Parse Factor in R

Parse Factor in R

By : Babu110
Date : November 22 2020, 04:01 AM
help you fix your problem I have a column of ID data (class factor) in the following format: 01-001 etc. , Looks like you want substr:
code :
substr(c("01-001", "12-121"), 0, 2)
# [1] "01" "12"
as.numeric(substr(c("01-001", "12-121"), 0, 2))
# [1]  1 12

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How to parse time factor regular expression

How to parse time factor regular expression

By : Maxwell
Date : March 29 2020, 07:55 AM
Any of those help Aside from the misuse of language pointed out by others, instead of this...
code :
var parts = currentVal.split(" ");
var parts = currentVal.split(":");
var parts = currentVal.split(/[ :]/);
parse out string, set it as a factor column in R data.table

parse out string, set it as a factor column in R data.table

By : Santino Guerrieri
Date : March 29 2020, 07:55 AM
I hope this helps you . You can vectorize this with regular expressions, e.g. by using gsub():
Set up the data:
code :
strings <- c("lorem pear ipsum", "apple ipsum lorem", "lorem ipsum plum")
fruit <- c("pear", "apple", "plum")
ptn <- paste0(".*(", paste(fruit, collapse="|"), ").*")
gsub(ptn, "\\1", strings)
[1] "pear"  "apple" "plum" 
[1] ".*(pear|apple|plum).*"
DT <- data.table(name=strings, value=c(4, 2, 6))
DT[, factor:=gsub(ptn, "\\1", strings)]

                name value factor
1:  lorem pear ipsum     4   pear
2: apple ipsum lorem     2  apple
3:  lorem ipsum plum     6   plum
How to parse factor/dataframe with Rembedded

How to parse factor/dataframe with Rembedded

By : Ahmed Mansour
Date : March 29 2020, 07:55 AM
will help you If you're converting a factor to character, you want Rf_asCharacterFactor.
split dataframe by factor and name new df by the factor and addidtional description like "new_dataframe(factor)&quo

split dataframe by factor and name new df by the factor and addidtional description like "new_dataframe(factor)&quo

By : user2182087
Date : November 12 2020, 04:01 AM
hope this fix your issue cluster_list <- split(df, f = df$clust)
builds the cluster list, to manipulate the list use: lapply.
Parse error when writing a factor function in Haskell

Parse error when writing a factor function in Haskell

By : Hotel Appartments
Date : March 29 2020, 07:55 AM
Any of those help Because you're using mod function as an infix operator, you need to surround it in backquotes. The operator to compare equality is not = but == . The error free rewrite of your code is:
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