In kableExtra >= 0.8.0, the canonical way to insert a linebreak into text piped into a table from a kableExtra function such as add_header_above or pack_rows is to add an \n directly.
However, this appears not to work with the escape = FALSE argument, which is required if the text also contains LaTeX code.
How can one force linebreaks in kableExtra functions with escape = FALSE?
library(dplyr)
library(knitr)
library(kableExtra)
starwars %>%
filter(species == 'Gungan' | species == 'Droid') %>%
arrange(species) %>%
select(name, eye_color) %>%
kbl(booktabs = TRUE) %>%
pack_rows(
index = c(
'The droids: everybody\'s favourite' = 6,
'The Gungans: only beloved of \nthose aged under $3^2$' = 3),
escape = FALSE)
ISSUE
The issue at hand is that you wish to escape part of your header (i.e., the break) and not escape another part (i.e., the math code).
Further Complications
This core issue is further complicated by a number of factors:
when and how kableExtra is programmed to deal with escaping
a desire to have a solution that works for both html and LaTeX output
when and how R evaluates code
A SOLUTION
Here is a solution that will work for both html and LaTeX output, but it is not as clean and straight forward as your original code:
# a new version of `kableExtra::linebreak()` that takes into account what type
# of output is desired as well as how much escaping is necessary
linebreak2 <- function(x, double_escape = TRUE, ...) {
# if LaTeX insert text into a `\makecell[]{}` command and double escape
if(knitr::is_latex_output())
return(linebreak(x, double_escape = double_escape, ...))
# if html output just replace `\n`s with `<br/>`s
if(knitr::is_html_output())
return(gsub("\n", "<br/>", x))
# let x pass through for other types of output
return(x)
}
# build the index named vector outside the pipe flow
# in order to set the names using `linebreak2()`
index <- c(6, 3)
names(index) <- c(
'The droids: everybody\'s favourite',
linebreak2('The Gungans: only beloved of \nthose aged under $3^2$')
)
# proceed as before
starwars %>%
filter(species == 'Gungan' | species == 'Droid') %>%
arrange(species) %>%
select(name, eye_color) %>%
kbl(booktabs = TRUE) %>%
pack_rows(index = index, escape = FALSE)
PDF Output
HTML Output
You could use html line break tag <br/>:
starwars %>%
filter(species == 'Gungan' | species == 'Droid') %>%
arrange(species) %>%
select(name, eye_color) %>%
kbl(booktabs = TRUE) %>%
pack_rows(
index = c(
'The droids: everybody\'s favourite' = 6,
'The Gungans: only beloved of <br/> those aged under $3^2$' = 3),
escape = FALSE)
Related
Hello stackoverflow community,
I'd like to use the function for column renaming gt:cols_label() and globally define a list for this to input to the .list-argument of gt:cols_label(). The list should contain:
the new column names
the desired formatting options (using the md()-Function)
While providing both info in the ...-argument directly works:
mpg %>%
gt::gt() %>%
gt::cols_label(
manufacturer = md("*M*a**nufac***ture***r**")
)
This does not (and neither does the commented try):
rename_format_cols <- c(
"manufacturer" = md("*M*a**nufac***ture***r**")
) %>%
as.list()
# rename_format_cols <- c(
# "manufacturer" = "md("*M*a**nufac***ture***r**")"
# ) %>%
# as.list()
mpg %>%
gt::gt() %>%
gt::cols_label(
.list = rename_format_cols
)
I'd be very glad if you'd helped out.
Best regards, paulge
I have generated the following table using the Flextable package in R. I created a conditionally formatted column with LaTeX arrow symbols in it, however the symbols aren't displayed when I generate it as a flextable. Is there a way to fix this?
library(tidyverse)
library(flextable)
data.frame(one = c(10,20,30), two = c(30,20,6)) %>%
mutate(Trend = case_when(.[,2] == .[,1] ~ "$\\rightarrow$", .[,2] > .[,1] ~ "$\\nearrow$", TRUE ~ "$\\searrow$")) %>%
flextable()
It may be easier to do this with unicode values for the symbols
library(dplyr)
library(flextable)
data.frame(one = c(10,20,30), two = c(30,20,6)) %>%
mutate(Trend = ifelse(two == one, "\U2192", "\U2190")) %>%
flextable()
-output
I am currently using the Officer package to produce a Word document. I have been using body_add_par to add multiple chucks of blank lines throughout the document, but this method is already becoming tedious.
Is there a way to create a function, or somehow be able to write one line of code that is able to specify how many blank lines I want to insert?
Practice_R.docx = read_docx() %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par(paste("test")) %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par("") %>%
body_add_par(paste("test2"))%>%
body_add_break( pos = "after")
Try:
# repeat the addition of an empty paragraph n-times
body_add_par_n <- function(doc, n) {
i <- 1 # initialize counter
while (i<=n) { # control if the counter is less then desired n
doc <- body_add_par(doc, "") # add paragraph to the object
i <- i+1 # increment the counter
}
doc # return the object
}
Practice_R.docx = read_docx() %>%
body_add_par_n(3) %>%
body_add_par(paste("test")) %>%
body_add_par_n(6) %>%
body_add_par(paste("test2"))%>%
body_add_break( pos = "after")
I'm new to R and I don't know all basic concepts yet. The task is to produce a one merged table with multiple response sets. I am trying to do this using expss library and a loop.
