R Extract time from timestamp then save as a new column
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Please can someone advise where i am going wrong. I have the code below to extract hour, minutes, and seconds. It doesn't work. I am getting an error message:
Error in UseMethod("mutate_") : no applicable method for 'mutate_'
applied to an object of class "function"
lubridate::ymd_hms(as.character(d1$timestamp, tz = "UTC"))
df %>% mutate(hours=hour(strptime(d1$timestamp, %H:%M')) %>%
as.character() )
r
edited Jan 4 at 15:22
nick
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
add a comment |
Please can someone advise where i am going wrong. I have the code below to extract hour, minutes, and seconds. It doesn't work. I am getting an error message:
Error in UseMethod("mutate_") : no applicable method for 'mutate_'
applied to an object of class "function"
lubridate::ymd_hms(as.character(d1$timestamp, tz = "UTC"))
df %>% mutate(hours=hour(strptime(d1$timestamp, %H:%M')) %>%
as.character() )
r
edited Jan 4 at 15:22
nick
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
can you please show an example of the actual data?
– nick
Jan 4 at 15:12
add a comment |
Please can someone advise where i am going wrong. I have the code below to extract hour, minutes, and seconds. It doesn't work. I am getting an error message:
Error in UseMethod("mutate_") : no applicable method for 'mutate_'
applied to an object of class "function"
lubridate::ymd_hms(as.character(d1$timestamp, tz = "UTC"))
df %>% mutate(hours=hour(strptime(d1$timestamp, %H:%M')) %>%
as.character() )
r
edited Jan 4 at 15:22
nick
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
Please can someone advise where i am going wrong. I have the code below to extract hour, minutes, and seconds. It doesn't work. I am getting an error message:
Error in UseMethod("mutate_") : no applicable method for 'mutate_'
applied to an object of class "function"
lubridate::ymd_hms(as.character(d1$timestamp, tz = "UTC"))
df %>% mutate(hours=hour(strptime(d1$timestamp, %H:%M')) %>%
as.character() )
r
r
edited Jan 4 at 15:22
nick
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
edited Jan 4 at 15:22
nick
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
edited Jan 4 at 15:22
nick
6881619
edited Jan 4 at 15:22
nick
6881619
edited Jan 4 at 15:22
nick
6881619
6881619
asked Jan 4 at 15:02
user5576922user5576922
114
asked Jan 4 at 15:02
user5576922user5576922
114
asked Jan 4 at 15:02
user5576922user5576922
114
114
can you please show an example of the actual data?
– nick
Jan 4 at 15:12
add a comment |
can you please show an example of the actual data?
– nick
Jan 4 at 15:12
can you please show an example of the actual data?
– nick
Jan 4 at 15:12
can you please show an example of the actual data?
– nick
Jan 4 at 15:12
add a comment |
2 Answers
2
active
oldest
votes
Maybe this can help:
# some fake data
df <- data.frame(a = 1, timestamp = '2018-11-11 12:13:14')
# convert as data
library(lubridate)
df$timestamp <- ymd_hms(as.character(df$timestamp, tz = "UTC"))
# add the hour:minutes column
library(dplyr)
df %>% mutate(hour_ = format(df$timestamp,'%H:%M'))
a timestamp hour_
1 1 2018-11-11 12:13:14 12:13
Or in only one dplyr chain:
df %>%
mutate(hour_ = format(ymd_hms(as.character(df$timestamp, tz = "UTC")),'%H:%M'))
answered Jan 4 at 15:15
s_ts_t
3,74121133
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output ofdput(head(d1,5))editing the question? This line is going to print something likestructure(..... They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.
– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
add a comment |
I would check (try this function) if the package using that function is actually loaded - and in the right order.
If you're calling mutate from dplyr try dplyr::mutate instead of just mutate. If that works than it may just be that you have to check if a similar named function from another package is loaded after dplyr and change the order in which you load them if that is a suitable option.
answered Jan 4 at 15:09
Marc0Marc0
1116
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Maybe this can help:
# some fake data
df <- data.frame(a = 1, timestamp = '2018-11-11 12:13:14')
# convert as data
library(lubridate)
df$timestamp <- ymd_hms(as.character(df$timestamp, tz = "UTC"))
# add the hour:minutes column
library(dplyr)
df %>% mutate(hour_ = format(df$timestamp,'%H:%M'))
a timestamp hour_
1 1 2018-11-11 12:13:14 12:13
Or in only one dplyr chain:
df %>%
mutate(hour_ = format(ymd_hms(as.character(df$timestamp, tz = "UTC")),'%H:%M'))
answered Jan 4 at 15:15
s_ts_t
3,74121133
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output ofdput(head(d1,5))editing the question? This line is going to print something likestructure(..... They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.
– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
add a comment |
Maybe this can help:
# some fake data
df <- data.frame(a = 1, timestamp = '2018-11-11 12:13:14')
# convert as data
library(lubridate)
df$timestamp <- ymd_hms(as.character(df$timestamp, tz = "UTC"))
# add the hour:minutes column
library(dplyr)
df %>% mutate(hour_ = format(df$timestamp,'%H:%M'))
a timestamp hour_
1 1 2018-11-11 12:13:14 12:13
Or in only one dplyr chain:
df %>%
mutate(hour_ = format(ymd_hms(as.character(df$timestamp, tz = "UTC")),'%H:%M'))
answered Jan 4 at 15:15
s_ts_t
3,74121133
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output ofdput(head(d1,5))editing the question? This line is going to print something likestructure(..... They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.
– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
add a comment |
Maybe this can help:
# some fake data
df <- data.frame(a = 1, timestamp = '2018-11-11 12:13:14')
# convert as data
library(lubridate)
df$timestamp <- ymd_hms(as.character(df$timestamp, tz = "UTC"))
# add the hour:minutes column
library(dplyr)
df %>% mutate(hour_ = format(df$timestamp,'%H:%M'))
a timestamp hour_
1 1 2018-11-11 12:13:14 12:13
Or in only one dplyr chain:
df %>%
mutate(hour_ = format(ymd_hms(as.character(df$timestamp, tz = "UTC")),'%H:%M'))
answered Jan 4 at 15:15
s_ts_t
3,74121133
Maybe this can help:
# some fake data
df <- data.frame(a = 1, timestamp = '2018-11-11 12:13:14')
# convert as data
library(lubridate)
df$timestamp <- ymd_hms(as.character(df$timestamp, tz = "UTC"))
# add the hour:minutes column
library(dplyr)
df %>% mutate(hour_ = format(df$timestamp,'%H:%M'))
a timestamp hour_
1 1 2018-11-11 12:13:14 12:13
Or in only one dplyr chain:
df %>%
mutate(hour_ = format(ymd_hms(as.character(df$timestamp, tz = "UTC")),'%H:%M'))
answered Jan 4 at 15:15
s_ts_t
3,74121133
answered Jan 4 at 15:15
s_ts_t
3,74121133
answered Jan 4 at 15:15
s_ts_t
3,74121133
answered Jan 4 at 15:15
s_ts_t
3,74121133
3,74121133
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output ofdput(head(d1,5))editing the question? This line is going to print something likestructure(..... They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.
– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
add a comment |
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output ofdput(head(d1,5))editing the question? This line is going to print something likestructure(..... They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.
– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
@user5576922 you are welcome, if you share the data that creates the problem, posting a part of them editing the question, it's going to be easier to help.
– s_t
Jan 4 at 15:22
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
How do i define a data frame instead of a specific date, please?
– user5576922
Jan 4 at 15:26
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
d1 conatins my data as a data frame
– user5576922
Jan 4 at 15:27
Could you share the output of
dput(head(d1,5)) editing the question? This line is going to print something like structure(.... . They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.– s_t
Jan 4 at 15:29
Could you share the output of
dput(head(d1,5)) editing the question? This line is going to print something like structure(.... . They are the first 5 rows of your dataset. Post them editing the question, it's going to be easier and faster to help you having it.– s_t
Jan 4 at 15:29
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
Thanks again for your help. I am sorted now.
– user5576922
Jan 5 at 10:58
add a comment |
I would check (try this function) if the package using that function is actually loaded - and in the right order.
If you're calling mutate from dplyr try dplyr::mutate instead of just mutate. If that works than it may just be that you have to check if a similar named function from another package is loaded after dplyr and change the order in which you load them if that is a suitable option.
answered Jan 4 at 15:09
Marc0Marc0
1116
add a comment |
I would check (try this function) if the package using that function is actually loaded - and in the right order.
If you're calling mutate from dplyr try dplyr::mutate instead of just mutate. If that works than it may just be that you have to check if a similar named function from another package is loaded after dplyr and change the order in which you load them if that is a suitable option.
answered Jan 4 at 15:09
Marc0Marc0
1116
add a comment |
I would check (try this function) if the package using that function is actually loaded - and in the right order.
If you're calling mutate from dplyr try dplyr::mutate instead of just mutate. If that works than it may just be that you have to check if a similar named function from another package is loaded after dplyr and change the order in which you load them if that is a suitable option.
answered Jan 4 at 15:09
Marc0Marc0
1116
I would check (try this function) if the package using that function is actually loaded - and in the right order.
If you're calling mutate from dplyr try dplyr::mutate instead of just mutate. If that works than it may just be that you have to check if a similar named function from another package is loaded after dplyr and change the order in which you load them if that is a suitable option.
answered Jan 4 at 15:09
Marc0Marc0
1116
edited Jan 4 at 15:18
answered Jan 4 at 15:09
Marc0Marc0
1116
answered Jan 4 at 15:09
Marc0Marc0
1116
answered Jan 4 at 15:09
Marc0Marc0
1116
1116
add a comment |
add a comment |
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can you please show an example of the actual data?
– nick
Jan 4 at 15:12