R data.table create list column by group
I have a data.table below, and I would like to apply a function to column v2 group by v1 and order, the resulting column v3 should be a list of vectors below.
How do I write this function such that it will return a vector of 0s for each group's first row (order == 1 & v1 %in% c(1, 2)). For each subsequent row in the group, the vector will append previous row's v2 value to the right of the vector while bumping off one 0 from the left.
Initial data.table
t3 <- data.table(
v1 = rep(1:2, each = 5),
order = rep(1:5, 2),
v2 = c(6, 9, 6, 8, 6, 2, 5, 7, 8, 2)
)
v1 order v2
1: 1 1 6
2: 1 2 9
3: 1 3 6
4: 1 4 8
5: 1 5 6
6: 2 1 2
7: 2 2 5
8: 2 3 7
9: 2 4 8
10: 2 5 2
applying the function...
output:
t3[, v3 := list(c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 6),
c(0, 0, 0, 6, 9),
c(0, 0, 6, 9, 6),
c(0, 6, 9, 6, 8),
c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 2),
c(0, 0, 0, 2, 5),
c(0, 0, 2, 5, 7),
c(0, 2, 5, 7, 8))]
v1 order v2 v3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
r data.table
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
add a comment |
I have a data.table below, and I would like to apply a function to column v2 group by v1 and order, the resulting column v3 should be a list of vectors below.
How do I write this function such that it will return a vector of 0s for each group's first row (order == 1 & v1 %in% c(1, 2)). For each subsequent row in the group, the vector will append previous row's v2 value to the right of the vector while bumping off one 0 from the left.
Initial data.table
t3 <- data.table(
v1 = rep(1:2, each = 5),
order = rep(1:5, 2),
v2 = c(6, 9, 6, 8, 6, 2, 5, 7, 8, 2)
)
v1 order v2
1: 1 1 6
2: 1 2 9
3: 1 3 6
4: 1 4 8
5: 1 5 6
6: 2 1 2
7: 2 2 5
8: 2 3 7
9: 2 4 8
10: 2 5 2
applying the function...
output:
t3[, v3 := list(c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 6),
c(0, 0, 0, 6, 9),
c(0, 0, 6, 9, 6),
c(0, 6, 9, 6, 8),
c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 2),
c(0, 0, 0, 2, 5),
c(0, 0, 2, 5, 7),
c(0, 2, 5, 7, 8))]
v1 order v2 v3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
r data.table
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
add a comment |
I have a data.table below, and I would like to apply a function to column v2 group by v1 and order, the resulting column v3 should be a list of vectors below.
How do I write this function such that it will return a vector of 0s for each group's first row (order == 1 & v1 %in% c(1, 2)). For each subsequent row in the group, the vector will append previous row's v2 value to the right of the vector while bumping off one 0 from the left.
Initial data.table
t3 <- data.table(
v1 = rep(1:2, each = 5),
order = rep(1:5, 2),
v2 = c(6, 9, 6, 8, 6, 2, 5, 7, 8, 2)
)
v1 order v2
1: 1 1 6
2: 1 2 9
3: 1 3 6
4: 1 4 8
5: 1 5 6
6: 2 1 2
7: 2 2 5
8: 2 3 7
9: 2 4 8
10: 2 5 2
applying the function...
output:
t3[, v3 := list(c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 6),
c(0, 0, 0, 6, 9),
c(0, 0, 6, 9, 6),
c(0, 6, 9, 6, 8),
c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 2),
c(0, 0, 0, 2, 5),
c(0, 0, 2, 5, 7),
c(0, 2, 5, 7, 8))]
v1 order v2 v3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
r data.table
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
I have a data.table below, and I would like to apply a function to column v2 group by v1 and order, the resulting column v3 should be a list of vectors below.
How do I write this function such that it will return a vector of 0s for each group's first row (order == 1 & v1 %in% c(1, 2)). For each subsequent row in the group, the vector will append previous row's v2 value to the right of the vector while bumping off one 0 from the left.
Initial data.table
t3 <- data.table(
v1 = rep(1:2, each = 5),
order = rep(1:5, 2),
v2 = c(6, 9, 6, 8, 6, 2, 5, 7, 8, 2)
)
v1 order v2
1: 1 1 6
2: 1 2 9
3: 1 3 6
4: 1 4 8
5: 1 5 6
6: 2 1 2
7: 2 2 5
8: 2 3 7
9: 2 4 8
10: 2 5 2
applying the function...
output:
t3[, v3 := list(c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 6),
c(0, 0, 0, 6, 9),
c(0, 0, 6, 9, 6),
c(0, 6, 9, 6, 8),
c(0, 0, 0, 0, 0),
c(0, 0, 0, 0, 2),
c(0, 0, 0, 2, 5),
c(0, 0, 2, 5, 7),
c(0, 2, 5, 7, 8))]
v1 order v2 v3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
r data.table
r data.table
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
asked Nov 21 '18 at 15:55
EKtheSageEKtheSage
796
796
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We could try
t3[order(order), .(order, v2, shift(v2, 5:1, fill = 0)), by = v1]
Output:
v1 order v2 V3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
This is excellent answer! I didn't know you could create a list usingshift. Thank you, thank you, thank you!!
– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
We could try
t3[order(order), .(order, v2, shift(v2, 5:1, fill = 0)), by = v1]
Output:
v1 order v2 V3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
This is excellent answer! I didn't know you could create a list usingshift. Thank you, thank you, thank you!!
– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
add a comment |
We could try
t3[order(order), .(order, v2, shift(v2, 5:1, fill = 0)), by = v1]
Output:
v1 order v2 V3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
This is excellent answer! I didn't know you could create a list usingshift. Thank you, thank you, thank you!!
– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
add a comment |
We could try
t3[order(order), .(order, v2, shift(v2, 5:1, fill = 0)), by = v1]
Output:
v1 order v2 V3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
We could try
t3[order(order), .(order, v2, shift(v2, 5:1, fill = 0)), by = v1]
Output:
v1 order v2 V3
1: 1 1 6 0,0,0,0,0
2: 1 2 9 0,0,0,0,6
3: 1 3 6 0,0,0,6,9
4: 1 4 8 0,0,6,9,6
5: 1 5 6 0,6,9,6,8
6: 2 1 2 0,0,0,0,0
7: 2 2 5 0,0,0,0,2
8: 2 3 7 0,0,0,2,5
9: 2 4 8 0,0,2,5,7
10: 2 5 2 0,2,5,7,8
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
answered Nov 21 '18 at 16:59
J.R.J.R.
3,29811219
3,29811219
This is excellent answer! I didn't know you could create a list usingshift. Thank you, thank you, thank you!!
– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
add a comment |
This is excellent answer! I didn't know you could create a list usingshift. Thank you, thank you, thank you!!
– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
This is excellent answer! I didn't know you could create a list using
shift. Thank you, thank you, thank you!!– EKtheSage
Nov 22 '18 at 1:14
This is excellent answer! I didn't know you could create a list using
shift. Thank you, thank you, thank you!!– EKtheSage
Nov 22 '18 at 1:14
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
You are welcome. Matt & Company have done an excellent job, ´data.table´ is awesome!
– J.R.
Nov 22 '18 at 9:56
add a comment |
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