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Erase minutes and seconds from character dates in R
The Next CEO of Stack OverflowSelect every other element from a vectorIn R, split a character vector by a specific character; save 3rd piece in new vectorHow to Correctly Use Lists in R?Identifying unique terms from list of character vectorsConvert data.frame columns from factors to charactersRemove quotes from a character vector in RReplace specific characters within stringsdata.table vs dplyr: can one do something well the other can't or does poorly?In R, how do I extract minutes and seconds from charactersCalculate character string “days, hours, minutes, seconds” to numeric total daysFrom sessions to minute by minute summaryConvert character vector to a new vector of substrings
2
I have this timestamp vector:
c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
I'd like to strip off the minutes and seconds from the character vector so that I just have 01/09/2019 9 and 01/09/2019 10
What is the most efficient method to do so?
r
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
add a comment |
2
I have this timestamp vector:
c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
I'd like to strip off the minutes and seconds from the character vector so that I just have 01/09/2019 9 and 01/09/2019 10
What is the most efficient method to do so?
r
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
I'm pretty biased towardstidyversepackages :)
– JasonBaik
Mar 22 at 14:40
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07
add a comment |
2
2
2
0
I have this timestamp vector:
c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
I'd like to strip off the minutes and seconds from the character vector so that I just have 01/09/2019 9 and 01/09/2019 10
What is the most efficient method to do so?
r
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
I have this timestamp vector:
c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
I'd like to strip off the minutes and seconds from the character vector so that I just have 01/09/2019 9 and 01/09/2019 10
What is the most efficient method to do so?
r
r
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
asked Mar 21 at 1:02
JasonBaikJasonBaik
406114
406114
Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
I'm pretty biased towardstidyversepackages :)
– JasonBaik
Mar 22 at 14:40
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07
add a comment |
Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
I'm pretty biased towardstidyversepackages :)
– JasonBaik
Mar 22 at 14:40
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07
Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
I'm pretty biased towards
tidyverse packages :)– JasonBaik
Mar 22 at 14:40
I'm pretty biased towards
tidyverse packages :)– JasonBaik
Mar 22 at 14:40
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07
add a comment |
6 Answers
6
active
oldest
votes
1
You could also use str_extract from stringr:
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
str_extract(date_strings, ".+(?=:.+:)")
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10"
[5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[9] "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 17:06
Ben GBen G
911321
add a comment |
3
Here's one.
datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")
# Result
[1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
add a comment |
2
What is your desired output class? How about this one:
v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
strptime(v, "%m/%d/%Y %H")
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
add a comment |
2
This seems nice,
unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]
(Made with help from here)
Alternative could be,
sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)
Using some benchmarks and recent comments by Ronak, that fixed=TRUE makes the methods a lot faster, we see that method four (the above method) is fastest,
mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
three = strptime(mystring, "%m/%d/%Y %H"),
four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"),
six = gsub("(.*?):.*", "\1", mystring),
seven = str_extract(mystring, ".+(?=:.+:)"),
times = 100000)
Unit: microseconds
expr min lq mean median uq max neval
one 42.792 49.471 85.63742 52.572 57.1310 669280.96 1e+05
two 64.637 70.618 114.16364 73.252 77.6840 582466.94 1e+05
three 129.456 134.771 156.82308 136.188 139.2030 339715.94 1e+05
four 12.860 15.641 22.75699 17.254 18.5440 305703.52 1e+05
five 482.888 505.647 633.15388 512.880 552.1155 551274.28 1e+05
six 37.889 43.121 52.79030 45.567 49.1880 32954.59 1e+05
seven 53.432 59.051 88.05015 62.326 69.9320 1180361.17 1e+05
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
yep. I thinkfixed = TRUEmakes it a lot faster.
– Ronak Shah
Mar 21 at 1:40
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
add a comment |
0
Another one:
dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))
gives
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10" "01/09/2019 10"
[6] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
add a comment |
0
Here is another one using gsub
Capture the pattern by () and \1 to refer to the captured group, need ?to make regex lazy as there are multiple :.
gsub("(.*?):.*", "\1", dates)
answered Mar 21 at 1:56
MKaMKa
358412
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You could also use str_extract from stringr:
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
str_extract(date_strings, ".+(?=:.+:)")
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10"
[5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[9] "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 17:06
Ben GBen G
911321
add a comment |
1
You could also use str_extract from stringr:
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
str_extract(date_strings, ".+(?=:.+:)")
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10"
[5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[9] "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 17:06
Ben GBen G
911321
add a comment |
1
1
1
You could also use str_extract from stringr:
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
str_extract(date_strings, ".+(?=:.+:)")
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10"
[5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[9] "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 17:06
Ben GBen G
911321
You could also use str_extract from stringr:
date_strings <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
str_extract(date_strings, ".+(?=:.+:)")
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10"
[5] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[9] "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 17:06
Ben GBen G
911321
answered Mar 21 at 17:06
Ben GBen G
911321
answered Mar 21 at 17:06
Ben GBen G
911321
answered Mar 21 at 17:06
Ben GBen G
911321
911321
add a comment |
add a comment |
3
Here's one.
datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")
# Result
[1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
add a comment |
3
Here's one.
datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")
# Result
[1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
add a comment |
3
3
3
Here's one.
datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")
# Result
[1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
Here's one.
datevec <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
format(as.POSIXct(datevec, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H")
# Result
[1] "01/09/2019 09" "01/09/2019 09" "01/09/2019 09" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
[7] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
edited Mar 21 at 1:32
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
answered Mar 21 at 1:09
hmhensenhmhensen
7621718
7621718
add a comment |
add a comment |
2
What is your desired output class? How about this one:
v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
strptime(v, "%m/%d/%Y %H")
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
add a comment |
2
What is your desired output class? How about this one:
v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
strptime(v, "%m/%d/%Y %H")
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
add a comment |
2
2
2
What is your desired output class? How about this one:
v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
strptime(v, "%m/%d/%Y %H")
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
What is your desired output class? How about this one:
v <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
strptime(v, "%m/%d/%Y %H")
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
edited Mar 21 at 1:30
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
answered Mar 21 at 1:13
Ozan147Ozan147
2,0521519
2,0521519
add a comment |
add a comment |
2
This seems nice,
unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]
(Made with help from here)
Alternative could be,
sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)
Using some benchmarks and recent comments by Ronak, that fixed=TRUE makes the methods a lot faster, we see that method four (the above method) is fastest,
mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
three = strptime(mystring, "%m/%d/%Y %H"),
four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"),
six = gsub("(.*?):.*", "\1", mystring),
seven = str_extract(mystring, ".+(?=:.+:)"),
times = 100000)
Unit: microseconds
expr min lq mean median uq max neval
one 42.792 49.471 85.63742 52.572 57.1310 669280.96 1e+05
two 64.637 70.618 114.16364 73.252 77.6840 582466.94 1e+05
three 129.456 134.771 156.82308 136.188 139.2030 339715.94 1e+05
four 12.860 15.641 22.75699 17.254 18.5440 305703.52 1e+05
five 482.888 505.647 633.15388 512.880 552.1155 551274.28 1e+05
six 37.889 43.121 52.79030 45.567 49.1880 32954.59 1e+05
seven 53.432 59.051 88.05015 62.326 69.9320 1180361.17 1e+05
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
yep. I thinkfixed = TRUEmakes it a lot faster.
– Ronak Shah
Mar 21 at 1:40
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
add a comment |
2
This seems nice,
unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]
(Made with help from here)
Alternative could be,
sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)
Using some benchmarks and recent comments by Ronak, that fixed=TRUE makes the methods a lot faster, we see that method four (the above method) is fastest,
mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
three = strptime(mystring, "%m/%d/%Y %H"),
four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"),
six = gsub("(.*?):.*", "\1", mystring),
seven = str_extract(mystring, ".+(?=:.+:)"),
times = 100000)
Unit: microseconds
expr min lq mean median uq max neval
one 42.792 49.471 85.63742 52.572 57.1310 669280.96 1e+05
two 64.637 70.618 114.16364 73.252 77.6840 582466.94 1e+05
three 129.456 134.771 156.82308 136.188 139.2030 339715.94 1e+05
four 12.860 15.641 22.75699 17.254 18.5440 305703.52 1e+05
five 482.888 505.647 633.15388 512.880 552.1155 551274.28 1e+05
six 37.889 43.121 52.79030 45.567 49.1880 32954.59 1e+05
seven 53.432 59.051 88.05015 62.326 69.9320 1180361.17 1e+05
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
yep. I thinkfixed = TRUEmakes it a lot faster.
