I have a dataframe where I would like to remove specific rows. I would like to remove the row where there is the "Référence" word and the 3 rows under the "référence" row. See my example here.
I think I have to use grepl function.
Thank you for your help.
Max.
You should use grep, not grepl. When you use grep you get the row indexes that match the pattern, while with grepl you get a boolean vector. You could do:
rowIndexes = grep(x = df$col1, pattern = "refer")
df = df[-c(rowIndexes, rowIndexes+1, rowIndexes+2),]
Example:
> df
a b c d e
1 00100 44 5 69 fr
2 refer 34 35 7 df
3 thisalso 46 15 167 as
4 thistoo 46 15 167 as
5 00100 11 5 67 uu
6 00100 563 25 23 tt
7 00100 44 5 69 fr
8 refer 34 35 7 df
9 thisalso 46 15 167 as
10 thistoo 11 5 67 uu
11 00100 563 25 23 tt
12 00100 44 5 69 fr
13 refer 34 35 7 df
14 thisalso 46 15 167 as
15 thistoo 11 5 67 uu
16 00100 563 25 23 tt
17 00100 563 25 23 tt
18 00100 563 25 23 tt
> rowIndexes = grep(x = df$col1, pattern = "refer")
> df = df[-c(rowIndexes, rowIndexes+1, rowIndexes+2),]
> df
a b c d e
1 00100 44 5 69 fr
5 00100 11 5 67 uu
6 00100 563 25 23 tt
7 00100 44 5 69 fr
11 00100 563 25 23 tt
12 00100 44 5 69 fr
16 00100 563 25 23 tt
17 00100 563 25 23 tt
18 00100 563 25 23 tt
If you want to remove N lines after o before a set of specific lines, do:
rowIndexes = grep(x = df$col1, pattern = "refer")
N = 2
indexesToRemove = sapply(rowIndexes, function(x){ x + (0:N) })
df = df[-indexesToRemove, ]
where N is an integer. If N is positive it will remove N rows after the lines with "refer". If N is negative, this will remove N previous rows.