Example data.frame:
df = read.table(text = 'colA colB
2 7
2 7
2 7
2 7
1 7
1 7
1 7
89 5
89 5
89 5
88 5
88 5
70 5
70 5
70 5
69 5
69 5
44 4
44 4
44 4
43 4
42 4
42 4
41 4
41 4
120 1
100 1', header = TRUE)
I need to add an index col based on colA and colB where colB shows the exact number of rows to group but it can be duplicated. colB groups rows based on colA and colA -1.
Expected output:
colA colB index_col
2 7 1
2 7 1
2 7 1
2 7 1
1 7 1
1 7 1
1 7 1
89 5 2
89 5 2
89 5 2
88 5 2
88 5 2
70 5 3
70 5 3
70 5 3
69 5 3
69 5 3
44 4 4
44 4 4
44 4 4
43 4 4
42 4 5
42 4 5
41 4 5
41 4 5
120 1 6
100 1 7
UPDATE
How can I adapt the code that works for the above df for the same purpose but by looking at colB values grouped based on colA, colA -1 and colA -2? i.e. (instead of 2 days considering 3 days)
new_df = read.table(text = 'colA colB
3 10
3 10
3 10
2 10
2 10
2 10
2 10
1 10
1 10
1 10
90 7
90 7
89 7
89 7
89 7
88 7
88 7
71 7
71 7
70 7
70 7
70 7
69 7
69 7
44 5
44 5
44 5
43 5
42 5
41 5
41 5
41 5
40 5
40 5
120 1
100 1', header = TRUE)
Expected output:
colA colB index_col
3 10 1
3 10 1
3 10 1
2 10 1
2 10 1
2 10 1
2 10 1
1 10 1
1 10 1
1 10 1
90 7 2
90 7 2
89 7 2
89 7 2
89 7 2
88 7 2
88 7 2
71 7 3
71 7 3
70 7 3
70 7 3
70 7 3
69 7 3
69 7 3
44 5 4
44 5 4
44 5 4
43 5 4
42 5 4
41 5 5
41 5 5
41 5 5
40 5 5
40 5 5
120 1 6
100 1 7
Thanks
We can use rleid
library(data.table)
index_col <-setDT(df)[, if(colB[1L] < .N) ((seq_len(.N)-1) %/% colB[1L])+1
else as.numeric(colB), rleid(colB)][, rleid(V1)]
df[, index_col := index_col]
df
# colA colB index_col
# 1: 2 7 1
# 2: 2 7 1
# 3: 2 7 1
# 4: 2 7 1
# 5: 1 7 1
# 6: 1 7 1
# 7: 1 7 1
# 8: 70 5 2
# 9: 70 5 2
#10: 70 5 2
#11: 69 5 2
#12: 69 5 2
#13: 89 5 3
#14: 89 5 3
#15: 89 5 3
#16: 88 5 3
#17: 88 5 3
#18: 120 1 4
#19: 100 1 5
Or a one-liner would be
setDT(df)[, index_col := df[, ((seq_len(.N)-1) %/% colB[1L])+1, rleid(colB)][, as.integer(interaction(.SD, drop = TRUE, lex.order = TRUE))]]
Update
Based on the new update in the OP's post
setDT(new_df)[, index_col := cumsum(c(TRUE, abs(diff(colA))> 1))
][, colB := .N , index_col]
new_df
# colA colB index_col
# 1: 3 10 1
# 2: 3 10 1
# 3: 3 10 1
# 4: 2 10 1
# 5: 2 10 1
# 6: 2 10 1
# 7: 2 10 1
# 8: 1 10 1
# 9: 1 10 1
#10: 1 10 1
#11: 71 7 2
#12: 71 7 2
#13: 70 7 2
#14: 70 7 2
#15: 70 7 2
#16: 69 7 2
#17: 69 7 2
#18: 90 7 3
#19: 90 7 3
#20: 89 7 3
#21: 89 7 3
#22: 89 7 3
#23: 88 7 3
#24: 88 7 3
#25: 44 2 4
#26: 43 2 4
#27: 120 1 5
#28: 100 1 6
An approach in base R:
df$idxcol <- cumsum(c(1,abs(diff(df$colA)) > 1) + c(0,diff(df$colB) != 0) > 0)
which gives:
> df colA colB idxcol 1 2 7 1 2 2 7 1 3 2 7 1 4 2 7 1 5 1 7 1 6 1 7 1 7 1 7 1 8 70 5 2 9 70 5 2 10 70 5 2 11 69 5 2 12 69 5 2 13 89 5 3 14 89 5 3 15 89 5 3 16 88 5 3 17 88 5 3 18 120 1 4 19 100 1 5
On the updated example data, you need to adapt the approach to:
n <- 1
idx1 <- cumsum(c(1, diff(df$colA) < -n) + c(0, diff(df$colB) != 0) > 0)
idx2 <- ave(df$colA, cumsum(c(1, diff(df$colA) < -n)), FUN = function(x) c(0, cumsum(diff(x)) < -n ))
idx2[idx2==1 & c(0,diff(idx2))==0] <- 0
df$idxcol <- idx1 + cumsum(idx2)
which gives:
> df colA colB idxcol 1 2 7 1 2 2 7 1 3 2 7 1 4 2 7 1 5 1 7 1 6 1 7 1 7 1 7 1 8 89 5 2 9 89 5 2 10 89 5 2 11 88 5 2 12 88 5 2 13 70 5 3 14 70 5 3 15 70 5 3 16 69 5 3 17 69 5 3 18 44 4 4 19 44 4 4 20 44 4 4 21 43 4 4 22 42 4 5 23 42 4 5 24 41 4 5 25 41 4 5 26 120 1 6 27 100 1 7
For new_df just change n tot 2 and you will get the desired output for that as well.
来源：https://stackoverflow.com/questions/44245666/update-add-index-column-to-data-frame-based-on-two-columns