how to rebuild formula from tokens
Ex:
$formulaParser = new \PHPExcel_Calculation_FormulaParser('=IF(T1<X2,"foo", T1+X2)');
dump($formulaParser->getTokens());
I receive:
And I modify more value, ex: change X2 to T2 and now I need rebuild tokens to formula
Can anybody help me?
Tokenising a formula is a part of the parsing process, prior to execution of the formula; it's not intended as a method for changing a formula in any way; it's not intended to be reversible. - Mark Baker
So manually implement by replace string or implode array tokens like
https://github.com/PHPOffice/PHPExcel/issues/834#issuecomment-189158280
Related
I am noob with structured data implementation and don't have any code knowledge.
I have been looking for a week how to solve a warning with price in Google structured data testing tool.
My prices are with a comma which is not accepted by Google.
By checking the http://schema.org/price it tells me that "Use '.' (Unicode 'FULL STOP' (U+002E)) rather than ',' to indicate a decimal point. Avoid using these symbols as a readability separator."
I have a CSS variable element #PdtPrixRef named in a variable "Product-price" with a comma "12.5" but I can't find how to replace it in my structured data with the value "12.5"... Someone to help me?
Hereafter my actual script :
My actual GTM script
Should I add something to my script or making an VARIABLE (Custom Js)?
I think it's something like
value.replace(",", ".")
But I do't know how to write the full proper function from beginning to end...
Yes you can just create a Custom JavaScript Variable
Here is the code
function(){
var price = {{Product-price}};
return price.replace("," , ".");
}
Then using this variable to your JSON-LD script.
In my ML db, we have documents with distributor code like 'DIST:5012' (DIST:XXXX) XXXX is a four-digit number.
currently, in my TDE, the below code works well.
However instead of concat all the raw distributor codes, I want to simply concat the number part only. I used the fn:substring-after XQuery function. However, it won't work. It won't show that distributorCode column in the SQL View anymore. (Below code does not work.)
What is wrong? How to fix that?
Both fn:substring-after and fn:string-join is in TDE Dialect page.
https://docs.marklogic.com/9.0/guide/app-dev/TDE#id_99178
substring-after() expects a single string as input, not a sequence of strings.
To demonstrate, this will not work:
let $dist := ("DIST:5012", "DIST:5013")
return substring-after($dist, "DIST:")
This will:
for $dist in ("DIST:5012", "DIST:5013")
return substring-after($dist, "DIST:")
I need to double check what XPath expressions will work in a DTE, you might be able to change it to apply the substring-after() function in the last step:
fn:string-join( distributors/distributor/urn/substring-after(., 'DIST:'), ';')
Why can't R find this variable?
assign(paste0('my', '_var'), 2)
get(paste0('my', '_var')) ## isn't this returning an object?
save(get(paste0('my', '_var')), file = paste0('my', '_var.RDATA'))
This throws the error:
Error in save(paste0("my", "_var"), file = paste0("my", "_var.RDATA")) :
object ‘paste0("my", "_var")’ not found
From the help page, the save() function expects "the names of the objects to be saved (as symbols or character strings)." Those values are not evaulated, ie you can't put in functions that will eventually return strings or raw values themselves. Use the list= parameter if you want to call a function to return a string the the name of a variable.
save(list=paste0('my', '_var'), file = paste0('my', '_var.RDATA'))
Though using get/assign is often not a good practice in R. They are usually better ways so you might want to rethink your general approach.
And finally, if you are saving a single object, you might want to consider saveRDS() instead. Often that's the behavior people are expecting when they use the save() function.
The documentation for save says that ... should be
the names of the objects to be saved (as symbols or character strings).
And indeed if you type save into the console you can see that the source has the line
names <- as.character(substitute(list(...)))[-1L]
where substitute captures its argument and doesn't evaluate it. So as the error suggests, it is looking for an object with the name paste0('my', '_var'), not evaluating the expressions supplied.
I'm writing a simple function that takes two arguments (state, outcome). State is used to subset a dataframe later.
Having said that, part of the requirement is that state be a length 2 character vector. I need to write more code to ensure that the state that is passed conforms to this requirement.
So I wrote the following:
best <- function(state, outcome) {
outcome <- read.csv("outcome-of-care-measures.csv", colClasses = "character")
state <- vector(mode = "character", length = 2)
st.checkTbl <- outcome[8]
state
}
However, when I call the function and pass the arguments:
best("AXA") or best("FOO") or even best("TX") or best(AL)
All I get back is: "" ""
If I comment out the #state <- ... then it passes the argument just fine and it prints "FOO" or "AXA" or "TX", etc.
How can I ensure that the argument passed to the function is stored as a variable (state) in the function? Or, am I way overthinking this? Really I just wanted to test that what I am passing to the state argument can be printed for test purposes.
. Sorry for the 101 lesson.
You would generally read your data outside of any function, like so:
outcome.data <- read.csv("outcome-of-care-measures.csv", colClasses = "character")
Otherwise, since a function has its own namespace, all the variables defined inside of it will vanish upon its return, unless they themselves are returned by the function with return(...). Several objects can be returned by putting them in a list: return(list(item1=var1, item2=var2)).
Some functions, such as assign, have the envir parameter that can be set to .GlobalEnv to change this behavior. Altering an object can also be done inside a function using the <<- operator instead of <-, although this practice is generally recommended against.
As a side note, when using a function, you need to define clearly:
What are its inputs
What does it do
What does it return
It's not useful, for instance, to use outcome as a function parameter and then read into a variable named income the content of a csv file. Your argument is then useless as it will be written over. That's why you had to comment out the line defining your state variable inside the function to actually be able to use state as it was received by the function.
This surely won't answer all your questions, but hopefully it can help you clarify certain things. For the rest there are plenty of good tutorials to learn further on how to program in R and how/when to use functions. Best of luck and happy learning!
I am trying to use the hash package in R to replicate dictionary behavior in python. I have created it like this,
library(hash)
titles = hash(NAME = list("exact"=list('NAME','Age'), "partial"=list()),
Dt = list("exact"=list('Dt'), "partial"=list()),
CC = list("exact"=list(), "partial"=list()))
I can access the keys in the hash using keys(titles) , values using values(titles), and access values for a particular key using values(titles['Name']).
But how can I access the elements of the inner list? e.g. list('NAME','Age') ?
I need to access the elements based on its names, in this case - "exact" or else I need to know which element of the outer list this element belong to, whether its "exact" or "partial".
Simply:
titles[["NAME"]][["exact"]]
as hrbmstr wrote. There's nothing special about this whatsoever.
In your nested-list, "exact" and "partial" are simply two string keys. Again, there's no special magic significance to their names.
Also, this is in fact the recommended proper R syntax (esp. when the key is variable), it's not "bringing gosh-awful Python syntax".