R - How do I remove a varying number of digits from a date-vector2019 Community Moderator ElectionTidying Time Intervals for Plotting a Histogram in RWhich open-source sgdb for kind of large dataHow can I vectorize this code in R? Maybe with the apply() function?Successful ETL Automation: Libraries, Review papers, Use Casesk-Nearest Neighbours with time series data - how to obtain whole-time-period estimatorshow to remove unwanted characters from dataRunning multiple random forest and combining themR mice doesn't give a 'valid' sollutionMerging dataframes in Pandas is taking a surprisingly long time
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R - How do I remove a varying number of digits from a date-vector
2019 Community Moderator ElectionTidying Time Intervals for Plotting a Histogram in RWhich open-source sgdb for kind of large dataHow can I vectorize this code in R? Maybe with the apply() function?Successful ETL Automation: Libraries, Review papers, Use Casesk-Nearest Neighbours with time series data - how to obtain whole-time-period estimatorshow to remove unwanted characters from dataRunning multiple random forest and combining themR mice doesn't give a 'valid' sollutionMerging dataframes in Pandas is taking a surprisingly long time
$begingroup$
I want to remove a varying number of digits from a date vector. My date vectors looks like this:
I want to convert this vector into a date vector, but first I have to get rid of the number in front of it. This is where I struggle since this number can be (here from 88-106) varys in my data.frame from 1-50 000. Does anyone know how to remove these number without destroying the date? I tried the following, but it didn't work out:
I would really appreciate your help! Thanks!
r bigdata data data-cleaning
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
$endgroup$
add a comment |
$begingroup$
I want to remove a varying number of digits from a date vector. My date vectors looks like this:
I want to convert this vector into a date vector, but first I have to get rid of the number in front of it. This is where I struggle since this number can be (here from 88-106) varys in my data.frame from 1-50 000. Does anyone know how to remove these number without destroying the date? I tried the following, but it didn't work out:
I would really appreciate your help! Thanks!
r bigdata data data-cleaning
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
$endgroup$
add a comment |
$begingroup$
I want to remove a varying number of digits from a date vector. My date vectors looks like this:
I want to convert this vector into a date vector, but first I have to get rid of the number in front of it. This is where I struggle since this number can be (here from 88-106) varys in my data.frame from 1-50 000. Does anyone know how to remove these number without destroying the date? I tried the following, but it didn't work out:
I would really appreciate your help! Thanks!
r bigdata data data-cleaning
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
$endgroup$
I want to remove a varying number of digits from a date vector. My date vectors looks like this:
I want to convert this vector into a date vector, but first I have to get rid of the number in front of it. This is where I struggle since this number can be (here from 88-106) varys in my data.frame from 1-50 000. Does anyone know how to remove these number without destroying the date? I tried the following, but it didn't work out:
I would really appreciate your help! Thanks!
r bigdata data data-cleaning
r bigdata data data-cleaning
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
asked Jan 3 at 15:16
Yannik SuhreYannik Suhre
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
assuming your strings are of the form [NUMBER][WHITESPACE][STUFF-TO-KEEP]
you can use regex, for example with the stringr package.
# construct strings
strings <- paste(c(1,11,111,1111), "KEEP")
strings
[1] "1 KEEP" "11 KEEP" "111 KEEP" "1111 KEEP"
stringr::str_remove(strings, "[0-9].* (?=[a-zA-Z])")
[1] "KEEP" "KEEP" "KEEP" "KEEP"
the regex "[0-9].* (?=[a-zA-Z])" expained:
'[0-9].* ' # (including the whitespace!)
# matches any number of digigs from 0 to 9 followed by a whitespace
(?=[a-zA-Z])
# this is a lookahead group, so an additional requirement is,
# that a match is only used when the above match is followed
# by a letter in the alphabet, small(a-z) or large (A-Z)
of course, str_remove removes the matches string.
answered Feb 19 at 15:47
JaggeJagge
1
$endgroup$
add a comment |
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$begingroup$
assuming your strings are of the form [NUMBER][WHITESPACE][STUFF-TO-KEEP]
you can use regex, for example with the stringr package.
