concat a SPLIT variable in R - r

I've been trying to split a string in R and then joining it back together but none of the tricks have worked for what I need.
!!!Important !!! My question is not a duplicate:
saving a split result into a variable and then pasting, collapsing etc is not the same as just paste a vector like this
paste(c("bla", "bla"), collapse = " ")
> paste(c("The","birch", "canoe"), collapse = ' ')
[1] "The birch canoe"
> paste(s, collapse=" ")
[1] "c(\"The\", \"birch\", \"canoe\", \"slid\", \"on\", \"the\", \"smooth\", \"planks.\")"
Here's the code:
I take pre-saved sentences in R
sentences[1]
and split it
s <- str_split(sentences[1])
this is what I get:
[1] "The" "birch" "canoe" "slid" "on" "the" "smooth" "planks."
Now when I try to join this back together I get backslashes
toString(s)
"c(\"The\", \"birch\", \"canoe\", \"slid\", \"on\", \"the\", \"smooth\", \"planks.\")"
paste produces the same result:
> paste(s)
[1] "c(\"The\", \"birch\", \"canoe\", \"slid\", \"on\", \"the\", \"smooth\", \"planks.\")"
I tried using str_split_fixed and wrap it into a vector, but it joins the sentence back together with a comma, even if I ask it not to.
v <- as.vector(str_split_fixed(sentences[1], " ", 5))
toString(v, sep="")
[1] "The, birch, canoe, slid, on the smooth planks."
I thought maybe str_split_i or str_split_1 could solve it as according to the documentation in theory it should, but that's what I get when I try to use it
"could not find function "str_split_1" "
Are there any other ways to join back a string after splitting it without it producing commas or backslashes?..

See the difference between:
s <- list(c("The" , "birch" , "canoe" , "slid" , "on" , "the" , "smooth" , "planks."))
paste(s[1], collapse = " ")
#[1] "c(\"The\", \"birch\", \"canoe\", \"slid\", \"on\", \"the\", \"smooth\", \"planks.\")"
and
paste(s[[1]], collapse = " ")
#[1] "The birch canoe slid on the smooth planks."
This is because [[ will extract the vector, and [ and will keep the output as a list.

Related

Why does paste() concatenate list elements in the wrong order?

Given the following string:
my.str <- "I welcome you my precious dude"
One splits it:
my.splt.str <- strsplit(my.str, " ")
And then concatenates:
paste(my.splt.str[[1]][1:2], my.splt.str[[1]][3:4], my.splt.str[[1]][5:6], sep = " ")
The result is:
[1] "I you precious" "welcome my dude"
When not using the colon operator it returns the correct order:
paste(my.splt.str[[1]][1], my.splt.str[[1]][2], my.splt.str[[1]][3], my.splt.str[[1]][4], my.splt.str[[1]][5], my.splt.str[[1]][6], sep = " ")
[1] "I welcome you my precious dude"
Why is this happening?
paste is designed to work with vectors element-by-element. Say you did this:
names <- c('Alice', 'Bob', 'Charlie')
paste('Hello', names)
You'd want to result to be [1] "Hello Alice" "Hello Bob" "Hello Charlie", rather than "Hello Hello Hello Alice Bob Charlie".
To make it work like you want it to, rather than giving the different sections to paste as separate arguments, you could first combine them into a single vector with c:
paste(c(my.splt.str[[1]][1:2], my.splt.str[[1]][3:4], my.splt.str[[1]][5:6]), collapse = " ")
## [1] "I welcome you my precious dude"
We can use collapse instead of sep
paste(my.splt.str[[1]], collapse= ' ')
If we use the first approach by OP, it is pasteing the corresponding elements from each of the subset
If we want to selectively paste, first create an object because the [[ repeat can be avoided
v1 <- my.splt.str[[1]]
v1[3:4] <- toupper(v1[3:4])
paste(v1, collapse=" ")
#[1] "I welcome YOU MY precious dude"
When we have multiple arguments in paste, it is doing the paste on the corresponding elements of it
paste(v1[1:2], v1[3:4])
#[1] "I you" "welcome my"
If we use collapse, then it would be a single string, but still the order is different because the first element of v1[1:2] is pasteed with the first element of v1[3:4] and 2nd with the 2nd element
paste(v1[1:2], v1[3:4], collapse = ' ')
#[1] "I you welcome my"
It is documented in ?paste
paste converts its arguments (via as.character) to character strings, and concatenates them (separating them by the string given by sep). If the arguments are vectors, they are concatenated term-by-term to give a character vector result. Vector arguments are recycled as needed, with zero-length arguments being recycled to "".
Also, converting to uppercase can be done on a substring without splitting as well
sub("^(\\w+\\s+\\w+)\\s+(\\w+\\s+\\w+)", "\\1 \\U\\2", my.str, perl = TRUE)
#[1] "I welcome YOU MY precious dude"

