I am trying to write a GUI that will display the name of the sketch it was generated from using a simple text() command. However, I am running into trouble getting any of the general JS solutions to work for me. Many solutions I have found use the filename reserved word but that does not seem to be reserved in Processing 3.5.4. I have also tried parsing the strings using a similar method to what can be found here. I am very new to processing and this is only my 2nd attempt at using Processing.
Any advice would be greatly appreciated
You can get the path (as a string) to the sketch with sketchPath().
From there you could either parse the string (pull off everything after the last slash) to get the sketch name, or you can use sketchFile() to get a reference to the file itself and get the name from there:
String path = sketchPath();
File file = sketchFile(path);
String sketchName = file.getName();
println(sketchName);
You could combine this all into one line like so:
String sketchName = sketchFile(sketchPath()).getName();
Related
I need to read a text file with readLines() and I've already found this question, but the code in the answers always uses some variation of javaClass; it seems to work only inside a class, while I'm using just a simple Kotlin file with no declared classes. Writing it like this is correct syntax-wise but it looks really ugly and it always returns null, so it must be wrong:
val lines = object {}.javaClass.getResource("file.txt")?.toURI()?.toPath()?.readLines()
Of course I could just specify the raw path like this, but I wonder if there's a better way:
val lines = File("src/main/resources/file.txt").readLines()
Thanks to this answer for providing the correct way to read the file. Currently, reading files from resources without using javaClass or similar constructs doesn't seem to be possible.
// use this if you're inside a class
val lines = this::class.java.getResourceAsStream("file.txt")?.bufferedReader()?.readLines()
// use this otherwise
val lines = object {}.javaClass.getResourceAsStream("file.txt")?.bufferedReader()?.readLines()
According to other similar questions I've found, the second way might also work within a lambda but I haven't tested it. Notice the need for the ?. operator and the lines?.let {} syntax needed from this point onward, because getResourceAsStream() returns null if no resource is found with the given name.
Kotlin doesn't have its own means of getting a resource, so you have to use Java's method Class.getResource. You should not assume that the resource is a file (i.e. don't use toPath) as it could well be an entry in a jar, and not a file on the file system. To read a resource, it is easier to get the resource as an InputStream and then read lines from it:
val lines = this::class.java.getResourceAsStream("file.txt").bufferedReader().readLines()
I'm not sure if my response attempts to answer your exact question, but perhaps you could do something like this:
I'm guessing in the final use case, the file names would be dynamic - Not statically declared. In which case, if you have access to or know the path to the folder, you could do something like this:
// Create an extension function on the String class to retrieve a list of
// files available within a folder. Though I have not added a check here
// to validate this, a condition can be added to assert if the extension
// called is executed on a folder or not
fun String.getFilesInFolder(): Array<out File>? = with(File(this)) { return listFiles() }
// Call the extension function on the String folder path wherever required
fun retrieveFiles(): Array<out File>? = [PATH TO FOLDER].getFilesInFolder()
Once you have a reference to the List<out File> object, you could do something like this:
// Create an extension function to read
fun File.retrieveContent() = readLines()
// You can can further expand this use case to conditionally return
// readLines() or entire file data using a buffered reader or convert file
// content to a Data class through GSON/whatever.
// You can use Generic Constraints
// Refer this article for possibilities
// https://kotlinlang.org/docs/generics.html#generic-constraints
// Then simply call this extension function after retrieving files in the folder.
listOfFiles?.forEach { singleFile -> println(singleFile.retrieveContent()) }
In order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
The url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Solved, now working example!
I have a set of time series that is populated in a folder structure as follows:
TimeSeriesData\Config_000000\seed_001\Aggregate_Quantity.txt
…
TimeSeriesData\Config_000058\seed_010\Aggregate_Quantity.txt
And in each file there is a first line containing a character, followed by lines containing the data. E.g.
share
-21.75
-20.75
…
Now, I would like to import all those files (at best, without specifying ex post the number of configs and seeds, at worst with supplying it) into single time series of the like: “AggQuant_config_seed” where config relates to the config ID and seed to the seed id.
