I have a txt file with this data in it:
1 message («random_choice»)[5];
2 reply («принято»)[2][3];
3 regulate («random_choice»)[5];
4 Early reg («for instance»)[2][3][4];
4xx: Success (загрузка):
6 OK («fine»)[2][3];
I want to turn it into dataframe, consisting of three columns ID, message, comment.
I also want to remove unnecessary numbers at the end in square brackets.
And also some values in ID column have strings (usually xx). In these cases, column must be just empty.
So, desired result must look like this:
ID Message Comment
1 message random_choice
2 reply принято
3 regulate random_choice
4 Early reg for instance
Success загрузка
6 OK fine
How could i do that? Even when i try to read this txt file i get strange error:
df <- read.table("data_received.txt", header = TRUE)
error i get:
Error in read.table("data_received.txt", header = TRUE) :
more columns than column names
You can use strcapture for this.
Fake data, you'll likely do txt <- readLines("data_received.txt"). (Since my locale on windows is not being friendly to those strings, I'll replace with straight ascii, assuming it'll work just fine on your system.)
txt <- readLines(textConnection("1 message («random_choice»)[5];
# 2 reply («asdf»)[2][3];
# 3 regulate («random_choice»)[5];
# 4 Early reg («for instance»)[2][3][4];
# 4xx: Success (something):
# 6 OK («fine»)[2][3];"))
The breakout:
out <- strcapture("^(\\S+)\\s+([^(]+)\\s+\\((.*)\\).*$", txt,
proto = data.frame(ID=0L, Message="", Comment=""))
# Warning in fun(mat[, i]) : NAs introduced by coercion
out
# ID Message Comment
# 1 1 message «random_choice»
# 2 2 reply «asdf»
# 3 3 regulate «random_choice»
# 4 4 Early reg «for instance»
# 5 NA Success something
# 6 6 OK «fine»
The proto= argument indicates what type of columns are generated. Since I set the ID=0L, it assumes it'll be integer, so anything that does not convert to integer becomes NA (which satisfies your fifth row omission).
Explanation on the regex:
in general:
* means zero-or-more of the previous character (or character class)
+ means one-or-more
? (not used, but useful nonetheless) means zero or one
^ and $ mean the beginning and end of the string, respectively (a ^ within [..] is different)
(...) is a capture group: anything within the non-escaped parens is stored, anything not is discarded
[...] is a character group, any of the characters is a match; if this is instead [^..], then it is inverted: anything except what is listed
[[...]] is a character class
^(\\S+), start with (^) one or more (+) non-space characters (\\S);
\\s+ one or more space character (\\s) (discarded);
([^(]+) one or more character that is not a left-paren;
\\((.*)\\)$ a literal left-paren (\\() and then zero or more of anything (.*), all the way to a literal right-paren (\\)) and the end of the string ($).
It should be noted that \\s and \\S are non-POSIX regex characters, where it is generally suggested to use [^[:space:]] for \\S (no space chars) and [[:space:]] for \\s. Those are equivalent but I went with code-golf initially. With this replacement, it looks like
out <- strcapture("^([^[:space:]]+)[[:space:]]+([^(]+)[[:space:]]+\\((.*)\\).*$", txt,
proto = data.frame(ID=0L, Message="", Comment=""))
We can use {unglue}. Here we see you have two patterns, one contains "«" and ID, the other doesn't. {unglue} will use the first pattern that matches. any {foo} or {} expression matches the regex ".*?", and a data.frame is built from the names put between brackets.