This is the code in R without a loop (works fine):
#libraries
#blah, blah...
#path
df.path = "C:/dataset.sav"
#dataset load
df = read_sav(df.path)
#table
table_undropped1 = df %>%
tab_cells(mdset(q20s1i1 %to% q20s1i8)) %>%
tab_total_row_position("none") %>%
tab_stat_cpct() %>%
tab_pivot()
There are 10 multiple response sets therefore I need to create 10 tables in a manner shown above. Then I transpose those tables and merge. To simplify the code (and learn something new) I decided to produce tables using a loop. However nothing works. I'd looked for a solution and I think the most close to correct one is:
#this generates a message: '1' not found
for(i in 1:10) {
assign(paste0("table_undropped",i),1) = df %>%
tab_cells(mdset(assign(paste0("q20s",i,"i1"),1) %to% assign(paste0("q20s",i,"i8"),1)))
tab_total_row_position("none") %>%
tab_stat_cpct() %>%
tab_pivot()
}
Still it causes an error described above the code.
Alternatively, an SPSS macro for that would be (published only to better express the problem because I have to avoid SPSS):
define macro1 (x = !tokens (1)
/y = !tokens (1))
!do !i = !x !to !y.
mrsets
/mdgroup name = !concat($SET_,!i)
variables = !concat("q20s",!i,"i1") to !concat("q20s",!i,"i8")
value = 1.
ctables
/table !concat($SET_,!i) [colpct.responses.count pct40.0].
!doend
!enddefine.
*** MACRO CALL.
macro1 x = 1 y = 10.
In other words I am looking for a working substitute of !concat() in R.
%to% is not suited for parametric variable selection. There is a set of special functions for parametric variable selection and assignment. One of them is mdset_t:
for(i in 1:10) {
table_name = paste0("table_undropped",i)
..$table_name = df %>%
tab_cells(mdset_t("q20s{i}i{1:8}")) %>% # expressions in the curly brackets will be evaluated and substituted
tab_total_row_position("none") %>%
tab_stat_cpct() %>%
tab_pivot()
}
However, it is not good practice to store all tables as separate variables in the global environment. Better approach is to save all tables in the list:
all_tables = lapply(1:10, function(i)
df %>%
tab_cells(mdset_t("q20s{i}i{1:8}")) %>%
tab_total_row_position("none") %>%
tab_stat_cpct() %>%
tab_pivot()
)
UPDATE.
Generally speaking, there is no need to merge. You can do all your work with tab_*:
my_big_table = df %>%
tab_total_row_position("none")
for(i in 1:10) {
my_big_table = my_big_table %>%
tab_cells(mdset_t("q20s{i}i{1:8}")) %>% # expressions in the curly brackets will be evaluated and substituted
tab_stat_cpct()
}
my_big_table = my_big_table %>%
tab_pivot(stat_position = "inside_columns") # here we say that we need combine subtables horizontally
I am using R version 3.5.2.
I would like to evaluate a string in a kable function, but I am having some issues. Normally, I can pass a string through a for loop using the get function but in the kableExtra::add_header_above function I get the following error:
Error: unexpected '=' in:"print(kable(df4,"html", col.names = c("zero","one")) %>% add_header_above(c(get("string") ="
I have tried a handful of techniques like creating a string outside of the kable function and calling it, using page breaks and print statements in the knit loop and trying the eval function as well. I have also added result ="asis" as suggested here
Here is a reproducible example:
```{r results="asis"}
library("knitr")
library("kableExtra")
df1 <- mtcars %>% dplyr::select(am,vs)
df1a <- df1 %>% mutate(type = "A")
df1b <- df1 %>% mutate(type = "B")
df1c <- df1 %>% mutate(type = "C")
df2 <- rbind(df1a,df1b,df1c)
vector <- as.vector(unique(df2$type))
for (variable in vector) {
df3 <- df2 %>% filter(type == (variable))
df4 <- table(df3$am,df3$vs)
print(kable(df4,"html", col.names = c("zero","one")) %>%
add_header_above(c(get("string") = 3)))
}
```
Ideally, I would like the header of the table to have the string name from the column type. Below is an example of what I want it to look it:
print(kable(df4,"html", col.names = c("zero","one")) %>%
add_header_above(c("A" = 3)))
I understand that the knitr function needs to be treated differently than regular R when using loops as found in this solution but I am still struggling to get the string to be evaluated correctly. Perhaps because the function requires a vecotr input, it is not evalauting it as a string?
You have to define your header as a vector. The name of the header should be the names of the vector and the value of the vector would be the number of columns the header will use.
The loop in the code should look like this:
for (variable in vector) {
df3 <- df2 %>% filter(type == (variable))
df4 <- table(df3$am,df3$vs)
header_temp = 3
names(header_temp) = get("variable")
print(kable(df4,"html", col.names = c("zero","one")) %>%
add_header_above(header_temp))
}
So first I define the number of columns the of the header in the variable header_temp and then i assign a name to it.