Try this: df <- read.table(text = “v1 v2 v3 v4 v5 0 1 1 1 1 0 1 1 0 1 1 0 1 1 0”, skip = 1) df ll <- vector(mode = “list”, length = ncol(df)-1) ll <- lapply(2:ncol(df), function(ncols){ tmp <- t(apply(df, 1, function(rows) combn(x = rows, m = ncols, prod))) if(ncols … Read more
Something like this (assuming your data is named df) # get the column names that follow this pattern cols = grepl(“[0-9]+_parameter”, names(df)) # see if any of those columns have a 97 any(df[cols] == 97) A base R way to filter for rows that have a 97 in any of those columns is this: df[rowSums(df[cols] … Read more
I’m at least not quite sure whether I understood correct. Maybe a better question could improve the answers… DF <- data.frame(Tree = c(4382, 4383, 4384, 5385, 4386), a = c(21.88, 13.93, 19.69, 20.02, 11.07), b = c(9.59, 12.95, 9.78, 8.23, 23.20), k = c(0.0538, 0.0811, 0.0597, 0.0489, 0.1276)) DOY <- c(1:365) DF_new <- data.frame(sapply(1:length(DF$Tree), function(x)(DF$a[x]*exp(-exp(DF$b[x]-DF$k[x]*DOY))))) … Read more
comb <- combn(colnames(data),v) colnames(results) <- apply(comb,2,function(rows) paste0(rows, collapse = “”)) insted of paste use paste0 @henrik and @chargaff solved how paste function working in R? [duplicate]