how to remove duplicate words in a certain pattern from a string in R

I aim to remove duplicate words only in parentheses from string sets.
a = c( 'I (have|has|have) certain (words|word|worded|word) certain',
'(You|You|Youre) (can|cans|can) do this (works|works|worked)',
'I (am|are|am) (sure|sure|surely) you know (what|when|what) (you|her|you) should (do|do)' )
What I want to get is just like this
a
[1]'I (have|has) certain (words|word|worded) certain'
[2]'(You|Youre) (can|cans) do this (works|worked)'
[3]'I (am|are) pretty (sure|surely) you know (what|when) (you|her) should (do|)'
In order to get the result, I used a code like this
a = gsub('\\|', " | ", a)
a = gsub('\\(', "( ", a)
a = gsub('\\)', " )", a)
a = vapply(strsplit(a, " "), function(x) paste(unique(x), collapse = " "), character(1L))
However, it resulted in undesirable outputs.
a
[1] "I ( have | has ) certain words word worded"
[2] "( You | Youre ) can cans do this works worked"
[3] "I ( am | are ) sure surely you know what when her should do"
Why did my code remove parentheses located in the latter part of strings?
What should I do for the result I want?
We can use gsubfn. Here, the idea is to select the characters inside the brackets by matching the opening bracket (\\( have to escape the bracket as it is a metacharacter) followed by one or more characters that are not a closing bracket ([^)]+), capture it as a group within the brackets. In the replacement, we split the group of characters (x) with strsplit, unlist the list output, get the unique elements and paste it together
library(gsubfn)
gsubfn("\\(([^)]+)", ~paste0("(", paste(unique(unlist(strsplit(x,
"[|]"))), collapse="|")), a)
#[1] "I (have|has) certain (words|word|worded) certain"
#[2] "(You|Youre) (can|cans) do this (works|worked)"
#[3] "I (am|are) (sure|surely) you know (what|when) (you|her) should (do)"
Take the answer above. This is more straightforward, but you can also try:
library(stringi)
library(stringr)
a_new <- gsub("[|]","-",a) # replace this | due to some issus during the replacement later
a1 <- str_extract_all(a_new,"[(](.*?)[)]") # extract the "units"
# some magic using stringi::stri_extract_all_words()
a2 <- unlist(lapply(a1,function(x) unlist(lapply(stri_extract_all_words(x), function(y) paste(unique(y),collapse = "|")))))
# prepare replacement
names(a2) <- unlist(a1)
# replacement and finalization
str_replace_all(a_new, a2)
[1] "I (have|has) certain (words|word|worded) certain"
[2] "(You|Youre) (can|cans) do this (works|worked)"
[3] "I (am|are) (sure|surely) you know (what|when) (you|her) should (do)"
The idea is to extract the words within the brackets as unit. Then remove the duplicates and replace the old unit with the updated.
a longer but more elaborate try
a = c( 'I (have|has|have) certain (words|word|worded|word) certain',
'(You|You|Youre) (can|cans|can) do this (works|works|worked)',
'I (am|are|am) (sure|sure|surely) you know (what|when|what) (you|her|you) should (do|do)' )
trim <- function (x) gsub("^\\s+|\\s+$", "", x)
# blank output
new_a <- c()
for (sentence in 1:length(a)) {
split <- trim(unlist(strsplit(a[sentence],"[( )]")))
newsentence <- c()
for (i in split) {
j1 <- as.character(unique(trim(unlist(strsplit(gsub('\\|'," ",i)," ")))))
if( length(j1)==0) {
next
} else {
ifelse(length(j1)>1,
newsentence <- c(newsentence,paste("(",paste(j1,collapse="|"),")",sep="")),
newsentence <- c(newsentence,j1[1]))
}
}
newsentence <- paste(newsentence,collapse=" ")
print(newsentence)
new_a <- c(new_a,newsentence)}
# [1] "I (have|has) certain (words|word|worded) certain"
# [2] "(You|Youre) (can|cans) do this (works|worked)"
# [3] "I (am|are) (sure|surely) you know (what|when) (you|her) should do"

