Faster way to get year from large data set in R
0
Is it any faster way to get the year from large data set (around 1GB) in R?
Currently I used data$year <- format(as.Date(data$pickup_datatime), "%Y") to get the year, but it took very long time.
r date
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
add a comment |
0
Is it any faster way to get the year from large data set (around 1GB) in R?
Currently I used data$year <- format(as.Date(data$pickup_datatime), "%Y") to get the year, but it took very long time.
r date
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
Do both functions take a long time, or is it one or the other? You might trylubridate::year()instead offormat.
– Mako212
Nov 15 '18 at 20:18
3
Instead of parsing the dates, you may trysubstrorstringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.
– Henrik
Nov 15 '18 at 20:47
1
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10
add a comment |
0
0
0
Is it any faster way to get the year from large data set (around 1GB) in R?
Currently I used data$year <- format(as.Date(data$pickup_datatime), "%Y") to get the year, but it took very long time.
r date
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
Is it any faster way to get the year from large data set (around 1GB) in R?
Currently I used data$year <- format(as.Date(data$pickup_datatime), "%Y") to get the year, but it took very long time.
r date
r date
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
edited Nov 15 '18 at 20:57
Henrik
42.3k994110
42.3k994110
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
asked Nov 15 '18 at 20:13
Tomislav FridwaldTomislav Fridwald
262
262
Do both functions take a long time, or is it one or the other? You might trylubridate::year()instead offormat.
– Mako212
Nov 15 '18 at 20:18
3
Instead of parsing the dates, you may trysubstrorstringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.
– Henrik
Nov 15 '18 at 20:47
1
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10
add a comment |
Do both functions take a long time, or is it one or the other? You might trylubridate::year()instead offormat.
– Mako212
Nov 15 '18 at 20:18
3
Instead of parsing the dates, you may trysubstrorstringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.
– Henrik
Nov 15 '18 at 20:47
1
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10
Do both functions take a long time, or is it one or the other? You might try
lubridate::year() instead of format.– Mako212
Nov 15 '18 at 20:18
Do both functions take a long time, or is it one or the other? You might try
lubridate::year() instead of format.– Mako212
Nov 15 '18 at 20:18
3
3
Instead of parsing the dates, you may try
substr or stringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.– Henrik
Nov 15 '18 at 20:47
Instead of parsing the dates, you may try
substr or stringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.– Henrik
Nov 15 '18 at 20:47
1
1
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10
add a comment |
1 Answer
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the lubridate package has a built-in function to get the year from a date-like object. Here's the use for your case:
data$year <- lubridate::year(data$pickup_datatime)
answered Nov 15 '18 at 23:12
dmcadmca
4681515
add a comment |
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the lubridate package has a built-in function to get the year from a date-like object. Here's the use for your case:
data$year <- lubridate::year(data$pickup_datatime)
answered Nov 15 '18 at 23:12
dmcadmca
4681515
add a comment |
0
the lubridate package has a built-in function to get the year from a date-like object. Here's the use for your case:
data$year <- lubridate::year(data$pickup_datatime)
answered Nov 15 '18 at 23:12
dmcadmca
4681515
add a comment |
0
0
0
the lubridate package has a built-in function to get the year from a date-like object. Here's the use for your case:
data$year <- lubridate::year(data$pickup_datatime)
answered Nov 15 '18 at 23:12
dmcadmca
4681515
the lubridate package has a built-in function to get the year from a date-like object. Here's the use for your case:
data$year <- lubridate::year(data$pickup_datatime)
answered Nov 15 '18 at 23:12
dmcadmca
4681515
answered Nov 15 '18 at 23:12
dmcadmca
4681515
answered Nov 15 '18 at 23:12
dmcadmca
4681515
answered Nov 15 '18 at 23:12
dmcadmca
4681515
4681515
add a comment |
add a comment |
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Do both functions take a long time, or is it one or the other? You might try
lubridate::year()instead offormat.– Mako212
Nov 15 '18 at 20:18
3
Instead of parsing the dates, you may try
substrorstringi::stri_sub, as I did here when grabbing hour: Fastest way to extract hour from time (HH:MM). When posting a question about speed, it's good if you also provide easily reproducible data of sufficient size to try the code on. Cheers.– Henrik
Nov 15 '18 at 20:47
1
Possible duplicate of Fastest way to extract hour from time (HH:MM)
– Florian
Nov 15 '18 at 21:10
...or at least post some sample data.
– Gregor
Nov 15 '18 at 21:10