Latex expression replacement in R - r

Question

I'm working with a LaTex document in R and I need to change {#1 \over #2} to \frac{#1}{#2}.
With simple expressions like:
{1\over 2}
{x^2+y^2\over \lambda}
I can do it with stringr::str_replace() or gsub base functions and the regex \\{([\\^a-z0-9\\\\\\s\\+\\-\\*/\(\)]+)\\s*\\\\over\\s*([\\^a-z0-9\\\\\\s\\+\\-\\*/\(\)]+)\\} (I guess there has to be a better approach to do this. I tried with \\{(.+)\\s*\\\\over\\s*(.*)\\} but it captured more than I wanted.)
But when I work with expressions like:
{e^{2c} \over x-1}
{2yz\over 1+x^{2} }
or a more complicated expression:
\\item $Dom\\left(Q\\right)\\ne {\\rm R}^{2} $ y uno de los puntos no pertenecientes al dominio es $\\left({1\\over 2} ,{1\\over 2} \right).$
the above regex failed.
Is there a regex that can catch all the alternatives? Thanks

Answer 1

Given some sample strings:
> strings
[1] "{1\\over 2}" "{x^2+y^2\\over \\lambda}"
This monster:
> unlist(
lapply(
strsplit(
sub("\\}$","",
sub("^\\{","",strings)),"\\\\over"),
function(x){paste0("\\frac{",x[1],"}{",x[2],"}")}))
produces:
[1] "\\frac{1}{ 2}" "\\frac{x^2+y^2}{ \\lambda}"
This will break if there's more than one \over in the source string. And probably in many other cases too... Oh, it doesn't work if there's spaces before the first { or after the closing }.
On your other examples you get this:
in out
[1,] "{1\\over 2}" "\\frac{1}{ 2}"
[2,] "{x^2+y^2\\over \\lambda}" "\\frac{x^2+y^2}{ \\lambda}"
[3,] "{e^{2c} \\over x-1}" "\\frac{e^{2c} }{ x-1}"
[4,] "{2yz\\over 1+x^{2} }" "\\frac{2yz}{ 1+x^{2} }"

Answer 2

I rather enjoyed this question.
At some point you have to parse the document. parse_tex from TeXCheckR had LaTeX not plain TeX in mind but seems to do okay here. For multi-line instances of \over the script would need to be changed though the principle would be the same I think.
The challenge was for continued fractions.
library(data.table) # for shift
library(TeXCheckR) # for parse_tex
locate_over <- function(doc_parsed) {
lead <- function(x, n) data.table::shift(x, n = n, type = "lead", fill = "")
char <- .subset2(doc_parsed, "char")
which(char == "\\" &
lead(char == "o", 1L) &
lead(char == "v", 2L) &
lead(char == "e", 3L) &
lead(char == "r", 4L))
}
over2frac <- function(lines, verbose = FALSE) {
out <- lines
for (i in seq_along(lines)) {
if (grepl("\\over", lines[i], fixed = TRUE)) {
i_parsed <- parse_tex(lines[i])
# Find lhs
for (j in locate_over(i_parsed)) {
lhs_start <- max(which(.subset2(i_parsed, "char") %chin% c("$", "{") &
.subset2(i_parsed, "column") < j &
.subset2(i_parsed, "tex_group") == .subset2(i_parsed[j], "tex_group")))
rhs_end <- min(which(.subset2(i_parsed, "char") %chin% c("$", "}") &
.subset2(i_parsed, "column") > j + 4L &
.subset2(i_parsed, "tex_group") == .subset2(i_parsed[j], "tex_group")))
i_parsed[lhs_start, char := "{\\frac{"]
i_parsed[rhs_end, char := "}}"]
}
res <- paste0(i_parsed[["char"]], collapse = "")
res <- gsub("\\over", "}{", res, fixed = TRUE)
out[i] <- res
}
}
out
}
Test TeX document:
$5 \over 2$
This is another fraction: ${1 \over 2}$.
And another:
$$A = a \over b$$
What about:
$${{D \over C} \over H}$$
Finally:
$${e^{2c} \over x-1}$$
${2yz\over 1+x^{2} }$
$$\phi = 1 + {1 \over {1 + {1 \over {1 + {1 \over {1 + \ddots}}}}}}$$
\item $Dom\left(Q\right)\ne {\rm R}^{2} $ y uno de los puntos no pertenecientes al dominio es $\left({1\over 2} ,{1\over 2}\right).$
\bye
Resulting LaTeX document: with the necessary LaTeX-specific stuff, plus mandatory math mode for inline fractions. writeLines(over2frac(readLines("tex1.tex"), verbose = FALSE), "latex1.tex")
\documentclass{article}
\begin{document}
${\frac{5 }{ 2}}$
This is another fraction: ${\frac{1 }{ 2}}$.
And another:
${\frac{A = a }{ b}}$
What about:
$${\frac{{\frac{D }{ C}} }{ H}}$$
Finally:
$${\frac{e^{2c} }{ x-1}}$$
${\frac{2yz}{ 1+x^{2} }}$
$$\phi = 1 + {\frac{1 }{ {1 + {\frac{1 }{ {1 + {\frac{1 }{ {1 + \ddots}}}}}}}}}$$
\item $Dom\left(Q\right)\ne {\rm R}^{2} $ y uno de los puntos no pertenecientes al dominio es $\left({\frac{1}{ 2}} ,{\frac{1}{ 2}} \right).$
\end{document}

