How to use a shifting function in postgres to determine time difference by row?
0
I am looking to use some sort of shift function in postgresql to figure out time difference by row. Below i have a table of workers with start and stop times for each of two companies. I can do this very easily in R. Here is my data set print out and the dput for it :
print:
id start_time stop_time company time_diff
1: worker_1 2017-10-20 22:06:23 2017-10-20 22:09:48 company 1 NA
2: worker_1 2017-10-20 22:28:45 2017-10-20 23:16:56 company 2 0.31583333
3: worker_1 2017-10-20 22:37:07 2017-10-20 22:43:14 company 1 -0.66361111
4: worker_1 2017-10-20 22:59:30 2017-10-20 23:05:52 company 1 0.27111111
5: worker_1 2017-10-20 23:39:57 2017-10-20 23:43:00 company 1 0.56805556
6: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 0.18166667
7: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 -0.22583333
8: worker_2 2017-10-19 18:03:43 2017-10-19 18:04:48 company 1 NA
9: worker_2 2017-10-19 18:04:48 2017-10-19 18:05:08 company 1 0.00000000
10: worker_2 2017-10-19 18:05:51 2017-10-19 18:18:47 company 1 0.01194444
11: worker_2 2017-10-19 18:50:39 2017-10-19 21:01:14 company 1 0.53111111
dput:
structure(list(id = c("worker_1", "worker_1", "worker_1", "worker_1",
"worker_1", "worker_1", "worker_1", "worker_2", "worker_2", "worker_2",
"worker_2"), start_time = structure(c(1508551583, 1508552925,
1508553427, 1508554770, 1508557197, 1508558034, 1508558034, 1508450623,
1508450688, 1508450751, 1508453439), class = c("POSIXct", "POSIXt"
)), stop_time = structure(c(1508551788, 1508555816, 1508553794,
1508555152, 1508557380, 1508558847, 1508558847, 1508450688, 1508450708,
1508451527, 1508461274), class = c("POSIXct", "POSIXt")), company = c("company 1",
"company 2", "company 1", "company 1", "company 1", "company 1",
"company 1", "company 1", "company 1", "company 1", "company 1"
), time_diff = c(NA, 0.315833333333333, -0.663611111111111, 0.271111111111111,
0.568055555555556, 0.181666666666667, -0.225833333333333, NA,
0, 0.0119444444444444, 0.531111111111111)), row.names = c(NA,
-11L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000007a81ef0>)
I create the time_diff column using the following code with data.table package:
my_data[,time_diff:=as.numeric(
difftime(start_time
,shift(stop_time, type = "lag")
, units = "hours")
),.(id)]
How do i replicate this in postgresql?
r postgresql datetime difftime
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
add a comment |
0
I am looking to use some sort of shift function in postgresql to figure out time difference by row. Below i have a table of workers with start and stop times for each of two companies. I can do this very easily in R. Here is my data set print out and the dput for it :
print:
id start_time stop_time company time_diff
1: worker_1 2017-10-20 22:06:23 2017-10-20 22:09:48 company 1 NA
2: worker_1 2017-10-20 22:28:45 2017-10-20 23:16:56 company 2 0.31583333
3: worker_1 2017-10-20 22:37:07 2017-10-20 22:43:14 company 1 -0.66361111
4: worker_1 2017-10-20 22:59:30 2017-10-20 23:05:52 company 1 0.27111111
5: worker_1 2017-10-20 23:39:57 2017-10-20 23:43:00 company 1 0.56805556
6: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 0.18166667
7: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 -0.22583333
8: worker_2 2017-10-19 18:03:43 2017-10-19 18:04:48 company 1 NA
9: worker_2 2017-10-19 18:04:48 2017-10-19 18:05:08 company 1 0.00000000
10: worker_2 2017-10-19 18:05:51 2017-10-19 18:18:47 company 1 0.01194444
11: worker_2 2017-10-19 18:50:39 2017-10-19 21:01:14 company 1 0.53111111
dput:
structure(list(id = c("worker_1", "worker_1", "worker_1", "worker_1",