What is the equivalent to VBAs "&" in R?

In excel (and Excel VBA) it is really helpful to connect text and variable using "&":
a = 5
msgbox "The value is: " & a
will give
"The value is: 5"
How can I do this in R? I know there is a way to use "paste". However I wonder if there isn't any trick to do it as simple as in Excel VBA.
Thanks in advance.
This blog post suggests to define your own concatenation operator, which is similar to what VBA (and Javascript) has, but it retains the power of paste:
"%+%" <- function(...) paste0(..., sep = "")
"Concatenate hits " %+% "and this."
# [1] "Concatenate hits and this."
I am not a big fan of this solution though because it kind of obscures what paste does under the hood. For instance, is it intuitive to you that this would happen?
"Concatenate this string " %+% "with this vector: " %+% 1:3
# [1] "Concatenate this string with this vector: 1"
# [2] "Concatenate this string with this vector: 2"
# [3] "Concatenate this string with this vector: 3"
In Javascript for instance, this would give you Concatenate this string with this vector: 1,2,3, which is quite different. I cannot speak for Excel, but you should think about whether this solution is not more confusing to you than it is useful.
If you need Javascript-like solution, you can also try this:
"%+%" <- function(...) {
dots = list(...)
dots = rapply(dots, paste, collapse = ",")
paste(dots, collapse = "")
}
"Concatenate this string " %+% "with this string."
# [1] "Concatenate this string with this string."
"Concatenate this string " %+% "with this vector: " %+% 1:3
# [1] "Concatenate this string with this vector: 1,2,3"
But I haven't tested extensively, so be on lookout for unexpected results.
Another possibility is to use sprintf:
a <- 5
cat(sprintf("The value is %d\n",a))
## The value is 5
the %d denotes integer formatting (%f would give "The value is 5.000000"). The \n denotes a newline at the end of the string.
sprintf() can be more convenient than paste or paste0 when you want to put together a lot of pieces, e.g.
sprintf("The value of a is %f (95% CI: {%f,%f})",
a_est,a_lwr,a_upr)

Replacing strings in R

I am trying to replace strings in R in a large number of texts.
Essentially, this reproduces the format of the data from which I try to delete the '\n' parts.
document <- as.list(c("This is \\na try-out", "And it \\nfails"))
I can do this with a loop and gsub but it takes forever. I looked at this post for a solution. So I tried: temp <- apply(document, 2, function(x) gsub("\\n", " ", fixed=TRUE)). I also used lapply, but it also gives an error message. I can't figure this out, help!
use lapply if you want to return a list
document <- as.list(c("This is \\na try-out", "And it \\nfails"))
temp <- lapply(document, function(x) gsub("\\n", " ", x, fixed=TRUE))
##[[1]]
##[1] "This is a try-out"
##[[2]]
##[1] "And it fails"

