Mixing
flexible column comparison in data table r
I am trying to filter by comparing groups of columns with each other in a dynamic fashion. Suppose I have the data base.
###########
#Setup data
###########
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
The data table:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 1 2 9 0 0 9 0
2: 2 7 4 8 7 1 8
3: 3 5 10 9 5 10 9
4: 4 1 8 1 1 8 1
5: 5 6 0 0 6 0 1
6: 6 8 7 0 8 7 0
7: 7 0 0 6 0 0 2
By dynamic, I mean Tp can be an arbitrary integer.
I want to filter the following way:
For each t, e.g. prod1vint{t}, I want to compare to look at the "prm" version of it and check if it's not zero. If it is not zero, then I only want to keep rows for which all t'>t vint are less than or equal to the pre-prime values and all lower elements (t'<t) are the SAME, e.g.
For each t.., if prod1vint{t}!=0, then
1. prod1vint{t'}prm <= prod1vint{t'} for t'>t
2. prod1vint{t'}prm == prod1vint{t'} for t'<t
For example, the following output should be displayed:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 3 5 10 9 5 10 9
2: 4 1 8 1 1 8 1
3: 6 8 7 0 8 7 0
4: 7 0 0 6 0 0 2
(In case the X-Y problem...this might help... I am trying to make sure each vector (prod1vint1,prod1vint2,prod1vint3) is LIFO decreasing to it's prime. Ignore this bit if it doesn't help. My attempted solution involves coding various conditions such as the one above, which I am stuck on.)
r function data.table
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
add a comment |
I am trying to filter by comparing groups of columns with each other in a dynamic fashion. Suppose I have the data base.
###########
#Setup data
###########
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
The data table:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 1 2 9 0 0 9 0
2: 2 7 4 8 7 1 8
3: 3 5 10 9 5 10 9
4: 4 1 8 1 1 8 1
5: 5 6 0 0 6 0 1
6: 6 8 7 0 8 7 0
7: 7 0 0 6 0 0 2
By dynamic, I mean Tp can be an arbitrary integer.
I want to filter the following way:
For each t, e.g. prod1vint{t}, I want to compare to look at the "prm" version of it and check if it's not zero. If it is not zero, then I only want to keep rows for which all t'>t vint are less than or equal to the pre-prime values and all lower elements (t'<t) are the SAME, e.g.
For each t.., if prod1vint{t}!=0, then
1. prod1vint{t'}prm <= prod1vint{t'} for t'>t
2. prod1vint{t'}prm == prod1vint{t'} for t'<t
For example, the following output should be displayed:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 3 5 10 9 5 10 9
2: 4 1 8 1 1 8 1
3: 6 8 7 0 8 7 0
4: 7 0 0 6 0 0 2
(In case the X-Y problem...this might help... I am trying to make sure each vector (prod1vint1,prod1vint2,prod1vint3) is LIFO decreasing to it's prime. Ignore this bit if it doesn't help. My attempted solution involves coding various conditions such as the one above, which I am stuck on.)
r function data.table
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
1
i think you should reshape your data withmeltto facilitate this.
– MichaelChirico
Jan 1 at 23:17
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01
add a comment |
I am trying to filter by comparing groups of columns with each other in a dynamic fashion. Suppose I have the data base.
###########
#Setup data
###########
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
The data table:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 1 2 9 0 0 9 0
2: 2 7 4 8 7 1 8
3: 3 5 10 9 5 10 9
4: 4 1 8 1 1 8 1
5: 5 6 0 0 6 0 1
6: 6 8 7 0 8 7 0
7: 7 0 0 6 0 0 2
By dynamic, I mean Tp can be an arbitrary integer.
I want to filter the following way:
For each t, e.g. prod1vint{t}, I want to compare to look at the "prm" version of it and check if it's not zero. If it is not zero, then I only want to keep rows for which all t'>t vint are less than or equal to the pre-prime values and all lower elements (t'<t) are the SAME, e.g.
For each t.., if prod1vint{t}!=0, then
1. prod1vint{t'}prm <= prod1vint{t'} for t'>t
2. prod1vint{t'}prm == prod1vint{t'} for t'<t
For example, the following output should be displayed:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 3 5 10 9 5 10 9
2: 4 1 8 1 1 8 1
3: 6 8 7 0 8 7 0
4: 7 0 0 6 0 0 2
(In case the X-Y problem...this might help... I am trying to make sure each vector (prod1vint1,prod1vint2,prod1vint3) is LIFO decreasing to it's prime. Ignore this bit if it doesn't help. My attempted solution involves coding various conditions such as the one above, which I am stuck on.)
r function data.table
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
I am trying to filter by comparing groups of columns with each other in a dynamic fashion. Suppose I have the data base.
