Create column in R

Quite strange this way of saving the database, but anyway, it follows code that does what you need:

df <- structure(list(Potreiro = structure(c(3L, 3L, 3L, 3L, 3L, 4L, 3L, 4L, 3L, 4L), .Label = c("1A", "6B", "7A", "7B"), class = "factor"), Code = structure(c(4L, 1L, 8L, 3L, 2L, 4L, 6L, 5L, 8L, 7L ), .Label = c("2", "3", "4", "5", "50%", "70%", "ac", "ad", "av", "cd", "de", "Dem"), class = "factor")), .Names = c("Potreiro", "Code"), row.names = c(NA, 10L), class = "data.frame")
df$Code <- as.character(df$Code)
df$numero <- as.numeric(gsub("%", "", df$Code))
df$char <- ifelse(grepl("[[:alpha:]]", df$Code), df$Code, NA)
df$perc <- factor(grepl("%", df$Code), labels = c("Não", "Sim"))

Note that it was necessary to use regular expression to first fetch and remove % , because only then will a numeric column be created.

Next, check if there was any character in the column to create the character column.

As 70 and 50 are percentages, I created the count perc that signals if the number column is a Percentage.

See if this is what you needed:

   Potreiro Code numero char perc
1        7A    5      5 <NA>  Não
2        7A    2      2 <NA>  Não
3        7A   ad     NA   ad  Não
4        7A    4      4 <NA>  Não
5        7A    3      3 <NA>  Não
6        7B    5      5 <NA>  Não
7        7A  70%     70 <NA>  Sim
8        7B  50%     50 <NA>  Sim
9        7A   ad     NA   ad  Não
10       7B   ac     NA   ac  Não