"worker_1", "worker_1", "worker_1", "worker_2", "worker_2", "worker_2",
"worker_2"), start_time = structure(c(1508551583, 1508552925,
1508553427, 1508554770, 1508557197, 1508558034, 1508558034, 1508450623,
1508450688, 1508450751, 1508453439), class = c("POSIXct", "POSIXt"
)), stop_time = structure(c(1508551788, 1508555816, 1508553794,
1508555152, 1508557380, 1508558847, 1508558847, 1508450688, 1508450708,
1508451527, 1508461274), class = c("POSIXct", "POSIXt")), company = c("company 1",
"company 2", "company 1", "company 1", "company 1", "company 1",
"company 1", "company 1", "company 1", "company 1", "company 1"
), time_diff = c(NA, 0.315833333333333, -0.663611111111111, 0.271111111111111,
0.568055555555556, 0.181666666666667, -0.225833333333333, NA,
0, 0.0119444444444444, 0.531111111111111)), row.names = c(NA,
-11L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000007a81ef0>)
I create the time_diff column using the following code with data.table package:
my_data[,time_diff:=as.numeric(
difftime(start_time
,shift(stop_time, type = "lag")
, units = "hours")
),.(id)]
How do i replicate this in postgresql?
r postgresql datetime difftime
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
looks likeselect start_time - lag(stop_time) over(partition by id) from df
– IceCreamToucan
Nov 20 '18 at 21:15
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43
add a comment |
0
0
0
I am looking to use some sort of shift function in postgresql to figure out time difference by row. Below i have a table of workers with start and stop times for each of two companies. I can do this very easily in R. Here is my data set print out and the dput for it :
print:
id start_time stop_time company time_diff
1: worker_1 2017-10-20 22:06:23 2017-10-20 22:09:48 company 1 NA
2: worker_1 2017-10-20 22:28:45 2017-10-20 23:16:56 company 2 0.31583333
3: worker_1 2017-10-20 22:37:07 2017-10-20 22:43:14 company 1 -0.66361111
4: worker_1 2017-10-20 22:59:30 2017-10-20 23:05:52 company 1 0.27111111
5: worker_1 2017-10-20 23:39:57 2017-10-20 23:43:00 company 1 0.56805556
6: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 0.18166667
7: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 -0.22583333
8: worker_2 2017-10-19 18:03:43 2017-10-19 18:04:48 company 1 NA
9: worker_2 2017-10-19 18:04:48 2017-10-19 18:05:08 company 1 0.00000000
10: worker_2 2017-10-19 18:05:51 2017-10-19 18:18:47 company 1 0.01194444
11: worker_2 2017-10-19 18:50:39 2017-10-19 21:01:14 company 1 0.53111111
dput:
structure(list(id = c("worker_1", "worker_1", "worker_1", "worker_1",
"worker_1", "worker_1", "worker_1", "worker_2", "worker_2", "worker_2",
"worker_2"), start_time = structure(c(1508551583, 1508552925,
1508553427, 1508554770, 1508557197, 1508558034, 1508558034, 1508450623,
1508450688, 1508450751, 1508453439), class = c("POSIXct", "POSIXt"
)), stop_time = structure(c(1508551788, 1508555816, 1508553794,
1508555152, 1508557380, 1508558847, 1508558847, 1508450688, 1508450708,
1508451527, 1508461274), class = c("POSIXct", "POSIXt")), company = c("company 1",
"company 2", "company 1", "company 1", "company 1", "company 1",
"company 1", "company 1", "company 1", "company 1", "company 1"
), time_diff = c(NA, 0.315833333333333, -0.663611111111111, 0.271111111111111,
0.568055555555556, 0.181666666666667, -0.225833333333333, NA,
0, 0.0119444444444444, 0.531111111111111)), row.names = c(NA,
-11L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000007a81ef0>)
I create the time_diff column using the following code with data.table package:
my_data[,time_diff:=as.numeric(
difftime(start_time
,shift(stop_time, type = "lag")
, units = "hours")
),.(id)]
How do i replicate this in postgresql?
