One way to do it is to use the magic n() function within filter: library(dplyr) my_data <- data.frame(Year=1996, Site=”A”, Brood=c(1,1,2,2,2)) my_data %>% group_by(Year, Site, Brood) %>% filter(n() >= 3) The n() function gives the number of rows in the current group (or the number of rows total if there is no grouping).
You can use rowSums. df$no_calls <- rowSums(df == “nc”) df # rsID sample1 sample2 sample3 sample1304 no_calls #1 abcd aa bb nc nc 2 #2 efgh nc nc nc nc 4 #3 ijkl aa ab aa nc 1 Or, as pointed out by MrFlick, to exclude the first column from the row sums, you can … Read more
You really should not be storing data like this. Fortunately, there is a way to undo the damage with recursive SQL, something along these lines: WITH unpivot (lvl, id, fk_ref, reference, tail) AS ( SELECT 1, id, fk_ref, CASE WHEN LOCATE(‘,’,reference) > 0 THEN TRIM(LEFT(reference, LOCATE(‘,’,reference)-1)) ELSE TRIM(reference) END, CASE WHEN LOCATE(‘,’,reference) > 0 THEN … Read more
Is it even possible? tldr; Yes. Codepen Demo #1 Codepen Demo # 2 (Uses SASS and is configurable) The difficulty here is that each article is a grid within itself and therefore any one article has no knowledge about another. Because of this, there is no way for a component of one article like a … Read more
Update Not knowing what you are trying to do, I’ll share one more suggestion: Preallocate vectors of the type you want for each column, insert values into those vectors, and then, at the end, create your data.frame. Continuing with Julian’s f3 (a preallocated data.frame) as the fastest option so far, defined as: # pre-allocate space … 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
There are a few different ways of doing this. I prefer using apply, since it’s easily extendable: ##Generate some data dd = data.frame(a = 1:4, b= 1:0, c=0:3) ##Go through each row and determine if a value is zero row_sub = apply(dd, 1, function(row) all(row !=0 )) ##Subset as usual dd[row_sub,]
One possible solution to your problem would be to use merge. Checking if any row (all columns) from another dataframe (df2) are present in df1 is equivalent to determining the intersection of the the two dataframes. This can be accomplished using the following function: pd.merge(df1, df2, on=[‘A’, ‘B’, ‘C’, ‘D’], how=’inner’) For example, if df1 … Read more
You can use table like this: df[df$names %in% names(table(df$names))[table(df$names) >= 5],]
You could do something like this: summed <- rowSums(zscore[, c(1, 2, 3, 5)])