I am writing my function and want to use dplyr's filter() function to select rows of my data frame that satisfy a condition. This is my code:
library(tidyverse)
df <-data.frame(x = sample(1:100, 50), y = rnorm(50), z = sample(1:100,50), w = sample(1:100, 50),
p = sample(1:100,50))
new <- function(ang,brad,drau){
df%>%filter(!!drau %in% 1:50)%>%select(ang,brad) -> A
return(A)
}
brand <- c("z","w","p")
lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()
Anytime I run this function, it looks like filter
doesn't select any rows that satisfy the condition.
How can I make this work?
Update
For some reason, this works when I don't use `%in%, as in;
new <- function(ang,brad,drau){
df%>%filter(!!drau > 50)%>%select(ang,brad) -> A
return(A)
}
lapply(1:3, function(i) new(ang = "x", brad = "y", drau = brand[i]))%>%bind_rows()
However, the results are the same for every loop. Why is this so? and also why can't I use %in%
.
The filter() function is used to subset a data frame, retaining all rows that satisfy your conditions. To be retained, the row must produce a value of TRUE for all conditions. Note that when a condition evaluates to NA the row will be dropped, unlike base subsetting with [ .
The filter() method in R is used to subset a data frame based on a provided condition. If a row satisfies the condition, it must produce TRUE . Otherwise, non-satisfying rows will return NA values. Hence, the row will be dropped.
The filter() method in R programming language can be applied to both grouped and ungrouped data. The expressions include comparison operators (==, >, >= ) , logical operators (&, |, !, xor()) , range operators (between(), near()) as well as NA value check against the column values.
%>% is called the forward pipe operator in R. It provides a mechanism for chaining commands with a new forward-pipe operator, %>%. This operator will forward a value, or the result of an expression, into the next function call/expression. It is defined by the package magrittr (CRAN) and is heavily used by dplyr (CRAN).
I agree with @hrbrmstr's standard evaluation solution. As suggested by @hadley today here's NSE solution:
library(tidyverse)
df <-data.frame(x = sample(1:100, 50),
y = rnorm(50),
z = sample(1:100,50),
w = sample(1:100, 50),
p = sample(1:100,50))
new <- function(ang, brad, drau){
ang <- enquo(ang)
brad <- enquo(brad)
drau <- enquo(drau)
df %>% filter(UQ(drau) %in% 1:50) %>%
select(UQ(ang),UQ(brad))
}
brand <- c("z","w","p")
brand <- rlang::syms(brand)
map_df(brand, ~new(ang = x, brad = y, drau = UQ(.x)))
This appears to do what you want (but it needs confirmation by you):
library(tidyverse)
library(rlang)
set.seed(1492)
xdf <- data_frame(
x = sample(1:100, 50),
y = rnorm(50),
z = sample(1:100,50),
w = sample(1:100, 50),
p = sample(1:100,50)
)
new_df <- function(ang, brad, drau) {
drau <- sym(drau)
filter(xdf, UQE(drau) %in% 1:50) %>%
select(ang, brad)
}
brand <- c("z", "w", "p")
map_df(brand, ~new_df(ang = "x", brad = "y", drau = .x))
Despite there being a plethora of "official" "tidyverse" examples using df
, it's a function in the stats
pkg and I try to avoid using it anymore.
Since you're using the tidyverse, might as well take advantage of map_df()
from purrr
.
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