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I have a simple data frame that looks like this:
x <- c("aa", "aa", "aa", "bb", "cc", "cc", "cc") y <- c(101, 102, 113, 201, 202, 344, 407) df = data.frame(x, y) x y 1 aa 101 2 aa 102 3 aa 113 4 bb 201 5 cc 202 6 cc 344 7 cc 407 I would like to use a dplyr::filter() and a RegEx to filter out all the y observations that start with the number 1
I'm imagining that the code will look something like this:
df %>% filter(y != grep("^1")) But I am getting an Error in grep("^1") : argument "x" is missing, with no default
732
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 Regex Filter transform filters messages in the data stream according to a regular expression (regex) pattern, which you can define. You also define the Regex Filter to either accept or deny incoming messages based on the regular expression.
In this, first, pass your dataframe object to the filter function, then in the condition parameter write the column name in which you want to filter multiple values then put the %in% operator, and then pass a vector containing all the string values which you want in the result.
You need to double check the documentations for grepl and filter.
For grep/grepl you have to also supply the vector that you want to check in (y in this case) and filter takes a logical vector (i.e. you need to use grepl). If you want to supply an index vector (from grep) you can use slice instead.
df %>% filter(!grepl("^1", y)) Or with an index derived from grep:
df %>% slice(grep("^1", y, invert = TRUE)) But you can also just use substr because you are only interested in the first character:
df %>% filter(substr(y, 1, 1) != 1) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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