r postgresql datetime difftime
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
I am looking to use some sort of shift function in postgresql to figure out time difference by row. Below i have a table of workers with start and stop times for each of two companies. I can do this very easily in R. Here is my data set print out and the dput for it :
print:
id start_time stop_time company time_diff
1: worker_1 2017-10-20 22:06:23 2017-10-20 22:09:48 company 1 NA
2: worker_1 2017-10-20 22:28:45 2017-10-20 23:16:56 company 2 0.31583333
3: worker_1 2017-10-20 22:37:07 2017-10-20 22:43:14 company 1 -0.66361111
4: worker_1 2017-10-20 22:59:30 2017-10-20 23:05:52 company 1 0.27111111
5: worker_1 2017-10-20 23:39:57 2017-10-20 23:43:00 company 1 0.56805556
6: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 0.18166667
7: worker_1 2017-10-20 23:53:54 2017-10-21 00:07:27 company 1 -0.22583333
8: worker_2 2017-10-19 18:03:43 2017-10-19 18:04:48 company 1 NA
9: worker_2 2017-10-19 18:04:48 2017-10-19 18:05:08 company 1 0.00000000
10: worker_2 2017-10-19 18:05:51 2017-10-19 18:18:47 company 1 0.01194444
11: worker_2 2017-10-19 18:50:39 2017-10-19 21:01:14 company 1 0.53111111
dput:
structure(list(id = c("worker_1", "worker_1", "worker_1", "worker_1",
"worker_1", "worker_1", "worker_1", "worker_2", "worker_2", "worker_2",
"worker_2"), start_time = structure(c(1508551583, 1508552925,
1508553427, 1508554770, 1508557197, 1508558034, 1508558034, 1508450623,
1508450688, 1508450751, 1508453439), class = c("POSIXct", "POSIXt"
)), stop_time = structure(c(1508551788, 1508555816, 1508553794,
1508555152, 1508557380, 1508558847, 1508558847, 1508450688, 1508450708,
1508451527, 1508461274), class = c("POSIXct", "POSIXt")), company = c("company 1",
"company 2", "company 1", "company 1", "company 1", "company 1",
"company 1", "company 1", "company 1", "company 1", "company 1"
), time_diff = c(NA, 0.315833333333333, -0.663611111111111, 0.271111111111111,
0.568055555555556, 0.181666666666667, -0.225833333333333, NA,
0, 0.0119444444444444, 0.531111111111111)), row.names = c(NA,
-11L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000007a81ef0>)
I create the time_diff column using the following code with data.table package:
my_data[,time_diff:=as.numeric(
difftime(start_time
,shift(stop_time, type = "lag")
, units = "hours")
),.(id)]
How do i replicate this in postgresql?
r postgresql datetime difftime
r postgresql datetime difftime
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
asked Nov 20 '18 at 20:57
LoF10LoF10
154214
154214
Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
looks likeselect start_time - lag(stop_time) over(partition by id) from df
– IceCreamToucan
Nov 20 '18 at 21:15
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43
add a comment |
Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
looks likeselect start_time - lag(stop_time) over(partition by id) from df
– IceCreamToucan
Nov 20 '18 at 21:15
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43
Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
looks like
select start_time - lag(stop_time) over(partition by id) from df– IceCreamToucan
Nov 20 '18 at 21:15
looks like
select start_time - lag(stop_time) over(partition by id) from df– IceCreamToucan
Nov 20 '18 at 21:15
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43
add a comment |
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Related: stackoverflow.com/questions/17301816/lag-function-and-group-by
– IceCreamToucan
Nov 20 '18 at 20:58
looks like
select start_time - lag(stop_time) over(partition by id) from df– IceCreamToucan
Nov 20 '18 at 21:15
this was perfect, thank you
– LoF10
Nov 20 '18 at 21:43