Web2 If you have several 'x' groups, one option would be to use mapply. We split the 'y' using 'x' as grouping variable, create the vector of values to compare against ( c (15,5)) and use mapply to get the logical index for subsetting the 'df'. df [unlist (mapply ('>', split (df$y, df$x), c (15,5))),] # x y #1 1 30 #4 2 10 #5 2 18 Share WebJun 27, 2016 · Need to filter out rows that fall above 90 percentile in 'total_transfered_amount' column for every id seperately using dplyr package preferabely , for example I need to filter out following rows: 2 40000 2 3 30000 3 r dataframe dplyr data-analysis percentile Share Improve this question Follow edited Jun 27, 2016 at 9:48 …
How to filter R DataFrame by values in a column?
WebFeb 3, 2024 · Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question.Provide details and share your research! But avoid …. Asking for help, clarification, or responding to other answers. WebMay 31, 2024 · Filter To Show Rows Starting with a Specific Letter. Similarly, you can select only dataframe rows that start with a specific letter. For example, if you only wanted to select rows where the region starts with 'E', you could write: e = df[df['Region'].str[0] == 'E'] print(e.shape) # Returns: (411, 5) Select Dataframe Rows based on List of Values the other son full movie
R filter Data Frame by Multiple Conditions
WebNov 29, 2014 · library (dplyr) df <- data.frame (this = c (1, 2, 2), that = c (1, 1, 2)) column <- "this" df %>% filter (!!as.symbol (column) == 1) # this that # 1 1 1 Using alternative solutions Other ways to refer to the value "this" of the variable column inside dplyr::filter () that don't rely on rlang's injection paradigm include: WebAug 2, 2016 · 3 Answers Sorted by: 18 Using dplyr, you can try the following, assuming your table is df: library (dplyr) library (stringr) animalList <- c ("cat", "dog") filter (df, str_detect (animal, paste (animalList, collapse=" "))) I personally find the use of dplyr and stringr to be easier to read months later when reviewing my code. Share WebAug 10, 2024 · 3 Answers Sorted by: 4 Your combination of %in% and all sounds promising, in base R you could use those to your advantage as follows: to_keep = sapply (lapply (split (x,x$ID),function (x) {unique (x$Hour)}), function (x) {all (testVector %in% x)}) x = x [x$ID %in% names (to_keep) [to_keep],] the other sociology