I tried the following (using the non-preferred way), but “parsing” the “path” does not work / I do not know how to do it.
string base = "TimeSeriesData/Config_"
string middle = "/seed_"
string endd = "/Aggregate_Quantity.txt"
string path = ""
loop for (i=0;i<=58;i+=1)
loop for (j=1;j<=10;j+=1)
path = ""
sprintf path "%s%06d%s%03d%s",base,i,middle,j,endd
append #path #will be named share
rename share Agg_Q_$i_$j #rename
endloop
endloop
To sum up the problem, the following does not work:
string path="somwhere/a_file.txt" #string holding path
#wrong: append $path #use append on string
append #path #works!
And if possible, a way to search recursively through a set of folders, using the folder information together with file-information for the variable name, would be nice. Is that possible within gretl?
I realize that in many cases I would like to refer to a specific “help” section like those for functions and commands, but for operators instead (like “$”, which is obviously wrong here).
Problem:
Hello, I have been struggling recently in my programming endeavours. I have managed to receive the output below from Google Speech to Text, but I cannot figure out how draw data from this block.
Excerpt 1:
[VoiceMain]: Successfully initialized
{"result":[]}
{"result":[{"alternative":[{"transcript":"hello","confidence":0.46152416},{"transcript":"how low"},{"transcript":"how lo"},{"transcript":"how long"},{"transcript":"Polo"}],"final":true}],"result_index":0}
[VoiceMain]: Successfully initialized
{"result":[]}
{"result":[{"alternative":[{"transcript":"hello"},{"transcript":"how long"},{"transcript":"how low"},{"transcript":"howlong"}],"final":true}],"result_index":0}
Objective:
My goal is to extract the string "hello" (without the quotation marks) from the first transcript of each block and set it equal to a variable. The problem arises when I do not know what the phrase will be. Instead of "hello", the phrase may be a string of any length. Even if it is a different string, I would still like to set it to the same variable to which the phrase "hello" would have been set to.
Furthermore, I would like to extract the number after the word "confidence". In this case, it is 0.46152416. Data type does not matter for the confidence variable. The confidence variable appears to be more difficult to extract from the blocks because it may or may not be present. If it is not present, it must be ignored. If it is present however, it must be detected and stored as a variable.
Also please note that this text block is stored within a file named "CurlOutput.txt".
All help or advice related to solving this problem is greatly appreciated.
You could do this with regex, but then I am assuming you will want to use this as a dict later in your code. So here is a python approach to building this result as a dictionary.
import json
with open('CurlOutput.txt') as f:
lines = f.read().splitlines()
flag = '{"result":[]} '
for line in lines: # Loop through each lin in file
if flag in line: # check if this is a line with data on it
results = json.loads(line.replace(flag, ''))['result'] # Load data as a dict
# If you just want to change first index of alternative
# results[0]['alternative'][0]['transcript'] = 'myNewString'
# If you want to check all alternative for confidence and transcript
for result in results[0]['alternative']: # Loop over each alternative
transcript = result['transcript']
confidence = None
if 'confidence' in result:
confidence = result['confidence']
# now do whatever you want with confidence and transcript.
I would like to get all the files that a sub-folder holds in a string array.
So, I have tried something like the following:
var IOstore = IsolatedStorageFile.GetUserStoreForApplication();
string searchpath = System.IO.Path.Combine("product", ProductName);
string filesInSubDirs[] = IOstore.GetFileNames(searchpath);
But I got all the files in the "product" folder. I have also tried with "productname" only as the parameter.
Thanks for your help.