Simple Comparing of two texts in R

I want to compare two texts to similarity, therefore i need a simple function to list clearly and chronologically the words and phrases occurring in both texts. these words/sentences should be highlighted or underlined for better visualization)
on the base of #joris Meys ideas, i added an array to divide text into sentences and subordinate sentences.
this is how it looks like:
textparts <- function (text){
textparts <- c("\\,", "\\.")
i <- 1
while(i<=length(textparts)){
text <- unlist(strsplit(text, textparts[i]))
i <- i+1
}
return (text)
}
textparts1 <- textparts("This is a complete sentence, whereas this is a dependent clause. This thing works.")
textparts2 <- textparts("This could be a sentence, whereas this is a dependent clause. Plagiarism is not cool. This thing works.")
commonWords <- intersect(textparts1, textparts2)
commonWords <- paste("\\<(",commonWords,")\\>",sep="")
for(x in commonWords){
textparts1 <- gsub(x, "\\1*", textparts1,ignore.case=TRUE)
textparts2 <- gsub(x, "\\1*", textparts2,ignore.case=TRUE)
}
return(list(textparts1,textparts2))
However, sometimes it works, sometimes it doesn't.
I WOULD like to have results like these:
> return(list(textparts1,textparts2))
[[1]]
[1] "This is a complete sentence" " whereas this is a dependent clause*" " This thing works*"
[[2]]
[1] "This could be a sentence" " whereas this is a dependent clause*" " Plagiarism is not cool" " This thing works*"
whereas i get none results.
There are some problems with the answer of #Chase :
differences in capitalization are not taken into account
interpunction can mess up results
if there is more than one word similar, then you get a lot of warnings due to the gsub call.
Based on his idea, there is the following solution that makes use of tolower() and some nice functionalities of regular expressions :
compareSentences <- function(sentence1, sentence2) {
# split everything on "not a word" and put all to lowercase
x1 <- tolower(unlist(strsplit(sentence1, "\\W")))
x2 <- tolower(unlist(strsplit(sentence2, "\\W")))
commonWords <- intersect(x1, x2)
#add word beginning and ending and put words between ()
# to allow for match referencing in gsub
commonWords <- paste("\\<(",commonWords,")\\>",sep="")
for(x in commonWords){
# replace the match by the match with star added
sentence1 <- gsub(x, "\\1*", sentence1,ignore.case=TRUE)
sentence2 <- gsub(x, "\\1*", sentence2,ignore.case=TRUE)
}
return(list(sentence1,sentence2))
}
This gives following result :
text1 <- "This is a test. Weather is fine"
text2 <- "This text is a test. This weather is fine. This blabalba This "
compareSentences(text1,text2)
[[1]]
[1] "This* is* a* test*. Weather* is* fine*"
[[2]]
[1] "This* text is* a* test*. This* weather* is* fine*. This* blabalba This* "
I am sure that there are far more robust functions on the natural language processing page, but here's one solution using intersect() to find the common words. The approach is to read in the two sentences, identify the common words and gsub() them with a combination of the word and a moniker of our choice. Here I chose to use *, but you could easily change that, or add something else.
sent1 <- "I shot the sheriff."
sent2 <- "Dick Cheney shot a man."
compareSentences <- function(sentence1, sentence2) {
sentence1 <- unlist(strsplit(sentence1, " "))
sentence2 <- unlist(strsplit(sentence2, " "))
commonWords <- intersect(sentence1, sentence2)
return(list(
sentence1 = paste(gsub(commonWords, paste(commonWords, "*", sep = ""), sentence1), collapse = " ")
, sentence2 = paste(gsub(commonWords, paste(commonWords, "*", sep = ""), sentence2), collapse = " ")
))
}
> compareSentences(sent1, sent2)
$sentence1
[1] "I shot* the sheriff."
$sentence2
[1] "Dick Cheney shot* a man."

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