# construct strings
strings <- paste(c(1,11,111,1111), "KEEP")
strings
[1] "1 KEEP" "11 KEEP" "111 KEEP" "1111 KEEP"
stringr::str_remove(strings, "[0-9].* (?=[a-zA-Z])")
[1] "KEEP" "KEEP" "KEEP" "KEEP"
the regex "[0-9].* (?=[a-zA-Z])" expained:
'[0-9].* ' # (including the whitespace!)
# matches any number of digigs from 0 to 9 followed by a whitespace
(?=[a-zA-Z])
# this is a lookahead group, so an additional requirement is,
# that a match is only used when the above match is followed
# by a letter in the alphabet, small(a-z) or large (A-Z)
of course, str_remove removes the matches string.
answered Feb 19 at 15:47
JaggeJagge
1
$endgroup$
add a comment |
$begingroup$
assuming your strings are of the form [NUMBER][WHITESPACE][STUFF-TO-KEEP]
you can use regex, for example with the stringr package.
# construct strings
strings <- paste(c(1,11,111,1111), "KEEP")
strings
[1] "1 KEEP" "11 KEEP" "111 KEEP" "1111 KEEP"
stringr::str_remove(strings, "[0-9].* (?=[a-zA-Z])")
[1] "KEEP" "KEEP" "KEEP" "KEEP"
the regex "[0-9].* (?=[a-zA-Z])" expained:
'[0-9].* ' # (including the whitespace!)
# matches any number of digigs from 0 to 9 followed by a whitespace
(?=[a-zA-Z])
# this is a lookahead group, so an additional requirement is,
# that a match is only used when the above match is followed
# by a letter in the alphabet, small(a-z) or large (A-Z)
of course, str_remove removes the matches string.
answered Feb 19 at 15:47
JaggeJagge
1
$endgroup$
add a comment |
$begingroup$
assuming your strings are of the form [NUMBER][WHITESPACE][STUFF-TO-KEEP]
you can use regex, for example with the stringr package.
# construct strings
strings <- paste(c(1,11,111,1111), "KEEP")
strings
[1] "1 KEEP" "11 KEEP" "111 KEEP" "1111 KEEP"
stringr::str_remove(strings, "[0-9].* (?=[a-zA-Z])")
[1] "KEEP" "KEEP" "KEEP" "KEEP"
the regex "[0-9].* (?=[a-zA-Z])" expained:
'[0-9].* ' # (including the whitespace!)
# matches any number of digigs from 0 to 9 followed by a whitespace
(?=[a-zA-Z])
# this is a lookahead group, so an additional requirement is,
# that a match is only used when the above match is followed
# by a letter in the alphabet, small(a-z) or large (A-Z)
of course, str_remove removes the matches string.
answered Feb 19 at 15:47
JaggeJagge
1
$endgroup$
assuming your strings are of the form [NUMBER][WHITESPACE][STUFF-TO-KEEP]
you can use regex, for example with the stringr package.
# construct strings
strings <- paste(c(1,11,111,1111), "KEEP")
strings
[1] "1 KEEP" "11 KEEP" "111 KEEP" "1111 KEEP"
stringr::str_remove(strings, "[0-9].* (?=[a-zA-Z])")
[1] "KEEP" "KEEP" "KEEP" "KEEP"
the regex "[0-9].* (?=[a-zA-Z])" expained:
'[0-9].* ' # (including the whitespace!)
# matches any number of digigs from 0 to 9 followed by a whitespace
(?=[a-zA-Z])
# this is a lookahead group, so an additional requirement is,
# that a match is only used when the above match is followed
# by a letter in the alphabet, small(a-z) or large (A-Z)
of course, str_remove removes the matches string.
answered Feb 19 at 15:47
JaggeJagge
1
answered Feb 19 at 15:47
JaggeJagge
1
answered Feb 19 at 15:47
JaggeJagge
1
answered Feb 19 at 15:47
JaggeJagge
1
1
add a comment |
add a comment |
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