The search pattern for a sub-folder needs to include "*.*" at the end to pattern match any file, which would make your code something like the following:
var IOstore = IsolatedStorageFile.GetUserStoreForApplication();
string searchpath = System.IO.Path.Combine("product", ProductName);
searchpath = string.Format("{0}\\*.*", searchpath);
string filesInSubDirs[] = IOstore.GetFileNames(searchpath);
Something you might want to try. (this is sort of a left field answer, sorry). In my dropbox client http://sharpdropbox.codeplex.com/) I have a set of facades for System.IO.File, System.IO.FileInfo, System.IO.Directory, and System.IO.DirectoryInfo. They work pretty good and I have tested them.
Basically, you add a Using or Import for System.IO.IsolatedStorage and then PSFile, PSDirectory, PSFileInfo, or PSDirectoryInfo. It's saved me from having to remember all the nuances... for instance if you are querying a directory, it knows to add a slash, etc. BTW, the "PS" prefix stands for "Persisted Storage" which is what IsolatedStorage is sometimes called (starting them with an "I" implies they are interfaces.. and having no prefix makes things even more confusing).
Anyway, you can grab the code from source or I believe the last release had the DLLs for them (it's called something like "IsolatedStorageFacade-WP7")
I'm currently trying to attach image files to a model directly from a zip file (i.e. without first saving them on a disk). It seems like there should be a clearer way of converting a ZipEntry to a Tempfile or File that can be stored in memory to be passed to another method or object that knows what to do with it.
Here's my code:
def extract (file = nil)
Zip::ZipFile.open(file) { |zip_file|
zip_file.each { |image|
photo = self.photos.build
# photo.image = image # this doesn't work
# photo.image = File.open image # also doesn't work
# photo.image = File.new image.filename
photo.save
}
}
end
But the problem is that photo.image is an attachment (via paperclip) to the model, and assigning something as an attachment requires that something to be a File object. However, I cannot for the life of me figure out how to convert a ZipEntry to a File. The only way I've seen of opening or creating a File is to use a string to its path - meaning I have to extract the file to a location. Really, that just seems silly. Why can't I just extract the ZipEntry file to the output stream and convert it to a File there?
So the ultimate question: Can I extract a ZipEntry from a Zip file and turn it directly into a File object (or attach it directly as a Paperclip object)? Or am I stuck actually storing it on the hard drive before I can attach it, even though that version will be deleted in the end?
UPDATE
Thanks to blueberry fields, I think I'm a little closer to my solution. Here's the line of code that I added, and it gives me the Tempfile/File that I need:
photo.image = zip_file.get_output_stream image
However, my Photo object won't accept the file that's getting passed, since it's not an image/jpeg. In fact, checking the content_type of the file shows application/x-empty. I think this may be because getting the output stream seems to append a timestamp to the end of the file, so that it ends up looking like imagename.jpg20110203-20203-hukq0n. Edit: Also, the tempfile that it creates doesn't contain any data and is of size 0. So it's looking like this might not be the answer.
So, next question: does anyone know how to get this to give me an image/jpeg file?
UPDATE:
I've been playing around with this some more. It seems output stream is not the way to go, but rather an input stream (which is which has always kind of confused me). Using get_input_stream on the ZipEntry, I get the binary data in the file. I think now I just need to figure out how to get this into a Paperclip attachment (as a File object). I've tried pushing the ZipInputStream directly to the attachment, but of course, that doesn't work. I really find it hard to believe that no one has tried to cast an extracted ZipEntry as a File. Is there some reason that this would be considered bad programming practice? It seems to me like skipping the disk write for a temp file would be perfectly acceptable and supported in something like Zip archive management.
Anyway, the question still stands:
Is there a way of converting an Input Stream to a File object (or Tempfile)? Preferably without having to write to a disk.
Try this
Zip::ZipFile.open(params[:avatar].path) do |zipfile|
zipfile.each do |entry|
filename = entry.name
basename = File.basename(filename)
tempfile = Tempfile.new(basename)
tempfile.binmode
tempfile.write entry.get_input_stream.read
user = User.new
user.avatar = {
:tempfile => tempfile,
:filename => filename
}
user.save
end
end
Check out the get_input_stream and get_output_stream messages on ZipFile.