– Ronak Shah
Mar 21 at 1:40
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
add a comment |
2
2
2
This seems nice,
unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]
(Made with help from here)
Alternative could be,
sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)
Using some benchmarks and recent comments by Ronak, that fixed=TRUE makes the methods a lot faster, we see that method four (the above method) is fastest,
mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
three = strptime(mystring, "%m/%d/%Y %H"),
four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"),
six = gsub("(.*?):.*", "\1", mystring),
seven = str_extract(mystring, ".+(?=:.+:)"),
times = 100000)
Unit: microseconds
expr min lq mean median uq max neval
one 42.792 49.471 85.63742 52.572 57.1310 669280.96 1e+05
two 64.637 70.618 114.16364 73.252 77.6840 582466.94 1e+05
three 129.456 134.771 156.82308 136.188 139.2030 339715.94 1e+05
four 12.860 15.641 22.75699 17.254 18.5440 305703.52 1e+05
five 482.888 505.647 633.15388 512.880 552.1155 551274.28 1e+05
six 37.889 43.121 52.79030 45.567 49.1880 32954.59 1e+05
seven 53.432 59.051 88.05015 62.326 69.9320 1180361.17 1e+05
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
This seems nice,
unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)]
(Made with help from here)
Alternative could be,
sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1)
Using some benchmarks and recent comments by Ronak, that fixed=TRUE makes the methods a lot faster, we see that method four (the above method) is fastest,
mystring <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
microbenchmark(one = sapply(strsplit(mystring, split=':', fixed=TRUE), `[`, 1),
two = unlist(lapply(mystring,function(x) strsplit(x,":", fixed=TRUE)[[1]][1])),
three = strptime(mystring, "%m/%d/%Y %H"),
four = unlist(strsplit(mystring, split = ":", fixed=TRUE))[c(TRUE, FALSE,FALSE)],
five = format(as.POSIXct(mystring, format = "%d/%m/%Y %H:%M:%OS"), "%d/%m/%Y %H"),
six = gsub("(.*?):.*", "\1", mystring),
seven = str_extract(mystring, ".+(?=:.+:)"),
times = 100000)
Unit: microseconds
expr min lq mean median uq max neval
one 42.792 49.471 85.63742 52.572 57.1310 669280.96 1e+05
two 64.637 70.618 114.16364 73.252 77.6840 582466.94 1e+05
three 129.456 134.771 156.82308 136.188 139.2030 339715.94 1e+05
four 12.860 15.641 22.75699 17.254 18.5440 305703.52 1e+05
five 482.888 505.647 633.15388 512.880 552.1155 551274.28 1e+05
six 37.889 43.121 52.79030 45.567 49.1880 32954.59 1e+05
seven 53.432 59.051 88.05015 62.326 69.9320 1180361.17 1e+05
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
edited Mar 22 at 12:59
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
answered Mar 21 at 1:17
Hector HaffendenHector Haffenden
501114
501114
yep. I thinkfixed = TRUEmakes it a lot faster.
– Ronak Shah
Mar 21 at 1:40
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
add a comment |
yep. I thinkfixed = TRUEmakes it a lot faster.
– Ronak Shah
Mar 21 at 1:40
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
yep. I think
fixed = TRUE makes it a lot faster.– Ronak Shah
Mar 21 at 1:40
yep. I think
fixed = TRUE makes it a lot faster.– Ronak Shah
Mar 21 at 1:40
1
1
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
Very interesting, actually 4, with fixed=TRUE is fastest, changed the benchmark to show this.
– Hector Haffenden
Mar 21 at 1:43
add a comment |
0
Another one:
dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))
gives
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10" "01/09/2019 10"
[6] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
add a comment |
0
Another one:
dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))
gives
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10" "01/09/2019 10"
[6] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
add a comment |
0
0
0
Another one:
dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))
gives
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10" "01/09/2019 10"
[6] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
Another one:
dates <- c("01/09/2019 9:51:03", "01/09/2019 9:51:39", "01/09/2019 9:57:04",
"01/09/2019 10:01:41", "01/09/2019 10:06:06", "01/09/2019 10:09:36",
"01/09/2019 10:11:55", "01/09/2019 10:21:15", "01/09/2019 10:21:39",
"01/09/2019 10:52:20")
unlist(lapply(dates,function(x) strsplit(x,":")[[1]][1]))
gives
[1] "01/09/2019 9" "01/09/2019 9" "01/09/2019 9" "01/09/2019 10" "01/09/2019 10"
[6] "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10" "01/09/2019 10"
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
answered Mar 21 at 1:16
CIAndrewsCIAndrews
55849
55849
add a comment |
add a comment |
0
Here is another one using gsub
Capture the pattern by () and \1 to refer to the captured group, need ?to make regex lazy as there are multiple :.
gsub("(.*?):.*", "\1", dates)
answered Mar 21 at 1:56
MKaMKa
358412
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
add a comment |
0
Here is another one using gsub
Capture the pattern by () and \1 to refer to the captured group, need ?to make regex lazy as there are multiple :.
gsub("(.*?):.*", "\1", dates)
answered Mar 21 at 1:56
MKaMKa
358412
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
add a comment |
0
0
0
Here is another one using gsub
Capture the pattern by () and \1 to refer to the captured group, need ?to make regex lazy as there are multiple :.
gsub("(.*?):.*", "\1", dates)
answered Mar 21 at 1:56
MKaMKa
358412
Here is another one using gsub
Capture the pattern by () and \1 to refer to the captured group, need ?to make regex lazy as there are multiple :.
gsub("(.*?):.*", "\1", dates)
answered Mar 21 at 1:56
MKaMKa
358412
answered Mar 21 at 1:56
MKaMKa
358412
answered Mar 21 at 1:56
MKaMKa
358412
answered Mar 21 at 1:56
MKaMKa
358412
358412
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
add a comment |
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
Added to my benchmarking answer, hope thats okay.
– Hector Haffenden
Mar 21 at 2:17
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
All good, its great to see those benchmarks.
– MKa
Mar 21 at 2:55
add a comment |
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Interesting accepted answer, what is your definition of efficient?
– Hector Haffenden
Mar 22 at 12:53
I'm pretty biased towards
tidyversepackages :)– JasonBaik
Mar 22 at 14:40
ah I see, I don’t blame you :) next time you should probably put it in the question...
– Hector Haffenden
Mar 24 at 19:07