Split Character Columns and Get Names of Field in String

split character columns and get names of field in string

Using regex and the stringi packages:

setDT(myDT) # After creating data.table from structure()

library(stringi)

fields <- unique(unlist(stri_extract_all(regex = "[a-z]+(?==)", myDT$info)))
patterns <- sprintf("(?<=%s=)[^;]+", fields)
myDT[, (fields) := lapply(patterns, function(x) stri_extract(regex = x, info))]
myDT[, !"info"]

    chr  pos type end
1: chr1 <NA>    3   4
2: chr2 <NA> <NA>   6
3: chr4 TRUE    2   5

Edit: To get the correct type it seems (?) type.convert() can be used:

myDT[, (fields) := lapply(patterns, function(x) type.convert(stri_extract(regex = x, info), as.is = TRUE))]

R split column names with different occurrences of delimiter into strings and assign unique strings/string counts to a new dataframe

I think if you split at the "underscore, digit, underscore" it provides a solution to your statement above. This does eliminate the digit and the associated information. Does this matter?

names <- c("strainA_1_batch1", "strainA_2_batch2", "strainB_1_batch1", "strainC_1_batch2", "strainC_2_batch2", 
           "strainD_a_1_batch1", "strainD_b_1_batch1")

#split at the underscore, digit and underscore 
splitList <- strsplit(names, "_\\d_")

#convert to dataframe
df <-data.frame(t(as.data.frame.list(splitList)))

#clean up data.frame
rownames(df)<-NULL
names(df)<-c("Strain", "Batch")
df

#report
table(df$Strain)
table(df$Batch)

Another option is to replace the underscore on either side of the digit with a " " (or other character) and then split on the space.

names<-gsub("_(\\d)_", " \\1 ", names)

How to split a character column into multiple columns in R

You can get what you want with gsub:

gsub("^.* +- +([A-Za-z ]+) \\(.*$", "\\1", df$District)
[1] "North West" "North West" "North West" "North West" "North West" "North West"

The first argument to gsub ("^.* +- +([A-Za-z ]+) \(.*$") is a regular expression. It can be interpreted as follows:

From the the beginning of the string "^", match any characters ".*" followed by at least one space, a hyphen, and at least one space " +- +". Then capture the next text "()" that is made up of (at least one) letters and spaces "[A-Za-z ]+". Stop capturing when you reach a space followed by a parenthesis " \\(", then match everything until the end of the text ".*$".

The second argument of gsub, "\\1" says replace the text with the text that was captured by the parentheses.

To assign it to a variable:

df$name <- gsub("^.* +- +([A-Za-z ]+) \\(.*$", "\\1", df$District)

Split data frame string column into multiple columns

Use stringr::str_split_fixed

library(stringr)
str_split_fixed(before$type, "_and_", 2)

How to make a row the column names and split up a string into multiple rows

You could probably use this -

df = df[, c(1, 5)]

## Split on comma and add to dataframe
tmp = strsplit(df$molecules, ",")
df = cbind(df[, -2], do.call(rbind, tmp))

## Transpose the dataframe
df = t(df)
rownames(df) = NULL

How to split a column into multiple (non equal) columns in R

We could use cSplit from splitstackshape

library(splitstackshape)
cSplit(DF, "Col1",",")

-output

cSplit(DF, "Col1",",")
   Col1_1 Col1_2 Col1_3 Col1_4
1:      a      b      c   <NA>
2:      a      b   <NA>   <NA>
3:      a      b      c      d

Split an string by number of characters in a column of a data frame to create multiple columns in R?

We can use separate

library(tidyr)
separate(df, ID, into = c("Spl_1", "Spl_2"), sep = 4, remove = FALSE)
#           ID Spl_1  Spl_2 Var1 Var2
#1  0334KLM001  0334 KLM001   aa   xx
#2  1334HDM002  1334 HDM002  zvv   rr
#3  2334WEM003  2334 WEM003 qetr  qwe
#4  3334OKT004  3334 OKT004   ff  sdf
#5  4334WER005  4334 WER005   ee  sdf
#6  5334BBC006  5334 BBC006  qly  ssg
#7  6334QQQ007  6334 QQQ007   kk  htj
#8  7334AAA008  7334 AAA008   uu  yjy
#9  8334CBU009  8334 CBU009   ww wttt
#10 9334MLO010  9334 MLO010   aa   dg

If we want 3 columns, we can pass a vector in sep

separate(df, ID, into = c("Spl_1", "Spl_2", "Spl_3"), sep = c(4,8), remove = FALSE)
#           ID Spl_1 Spl_2 Spl_3 Var1 Var2
#1  0334KLM001  0334  KLM0    01   aa   xx
#2  1334HDM002  1334  HDM0    02  zvv   rr
#3  2334WEM003  2334  WEM0    03 qetr  qwe
#4  3334OKT004  3334  OKT0    04   ff  sdf
#5  4334WER005  4334  WER0    05   ee  sdf
#6  5334BBC006  5334  BBC0    06  qly  ssg
#7  6334QQQ007  6334  QQQ0    07   kk  htj
#8  7334AAA008  7334  AAA0    08   uu  yjy
#9  8334CBU009  8334  CBU0    09   ww wttt
#10 9334MLO010  9334  MLO0    10   aa   dg

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If the numbers at the beginning are not of fixed length, use extract

extract(df, ID, into = c("Spl_1", "Spl_2"), "^([0-9]+)(.*)", remove = FALSE)

and for 3 columns,

extract(df, ID, into = c("Spl_1", "Spl_2", "Spl_3"), "(.{4})(.{4})(.*)", remove = FALSE)

How to split a string into multiple columns by a given pattern?

If the strings are always in that same format, the following regular expression should work well:

library(stringr)
x <- "\r\n    \r\n        How to get a confirm ticket?\r\n        \r\n            I want to get a tatkal ticket confirm ..."
str_split(x, "(\r\n\\s*)+", simplify = TRUE)[, -1, drop = FALSE]
     [,1]                           [,2]                                       
[1,] "How to get a confirm ticket?" "I want to get a tatkal ticket confirm ..."

If your data actually comes from a table in a text file or from a web page, there are probably more convenient options.

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