txt <- c(
"1 message («random_choice»)[5];", "2 reply («asdf»)[2][3];",
"3 regulate («random_choice»)[5];", "4 Early reg («for instance»)[2][3][4];",
"4xx: Success (something):", "6 OK («fine»)[2][3];")
library(unglue)
patterns <-
c("{id} {Message} («{Comment}»){}",
"{} {Message} ({Comment}){}")
unglue_data(txt, patterns)
#> id Message Comment
#> 1 1 message random_choice
#> 2 2 reply asdf
#> 3 3 regulate random_choice
#> 4 4 Early reg for instance
#> 5 <NA> Success something
#> 6 6 OK fine
I am working in R for the first time and I have been having difficulty renaming column names in a dataframe (Grade.Data). I have a dataset imported from an csv file that has column names like this:
Student.ID
Grade
Interactive.Exercises.1..Health
Interactive.Exercises.2..Fitness
Quizzes.1..Week.1.Quiz
Quizzes.2..Week.2.Quiz
Case.Studies.1..Case.Study1
Case.Studies.2..Case.Study2
I would like to be able to change the variable names so that they are more simple, i.e. from Interactive.Exercises.1.Health to Interactive.Exercises.1 or Quizzes.1.Week.1.Quiz to Quizzes.1
So far, I have tried this:
grep(".*[0-9]", names(Grade.Data))
But I get this returned:
[1] 3 4 5 6 7 8 9 11 12 13 14 15 16 17 19 20 21 22 23 24 25
Can anyone help me figure out what is going on, and write a better regex expression? Thank you so much.
It seems you truncate column names after the first chunk of digits.
You may use the following sub solution:
names(Grade.Data) <- sub("^(.*?\\d+).*$", "\\1", names(Grade.Data))
See the regex demo
Details
^ - start of string
(.*?\\d+) - Group 1 (later referred with \1 from the replacement pattern) matching any 0+ chars as few as possible (.*?) and then 1 or more digits (\d+)
.* - any 0+ chars as many as possible
$ - end of string
There is nothing wrong with your regex itself. What you are looking for is probably the combination of regexpr - which gets the start and ending of your regex- and regmatches - which gets the actual string corresponding to the output of regexpr:
start_end <- regexpr(".*[0-9]", names(Grade.data))
regmatches(names(Grade.data), start_end)
# [1] "Interactive.Exercises.1" "Interactive.Exercises.2"
# [3] "Quizzes.1..Week.1" "Quizzes.2..Week.2"
# [5] "Case.Studies.1..Case.Study1"
Adding a question-mark behind the dot-star will make the regex match as few characters as possible, so it will stop after the first numeric value:
start_end <- regexpr(".*?[0-9]", names(Grade.data))
regmatches(names(Grade.data), start_end)
# [1] "Interactive.Exercises.1" "Interactive.Exercises.2"
# [3] "Quizzes.1" "Quizzes.2"
# [5] "Case.Studies.1"
you should use the function names, following I write a little example, the names string can be as long as you need.
names(x = Grade.Data) <- c("Col1_name", "Col2_name")
I've got some problems deleting duplicate elements in a string.
My data look similar to this:
idvisit path
1 1,16,23,59
2 2,14,14,19
3 5,19,23,19
4 10,10
5 23,23,27,29,23
I have a column containing an unique ID and a column containing a path for web page navigation.
The right column contains some cases, where pages just were reloaded and the page were tracked twice or even more.
The pages are separated with commas and are saved as factors.
My problem is, that I don't want to have multiple pages in a row, so the data should look like this.
idvisit path
1 1,16,23,59
2 2,14,19
3 5,19,23,19
4 10
5 23,27,29,23
The multiple pages next to each other should be removed. I know how to delete a specific multiple number using regexpressions, but I have about 20.000 different pages and can't do this for all of them.
Does anyone have a solution or a hint, for my problem?
