[Solved] Removing contents from a dataset in R
dataset <- dataset[!(dataset %in% c(‘A’,’B’))] solved Removing contents from a dataset in R
dataset <- dataset[!(dataset %in% c(‘A’,’B’))] solved Removing contents from a dataset in R
Are you asking something like the following? Let’s say your initial dataframe is df, which is the following: df A B C 1 2016-02-16 2016-03-21 2016-01-01 2 2016-07-07 2016-06-17 2016-01-31 3 2016-05-19 2016-09-10 2016-03-01 4 2016-01-14 2016-08-21 2016-04-01 5 2016-09-02 2016-06-15 2016-05-01 6 2016-05-09 2016-07-17 2016-05-31 7 2016-06-13 2016-06-23 2016-07-01 8 2016-09-17 2016-03-11 2016-07-31 9 … Read more
You can build a backtracking based solution to formulate all the possible sub-sequence for the given array. Sharing an ideone link for the same: https://ideone.com/V0gCDX all = [] def gen(A, idx = 0, cur = []): if idx >= len(A): if len(cur): all.append(cur) return gen(A, idx + 1, list(cur)) incl = list(cur) incl.append(A[idx]) gen(A, idx … Read more
try this one: import java.io.BufferedReader; import java.io.InputStreamReader; public class Rozwiazanie { public static void main(String[] args) { try { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] splittedLinia = br.readLine().split((char) 32 + “”);//moglaby byc ” ” ale tak na wszelki wypadek nie ma chuja zeby sie popierdolilo teraz nawet na linuxie int aktualnyWyraz = Integer.parseInt(splittedLinia[0]);//uwaga jakby … Read more
In R, a vector (a in your question) can be subset with a logical index vector (b in your question). If the corresponding members are TRUE in the logical vector, the elements (of a) are retained. a = c (3, 4, 7, 8) b = c(TRUE, TRUE, FALSE, FALSE) a[b] #[1] 3 4 a[b] will … Read more
Update: for multiple dictionaries Iterate over your dictionaries and for every one of them, check if the value of ‘mesic’ is in [6,7,8] and if so, get the corresponding dictionary values: d1 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 8, ‘teplotaC’: 26.88} d2 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 1, ‘teplotaC’: 26.88} … Read more
The duplicated function traverses its argument(s) sequentially and returns TRUE if there has been a prior value identical to the current value. It is a generic function, so it has a default definition (for vectors) but also a definition for other classes, such as objects of the data.frame class. The subset function treats expressions passed … Read more
maybe not the best way to do it, but will get the job done. vars_df = unique(df$x) for (i in 1:length(vars_df)) { assign(paste0(vars_df[i]), df %>% filter(x == vars_df[i]), envir = .GlobalEnv) } solved Automatically subset data frame by factor
I would use indexing # save all the “transactions” where iteamtype equals “a” to an object called F f <- df[ df$itemType %in% “a” , “transaction” ] #optional (look as f) print(f) print( unique(f)) # Subset to all the rows which equals one of the “transactions” stored in the object f df[ df$transaction %in% unique( … Read more
Tried to create your data using the following: df <- data.frame(col1 = c(1,1,2,2,1,1,2,2), col2 = c(34,34,342,23,34,34,342,23), col3 = c(3,4,3,4,3,4,3,4)) And, if you wish to subset based on only one column, you can use @kyle-marsh solution > df[df$col1 == 1, ] col1 col2 col3 1 1 34 3 2 1 34 4 5 1 34 3 … Read more
Well, there is a very easy and naive recursive solution to retrieving all subsets. You take away one element from your set, then find all subsets for this new, smaller set. Then you copy the result and add the element you previously removed to the copy. Add the results together and you’re done. For example: … Read more
We can try subset subset(dati, format(as.Date(Date),”%Y”)==2005) If we are trying to subset the data for each year, try split split(dati, format(as.Date(dati$Date), “%Y”)) solved R get subset from data frame filtering by year Date value
This is a workaround, a response to your #2 Looking at your code, there is a much easier way of subsetting data. Try this. Check if this solves your issue. library(dplyr) active<- clinic %>% filter(Days.since.injury.physio>20, Days.since.injury.physio<35, Days.since.injury.F.U.1>27, Days.since.injury.F.U.1<63 ) dplyr does wonders when it comes to subsetting and manipulation of data. The %>% symbol chains … Read more