###########
#Setup data
###########
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
The data table:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 1 2 9 0 0 9 0
2: 2 7 4 8 7 1 8
3: 3 5 10 9 5 10 9
4: 4 1 8 1 1 8 1
5: 5 6 0 0 6 0 1
6: 6 8 7 0 8 7 0
7: 7 0 0 6 0 0 2
By dynamic, I mean Tp can be an arbitrary integer.
I want to filter the following way:
For each t, e.g. prod1vint{t}, I want to compare to look at the "prm" version of it and check if it's not zero. If it is not zero, then I only want to keep rows for which all t'>t vint are less than or equal to the pre-prime values and all lower elements (t'<t) are the SAME, e.g.
For each t.., if prod1vint{t}!=0, then
1. prod1vint{t'}prm <= prod1vint{t'} for t'>t
2. prod1vint{t'}prm == prod1vint{t'} for t'<t
For example, the following output should be displayed:
n prod1vint1 prod1vint2 prod1vint3 prod1vint1prm prod1vint2prm prod1vint3prm
1: 3 5 10 9 5 10 9
2: 4 1 8 1 1 8 1
3: 6 8 7 0 8 7 0
4: 7 0 0 6 0 0 2
(In case the X-Y problem...this might help... I am trying to make sure each vector (prod1vint1,prod1vint2,prod1vint3) is LIFO decreasing to it's prime. Ignore this bit if it doesn't help. My attempted solution involves coding various conditions such as the one above, which I am stuck on.)
r function data.table
r function data.table
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
edited Jan 1 at 22:58
wolfsatthedoor
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
asked Jan 1 at 22:34
wolfsatthedoorwolfsatthedoor
1,627102959
1,627102959
1
i think you should reshape your data withmeltto facilitate this.
– MichaelChirico
Jan 1 at 23:17
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01
add a comment |
1
i think you should reshape your data withmeltto facilitate this.
– MichaelChirico
Jan 1 at 23:17
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01
1
1
i think you should reshape your data with
melt to facilitate this.– MichaelChirico
Jan 1 at 23:17
i think you should reshape your data with
melt to facilitate this.– MichaelChirico
Jan 1 at 23:17
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01
add a comment |
1 Answer
1
active
oldest
votes
Using melt as the comments suggest I would do it like this:
# this part is as replied from the question
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
# NEW CODE
fill.melt <- reshape2::melt(fill, id.vars = c('n'))
fill.melt$intpart <- sapply(fill.melt$variable,
function (x)
{stringr::str_extract(gsub('prod1','',x),
'\d')})
fill.melt$prmpart <- ifelse(grepl('prm', fill.melt$variable), 'prm','noprm')
fill.cast <- reshape2::dcast(fill.melt, n+intpart ~ prmpart , value.var = 'value')
fill.cast <- as.data.table(fill.cast)
t=3
tmp <- fill.cast[
((intpart >= t & prm <= noprm) | (intpart < t & prm == noprm)),]
ns <- unique(tmp$n)[table(tmp$n) == t]
fill[n %in% ns,]
answered Jan 2 at 16:31
karenkaren
487215
do you know thatdata.tablehas much faster and more functionalmelt/dcastfunctions?
– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
add a comment |
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1 Answer
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1 Answer
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votes
Using melt as the comments suggest I would do it like this:
# this part is as replied from the question
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
# NEW CODE
fill.melt <- reshape2::melt(fill, id.vars = c('n'))
fill.melt$intpart <- sapply(fill.melt$variable,
function (x)
{stringr::str_extract(gsub('prod1','',x),
'\d')})
fill.melt$prmpart <- ifelse(grepl('prm', fill.melt$variable), 'prm','noprm')
fill.cast <- reshape2::dcast(fill.melt, n+intpart ~ prmpart , value.var = 'value')
fill.cast <- as.data.table(fill.cast)
t=3
tmp <- fill.cast[
((intpart >= t & prm <= noprm) | (intpart < t & prm == noprm)),]
ns <- unique(tmp$n)[table(tmp$n) == t]
fill[n %in% ns,]
answered Jan 2 at 16:31
karenkaren
487215
do you know thatdata.tablehas much faster and more functionalmelt/dcastfunctions?
– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
add a comment |
Using melt as the comments suggest I would do it like this:
# this part is as replied from the question
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
# NEW CODE
fill.melt <- reshape2::melt(fill, id.vars = c('n'))
fill.melt$intpart <- sapply(fill.melt$variable,
function (x)
{stringr::str_extract(gsub('prod1','',x),
'\d')})
fill.melt$prmpart <- ifelse(grepl('prm', fill.melt$variable), 'prm','noprm')
fill.cast <- reshape2::dcast(fill.melt, n+intpart ~ prmpart , value.var = 'value')
fill.cast <- as.data.table(fill.cast)
t=3
tmp <- fill.cast[
((intpart >= t & prm <= noprm) | (intpart < t & prm == noprm)),]
ns <- unique(tmp$n)[table(tmp$n) == t]
fill[n %in% ns,]
answered Jan 2 at 16:31
karenkaren
487215
do you know thatdata.tablehas much faster and more functionalmelt/dcastfunctions?
– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
add a comment |
Using melt as the comments suggest I would do it like this:
# this part is as replied from the question
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
# NEW CODE
fill.melt <- reshape2::melt(fill, id.vars = c('n'))
fill.melt$intpart <- sapply(fill.melt$variable,
function (x)
{stringr::str_extract(gsub('prod1','',x),
'\d')})
fill.melt$prmpart <- ifelse(grepl('prm', fill.melt$variable), 'prm','noprm')
fill.cast <- reshape2::dcast(fill.melt, n+intpart ~ prmpart , value.var = 'value')
fill.cast <- as.data.table(fill.cast)
t=3
tmp <- fill.cast[
((intpart >= t & prm <= noprm) | (intpart < t & prm == noprm)),]
ns <- unique(tmp$n)[table(tmp$n) == t]
fill[n %in% ns,]
answered Jan 2 at 16:31
karenkaren
487215
Using melt as the comments suggest I would do it like this:
# this part is as replied from the question
set.seed(2)
fill = data.table(n=1:7)
Tp=3
for(t in 1:Tp){
set(x = fill, j = paste0('prod1vint',t), value = sample(0:10,7))
}
fill[1,paste0('prod1vint',3):=0]
fill[5,paste0('prod1vint',2):=0]
fill[5,paste0('prod1vint',3):=0]
for(t in 1:Tp){
fill[,paste0('prod1vint',t,'prm'):=get(paste0('prod1vint',t))]
}
fill[1,paste0('prod1vint',1,'prm'):=0]
fill[2,paste0('prod1vint',2,'prm'):=1]
fill[5,paste0('prod1vint',3,'prm'):=1]
fill[7,paste0('prod1vint',3,'prm'):=2]
# NEW CODE
fill.melt <- reshape2::melt(fill, id.vars = c('n'))
fill.melt$intpart <- sapply(fill.melt$variable,
function (x)
{stringr::str_extract(gsub('prod1','',x),
'\d')})
fill.melt$prmpart <- ifelse(grepl('prm', fill.melt$variable), 'prm','noprm')
fill.cast <- reshape2::dcast(fill.melt, n+intpart ~ prmpart , value.var = 'value')
fill.cast <- as.data.table(fill.cast)
t=3
tmp <- fill.cast[
((intpart >= t & prm <= noprm) | (intpart < t & prm == noprm)),]
ns <- unique(tmp$n)[table(tmp$n) == t]
fill[n %in% ns,]
answered Jan 2 at 16:31
karenkaren
487215
answered Jan 2 at 16:31
karenkaren
487215
answered Jan 2 at 16:31
karenkaren
487215
answered Jan 2 at 16:31
karenkaren
487215
487215
do you know thatdata.tablehas much faster and more functionalmelt/dcastfunctions?
– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
add a comment |
do you know thatdata.tablehas much faster and more functionalmelt/dcastfunctions?
– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
do you know that
data.table has much faster and more functional melt/dcast functions?– eddi
Jan 2 at 20:20
do you know that
data.table has much faster and more functional melt/dcast functions?– eddi
Jan 2 at 20:20
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
Absolutely, I am very fan of data.table, this is just a solution
– karen
Jan 3 at 8:11
add a comment |
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i think you should reshape your data with
meltto facilitate this.– MichaelChirico
Jan 1 at 23:17
Not sure how that helps
– wolfsatthedoor
Jan 2 at 0:01