Thanks
Sebastian
We can use tidyverse. Use the separate_rows to split the 'path' variable by the delimiter (,) to convert to a long format, then grouped by 'idvisit', we paste the run-length-encoding values
library(tidyverse)
separate_rows(df1, path) %>%
group_by(idvisit) %>%
summarise(path = paste(rle(path)$values, collapse=","))
# A tibble: 5 × 2
# idvisit path
# <int> <chr>
#1 1 1,16,23,59
#2 2 2,14,19
#3 3 5,19,23,19
#4 4 10
#5 5 23,27,29,23
Or a base R option is
df1$path <- sapply(strsplit(df1$path, ","), function(x) paste(rle(x)$values, collapse=","))
NOTE: If the 'path' column is factor class, convert to character before passing as argument to strsplit i.e. strsplit(as.character(df1$path), ",")
Using stringr package, with function: str_replace_all, I think it gets what you want using the following regular expression: ([0-9]+),\\1and then replace it with \\1 (we need to scape the \ special character):
library(stringr)
> str_replace_all("5,19,23,19", "([0-9]+),\\1", "\\1")
[1] "5,19,23,19"
> str_replace_all("10,10", "([0-9]+),\\1", "\\1")
[1] "10"
> str_replace_all("2,14,14,19", "([0-9]+),\\1", "\\1")
[1] "2,14,19"
You can use it in a array form: x <- c("5,19,23,19", "10,10", "2,14,14,19") then:
str_replace_all(x, "([0-9]+),\\1", "\\1")
[1] "5,19,23,19" "10" "2,14,19"
or using sapply:
result <- sapply(x, function(x) str_replace_all(x, "([0-9]+),\\1", "\\1"))
Then:
> result
5,19,23,19 10,10 2,14,14,19
"5,19,23,19" "10" "2,14,19"
Notes:
The first line is the attribute information:
> str(result)
Named chr [1:3] "5,19,23,19" "10" "2,14,19"
- attr(*, "names")= chr [1:3] "5,19,23,19" "10,10" "2,14,14,19"
If you don't want to see them (it does not affect the result), just do:
attributes(result) <- NULL
Then,
> result
[1] "5,19,23,19" "10" "2,14,19"
Explanation about the regular expression used: ([0-9]+),\\1
([0-9]+): Starts with a group 1 delimited by () and finds any digit (at least one)
,: Then comes a punctuation sign: , (we can include spaces here, but the original example only uses this character as delimiter)
\\1: Then comes an identical string to the group 1, i.e.: the repeated number. If that doesn't happen, then the pattern doesn't match.
Then if the pattern matches, it replaces it, with the value of the variable \\1, i.e. the first time the number appears in the pattern matched.
How to handle more than one duplicated number, for example 2,14,14,14,19?:
Just use this regular expression instead: ([0-9]+)(,\\1)+, then it matches when at least there is one repetition of the delimiter (right) and the number. You can try other possibilities using this regex101.com (in MHO it more user friendly than other online regular expression checkers).
I hope this would work for you, it is a flexible solution, you just need to adapt it with the pattern you need.
This is an example of my data.frame
no string
1 abc&URL_drf
2 abcdef&URL_efg
I need to replace word *&URL with "". So, I need a this result
no string
1 _drf
2 _efg
In case of Excel, I can easily make this result using '*&URL' in 'find and replace' function.
However, I cannot look for effective method in R.
In R, my approach is below.
First, I have split string using strsplit(df$string, "&URL") and then I have selected second column. I think that it is not a effective way.
Is there a any effective method?
# data
df <- read.table(text="no string
1 abc&URL_drf
2 abcdef&URL_efg", header=T, as.is=T)
# `gsub` function is to substitute the unwanted string with nothing,
# thus the `""`. The pattern of unwanted string was written in
# regular expressions.
df$string <- gsub("[a-z]+(&URL)", "", df$string)
# you get
no string
1 1 _drf
2 2 _efg
I suggest you use the grep function .
The grep function takes your regex as the first argument, and the input vector as the second argument.If you pass value=TRUE, then grep returns a vector with copies of the actual elements in the input vector that could be (partially) matched.
so in your case
grep("[a-z]+(&URL)", df$col, perl=TRUE, value=TRUE)
Another approach:
df <- transform(df, string = sub(".*&URL", "", string))
# no string
# 1 1 _drf
# 2 2 _efg
I want to extract alphanumeric characters from a partiular sentence in R.
I have tried the following:
aa=grep("[:alnum:]","abc")
.This should return integer(0),but it returns 1,which should not be the case as "abc" is not an alphanumeric.
What am I missing here?
Essentially I am looking for a function,that only searches for characters that are combinations of both alphabets and numbers,example:"ABC-0112","PCS12SCH"
Thanks in advance for your help.
[[:alnum:]] matches alphabets or digits. To match the string which contains the both then you should use,
x <- c("ABC", "ABc12", "--A-1", "abc--", "89=A")
grep("(.*[[:alpha:]].*[[:digit:]]|.*[[:digit:]].*[[:alpha:]])", x)
# [1] 2 3 5
or
which(grepl("[[:alpha:]]", x) & grepl("[[:digit:]]", x))
# [1] 2 3 5