id first middle last Age
1 Carol Jenny Smith 15
2 Sarah Carol Roberts 20
3 Josh David Richardson 22
I am t
Using tidyverse
library(tidyverse)
f1 <- function(data, wordToCompare, colsToCompare) {
wordToCompare <- enquo(wordToCompare)
data %>%
select(colsToCompare) %>%
mutate(!! wordToCompare := map(., ~
.x == as_label(wordToCompare)) %>%
reduce(`|`) %>%
as.integer)
}
f1(df1, Carol, c("first", 'middle', 'last'))
# first middle last Carol
#1 Carol Jenny Smith 1
#2 Sarah Carol Roberts 1
#3 Josh David Richardson 0
f1(df1, Sarah, c("first", 'middle', 'last'))
# first middle last Sarah
#1 Carol Jenny Smith 0
#2 Sarah Carol Roberts 1
#3 Josh David Richardson 0
Or this can also be done with pmap
df1 %>%
mutate(Carol = pmap_int(.[c('first', 'middle', 'last')],
~ +('Carol' %in% c(...))))
# id first middle last Age Carol
#1 1 Carol Jenny Smith 15 1
#2 2 Sarah Carol Roberts 20 1
#3 3 Josh David Richardson 22 0
which can be wrapped into a function
f2 <- function(data, wordToCompare, colsToCompare) {
wordToCompare <- enquo(wordToCompare)
data %>%
mutate(!! wordToCompare := pmap_int(.[colsToCompare],
~ +(as_label(wordToCompare) %in% c(...))))
}
f2(df1, Carol, c("first", 'middle', 'last'))
# id first middle last Age Carol
#1 1 Carol Jenny Smith 15 1
#2 2 Sarah Carol Roberts 20 1
#3 3 Josh David Richardson 22 0
NOTE: Both the tidyverse methods doesn't require any reshaping
With base R
, we can loop through the 'first', 'middle', 'last' column and use ==
for comparison to get a list
of logical vector
s, which we Reduce
to a single logical vector
with |
and coerce it to binary with +
df1$Carol <- +(Reduce(`|`, lapply(df1[2:4], `==`, 'Carol')))
df1
# id first middle last Age Carol
#1 1 Carol Jenny Smith 15 1
#2 2 Sarah Carol Roberts 20 1
#3 3 Josh David Richardson 22 0
NOTE: There are dupes for this post. For e.g. here
df1 <- structure(list(id = 1:3, first = c("Carol", "Sarah", "Josh"),
middle = c("Jenny", "Carol", "David"), last = c("Smith",
"Roberts", "Richardson"), Age = c(15L, 20L, 22L)),
class = "data.frame", row.names = c(NA,
-3L))
A solution using apply
family
df$Carol = lapply(1:nrow(df), function(x) any(df[x,]=="Carol))
We can use rowSums
df$Carol <- as.integer(rowSums(df[2:4] == "Carol") > 0)
df
# id first middle last Age Carol
#1 1 Carol Jenny Smith 15 1
#2 2 Sarah Carol Roberts 20 1
#3 3 Josh David Richardson 22 0
If we need it as a function
fun <- function(df, value) {
as.integer(rowSums(df[2:4] == value) > 0)
}
fun(df, "Carol")
#[1] 1 1 0
fun(df, "Sarah")
#[1] 0 1 0
but this assumes the columns you want to search are at position 2:4
.
To give more flexibility with column position
fun <- function(df, cols, value) {
as.integer(rowSums(df[cols] == value) > 0)
}
fun(df, c("first", "last","middle"), "Carol")
#[1] 1 1 0
fun(df, c("first", "last","middle"), "Sarah")
#[1] 0 1 0
Here's a tidyverse
option. We first reshape the data to long format, group by id
, and find levels of id
that have the desired name in at least one row. Then we reshape back to wide format.
library(tidyverse)
df %>%
gather(key, value, first:last) %>%
group_by(id) %>%
mutate(Carol = as.numeric(any(value=="Carol"))) %>%
spread(key, value)
id Age Carol first last middle 1 1 15 1 Carol Smith Jenny 2 2 20 1 Sarah Roberts Carol 3 3 22 0 Josh Richardson David
Or, as a function:
find.target = function(data, target) {
data %>%
gather(key, value, first:last) %>%
group_by(id) %>%
mutate(!!target := as.numeric(any(value==target))) %>%
spread(key, value) %>%
# Move new target column to end
select(-target, target)
}
find.target(df, "Carol")
find.target(df, "Sarah")
You could also do several at once. For example:
map(c("Sarah", "Carol", "David"), ~ find.target(df, .x)) %>%
reduce(left_join)
id Age first last middle Sarah Carol David 1 1 15 Carol Smith Jenny 0 1 0 2 2 20 Sarah Roberts Carol 1 1 0 3 3 22 Josh Richardson David 0 0 1
Another option using mutate
and if_else()
as you suggested:
library(tidyverse)
data = read_table(" id first middle last Age
1 Carol Jenny Smith 15
2 Sarah Carol Roberts 20
3 Josh David Richardson 22")
data %>%
mutate(carol = if_else(first == "Carol" | middle == "Carol" | last == "Carol",
"yes",
"no"))
Result:
# A tibble: 3 x 6
id first middle last Age carol
<dbl> <chr> <chr> <chr> <dbl> <chr>
1 1 Carol Jenny Smith 15 yes
2 2 Sarah Carol Roberts 20 yes
3 3 Josh David Richardson 22 no