I notice that you mention a function %like% in your current approach. I don’t know if that’s a reference to the %like% from “data.table”, but if it is, you can definitely use it as follows. Note that the object does not have to be a data.table (but also remember that subsetting approaches for data.frames and … Read more
This question was answered in well in the comments by @James, pointing to an excellent explanation by Hadley Wickham of the dangers of subset (and functions like it) [here]. Go read it! It’s a somewhat long read, so it may be helpful to record here the example that Hadley uses that most directly addresses the … Read more
First, two base alternatives. One relies on table, and the other on ave and length. Then, two data.table ways. 1. table tt <- table(df$name) df2 <- subset(df, name %in% names(tt[tt < 3])) # or df2 <- df[df$name %in% names(tt[tt < 3]), ] If you want to walk it through step by step: # count each … Read more
Have a look at ?”%in%”. dt[dt$fct %in% vc,] fct X 1 a 2 3 c 3 5 c 5 7 a 7 9 c 9 10 a 1 12 c 2 14 c 4 You could also use ?is.element: dt[is.element(dt$fct, vc),]
To select rows according to one ‘cell_type’ (e.g. ‘hesc’), use ==: expr[expr$cell_type == “hesc”, ] To select rows according to two or more different ‘cell_type’, (e.g. either ‘hesc’ or ‘bj fibroblast’), use %in%: expr[expr$cell_type %in% c(“hesc”, “bj fibroblast”), ]
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
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