Remove row if any column contains a specific string
You can read the data using na.strings = 'no_data'
to set them as NA
and then simply omit NAs (or take complete.cases
), i.e. (Using @akrun's data set)
d1 <- read.table(text = 'time speed wheels
1 1:00 30 no_data
2 2:00 no_data 18
3 no_data no_data no_data
4 3:00 50 18', na.strings = 'no_data', h=TRUE)
d1[complete.cases(d1),]
# time speed wheels
#4 3:00 50 18
#OR
na.omit(d1)
# time speed wheels
#4 3:00 50 18
We can use rowSums
to create a logical vector
and subset based on it
df1[rowSums(df1 == "no_data")==0, , drop = FALSE]
# time speed wheels
#4 3:00 50 18
data
df1 <- structure(list(time = c("1:00", "2:00", "no_data", "3:00"), speed = c("30",
"no_data", "no_data", "50"), wheels = c("no_data", "18", "no_data",
"18")), .Names = c("time", "speed", "wheels"), class = "data.frame",
row.names = c(NA, -4L))
akrun answer is quick, correct and simply as much is it can :) however if you like to make your life more complex you can also do:
dat
time speed wheels
1 1:00 30 no_data
2 2:00 no_data 18
3 no_data no_data no_data
4 3:00 50 18
dat$new <- apply(dat[,1:3], 1, function(x) any(x %in% c("no_data")))
dat <- dat[!(dat$new==TRUE),]
dat$new <- NULL
dat
time speed wheels
4 3:00 50 18
Two dplyr
options: (using Akrun's data from this answer)
library(dplyr)
## using the newer across()
df1 %>% filter(across(everything(), ~ !grepl("no_data", .)))
#> time speed wheels
#> 1 3:00 50 18
## with the superseded filter_all
df1 %>% filter_all(all_vars(!grepl("no_data", .)))
#> time speed wheels
#> 1 3:00 50 18
Caveat:
This only works if you want to remove all rows with that string. If you want to get all rows with this string, all_vars(grepl('no_data',.)
(without !
) would not be sufficient: This would only get the rows where all columns contain the string.
In this case, use filter_all(any_vars())
instead.