Answer 3

This gets you most of the way for your examples:
library(stringr)
s <- "Expression 1 is {1\\over 2}.
Expression 2 is {x^2+y^2\\over \\lambda}, yes it is.
Expression 3 is {e^{2c} \\over x-1}.
The last expression: {2yz\\over 1+x^{2} }, all done now."
s2 <- str_replace_all(s,
"\\{(.*?)\\s{0,}\\\\over\\s{0,}(.*?)\\}",
"\\frac\\{\\1\\}\\{\\2\\}")
s2
[1] "Expression 1 is frac{1}{2}.\n\nExpression 2 is frac{x^2+y^2}{\\lambda}, yes it is.\n\nExpression 3 is frac{e^{2c}}{x-1}.\n\nThe last expression: frac{2yz}{1+x^{2} }, all done now."
The only issue is that a space remains in the last expression, which may not be a problem since it existed in the original:
frac{2yz}{1+x^{2} }

Answer 4

x=c("{e^{2c} \\over x-1}","{2yz\\over 1+x^{2} },,dty{k^4e{-rpi/3}\\over\\sqrt{2pik}}")
gsub("\\{(.*?)\\\\over(.*?)\\}","\\\frac{\\1}{\\2}",x)
[1] "\frac{e^{2c} }{ x-1}"
[2] "\frac{2yz}{ 1+x^{2} },,dty\frac{k^4e{-rpi/3}}{\\sqrt{2pik}}"
Explanation:
\{(.*?)\\over(.*?)\\
List item{ matches the character { literally (case sensitive)
1st Capturing Group (.*?)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
\\ matches the
character \ literally (case sensitive) over matches the characters
over literally (case sensitive)
2nd Capturing Group (.*?)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
\\ matches the character \ literally (case sensitive)

Answer 5

This approach can handle:
multiple {...} containing \over on the same line
{...} not containing \over
other text before, after and between occurrences of {...}
lines not having any {...} with \over
For example, note in the example below the {jjj} on the second input line before the first occurrence of a {...} with \over works as expected.
It makes use of gsubfn which can handle balanced parentheses. First create a proto object p similar to the one in my answer here. p initializes a counter k to 0 and increments it for each { and decrements it for each }. It replaces any { for which k=1 with ! and also any } for which k=0 with !.
Then replace !...\over...! with \frac{...}{...} and replace any remaining !...! with {...}.
We have assumed that ! does not appear in the input but if it does choose a different character.
library(gsubfn)
library(magrittr)
# test input
s <- c("abc {1\\over 2} def {x^2+y^2\\over \\lambda} ghi { 12 } XYZ",
"X {jjj} A {e^{2c} \\over x-1} jkl {2yz\\over 1+x^{2} } Z")
# processing
p <- proto(
pre = function(.) .$k <- 0,
fun = function(., x) {
if (x == "{") .$k <- .$k + 1 else if (x == "}") .$k <- .$k - 1
if (x == "{" && .$k == 1) "!" else if (x == "}" && .$k == 0) "!" else x
})
s %>%
gsubfn("[{}]", p, .) %>%
gsub("!([^!]*)\\\\over ([^!]*)!", "\\\\frac{\\1}{\\2}", .) %>%
gsub("!([^!]*)!", "{\\1}", .)
giving this result:
[1] "abc \\frac{1}{2} def \\frac{x^2+y^2}{\\lambda} ghi { 12 } XYZ"
[2] "X {jjj} A \\frac{e^{2c} }{x-1} jkl \\frac{2yz}{1+x^{2} } Z"