0,1))
df
ID questionID correct
1 1 1 1
2 2 22 0
3 3 13 1
4 4 4 0
5 5 35 0
6 6 8 1
setDT(df)
dcast(df,ID~questionID,value.var="correct")
ID 1 4 8 13 22 35
1: 1 1 NA NA NA NA NA
2: 2 NA NA NA NA 0 NA
3: 3 NA NA NA 1 NA NA
4: 4 NA 0 NA NA NA NA
5: 5 NA NA NA NA NA 0
6: 6 NA NA 1 NA NA NA
# replace NA to what you want
df[is.na(df)]<- "-"