Both subset and with are designed for interactive use and warnings against their use within other functions will be found in their help pages. This stems from their strategy of evaluation arguments as expressions within an environment constructed from the names of their data arguments. These column/element names would otherwise not be “objects” in the … Read more
Recursion is your friend for this task. For each element – “guess” if it is in the current subset, and recursively invoke with the guess and a smaller superset you can select from. Doing so for both the “yes” and “no” guesses – will result in all possible subsets. Restraining yourself to a certain length … Read more
It’s pretty straightforward using [ to extract: grep will give you the position in which it matched your search pattern (unless you use value = TRUE). grep(“^G45”, My.Data$x) # [1] 2 Since you’re searching within the values of a single column, that actually corresponds to the row index. So, use that with [ (where you … Read more
If we decide to use subset function, then we need to watch out: For ordinary vectors, the result is simply ‘x[subset & !is.na(subset)]’. So only non-NA values will be retained. If you want to keep NA cases, use logical or condition to tell R not to drop NA cases: subset(df1, Height < 40 | is.na(Height)) … Read more
You can do it more concisely with n(): library(dplyr) dat %>% group_by(cat) %>% filter(n() == 5)
Here goes an implementation: subset([], []). subset([E|Tail], [E|NTail]):- subset(Tail, NTail). subset([_|Tail], NTail):- subset(Tail, NTail). It will generate all the subsets, though not in the order shown on your example. As per commenter request here goes an explanation: The first clause is the base case. It states that the empty list is a subset of the … Read more
You can try this: library(dplyr) ranges %>% arrange(start) %>% group_by(g = cumsum(cummax(lag(stop, default = first(stop))) < start)) %>% summarise(start = first(start), stop = max(stop)) # A tibble: 2 × 3 # g start stop # <int> <dbl> <dbl> #1 0 65.72000 87.75625 #2 1 89.61625 104.94062
You’ve gotten a handful of nice examples of how to do what you want. However, it’s also useful to understand the what’s happening and why things work the way they do. There are a few simple rules that will help you in the future. There’s a big difference between “fancy” indexing (i.e. using a list/sequence) … Read more
The easiest way I can think of is to key by id1 as well. eg setkey(DT1, id2,id1) DT2[, id1 := 3] setkey(DT2, id2, id1) # use i.v1 to reference v1 from the i component DT1[DT2, v1 := i.v1 ] DT1 id1 id2 v1 1: 2 e 0.7383247 2: 1 g 1.5952808 3: 2 j 0.3295078 … Read more
With data.table you could do: library(data.table) DT[, if(uniqueN(Day) >= 3) .SD, by = Group] which gives: Group Day 1: 1 1 2: 1 3 3: 1 5 4: 1 5 5: 3 1 6: 3 2 7: 3 3 Or with dplyr: library(dplyr) DT %>% group_by(Group) %>% filter(n_distinct(Day) >= 3) which gives the same result.