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Extracting Dates Using Regular Expression in R using grepl


Extracting Dates Using Regular Expression in R using grepl

By : VDOGamer
Date : November 22 2020, 10:31 AM
I wish did fix the issue. Regex101
This is a simple mistake:
code :
mystring = c("12-03-99", "A", "B")
date = grepl("[0-9]{2}-[0-9]{2}-[0-9]{2}", mystring)
date = grepl("\d{2}-\d{2}-\d{2}", mystring)
[0-9]{2}-[0-9]{2}-[0-9]{2}
[0-9]?[1-9]-[0-9]?[1-9]-[0-9]?[1-9]
\d?[1-9]-\d?[1-9]-\d?[1-9]
(\d?[1-9]-){2}\d?[1-9]


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Using grepl, searching for text with a leading paren, regular expression?

Using grepl, searching for text with a leading paren, regular expression?


By : A.Chap
Date : March 29 2020, 07:55 AM
it fixes the issue You need two backlashes to create a backslash in an R character. The reason you need (\\) is because it is interpreted twice, first as a character string and then as a regular expression.
code :
grepl('\\(3D', DVDTitles$PROJECT.NAME)
Regular expression parsed with grepl replacement

Regular expression parsed with grepl replacement


By : William Anfin
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Simple,,,, Just assign the string cat to the match elements. This will replace all the chars present in the element with cat
code :
> data <- c("cat 6kg","cat g250", "cat dog","cat 10 kg") 
> data[grepl("(^cat.[a-z][0-9])|(^cat.[0-9])",data)] <- "cat"
> data
[1] "cat"     "cat"     "cat dog" "cat" 
> data <- c("cat 6kg","cat g250", "cat dog","cat 10 kg") 
> data[grepl("^cat.[a-z]?[0-9]",data)] <- "cat"
> data
[1] "cat"     "cat"     "cat dog" "cat" 
What is the correct way to extract multiple required patterns from a column in the data using regular expression (grepl

What is the correct way to extract multiple required patterns from a column in the data using regular expression (grepl


By : Anchal Gupta
Date : March 29 2020, 07:55 AM
Any of those help I have a dataframe (can also use the CSV file which I generated) that has 7 columns and close to a million rows. , Here is a very crude method using dplyr and stringr.
code :
library(dplyr)
library(stringr)

df %>% # this assumes your data is in a data frame called 'df'
  mutate(small_id = str_extract(SOURCE, 'SMALL "[0-9]+"')) %>% 
  mutate(pubchem_id = str_extract(SOURCE, 'PUBCHEM "[0-9]+"')) %>% 
  mutate(spider_id = str_extract(SOURCE, 'SPIDER "[0-9]+"')) %>% 
  mutate(chebi_id = str_extract(SOURCE, 'CHEBI "[0-9]+"')) %>% 
  mutate(bas_id = str_extract(SOURCE, 'BAS "[0-9]+-[0-9]+-[0-9]+"')) %>% 
  mutate_at(vars(ends_with('_id')), ~gsub('[A-Z]|"|\\s+', '', .)) %>% 
  select(-SOURCE)

  `UNIQUE-NAME` TYPES     `COMMON-NAME`          FORMULA                          WEIGHT `SECONDARY-WEIGHT` small_id pubchem_id spider_id chebi_id bas_id    
  <chr>         <chr>     <chr>                  <chr>                             <dbl>              <dbl> <chr>    <chr>      <chr>     <chr>    <chr>     
1 CPD-12676     Compounds glycerophosphoglycerol (C 19);(H 29);(N 5);(O 10);(S 2)   29.8               27.6 35022    54758713   3315400   28643    55672-34-1
2 CPD-380       Enamines  UDP                    NA                                294.               287.  NA       NA         NA        NA       NA        
3 NAD           Steroids  prephenate             (O 4);(S 1)                       373.               381.  NA       87361293   NA        87100    91234-28-2
4 ADP           Rings     pyridine               (C 5);(H 5);(N 1)                  39.3               40.6 98234    NA         2311345   NA       NA        
5 CAD-392       Molecules pyrine                 (C 10);(H 9)                      393.               401.  NA       NA         5454678   NA       NA        
6 CAD-355       Groups    solution               (C 12);(H 12)                      32.5               40.7 NA       NA         NA        NA       NA        
7 CPD-39234     Compounds glycerophosphoglycerol (C 19);(H 29);(N 5);(O 10);(S 2)   92.4              101.  45465    NA         NA        33490    NA 
I get an error when I use grepl with paste. Invalid regular expression

I get an error when I use grepl with paste. Invalid regular expression


By : user2888149
Date : March 29 2020, 07:55 AM
With these it helps i am using the following code to match elements of two column, , You might be able to use %in% rather than grepl, i.e.
code :
test = articles[apply(articles, 1, function(i) any(i %in% dictionary)),]
test = articles[apply(articles, 1, function(i) any(stringr::str_detect(i, fixed(dictionary)))),]
mutiple regular expression using grepl in R

mutiple regular expression using grepl in R


By : user3438448
Date : March 29 2020, 07:55 AM
Hope this helps If I've understood right, you can combine with & and use | in a pattern to act as an OR command e.g.
code :
# not (start and (tmax or tmin or rain))
dat[, !(grepl(pattern="start", colnames(dat)) & 
          grepl(pattern="tmax|tmin|rain", colnames(dat)))] 
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