I am building a code that lets the user open some files.
reference = warndlg('Choose the files for analysis.');
uiwait(reference);
filenames2 = uigetfile('./*.txt','MultiSelect', 'on');
if ~iscell(filenames2)
filenames2 = {filenames2}; % force it to be a cell array of strings
end
numberOfFiles = numel(filenames2);
data = importdata(filenames2{i},delimiterIn,headerlinesIn);
When I run the code, the prompts show up, I press OK, and then nothing happens. The code just stops, telling me :
Error using importdata (line 137)
Unable to open file.
Error in FreqVSChampB_no_spec (line 119)
data=importdata(filenames2{1},delimiterIn,headerlinesIn);
I just don't have the opportunity to select a file. The cellarray stays empty as showed in the following image.
MATLAB can't find the file that you have selected. Your variable filenames2 contains only the name of the file, not its full path. If you don't provide the full path to importdata, it will search for whatever file name you provide on the MATLAB path, and if it can't find it it will error as you see.
Try something like this - I'm just doing it with single selection for ease of description, but you can do something similar with multiple selection.
[fileName, pathName] = uigetfile('*.txt');
fullNameWithPath = fullfile(pathName, fileName);
importdata(fullNameWithPath)
fullfile is useful, as it inserts the correct character between pathName and fileName (\ on Windows, / on Unix).
You can try to add
pause(0.1);
just after uiwait(reference);
For me it works. In fact I've noticed the active windows changes when we use uiwait and uigetfile.
Related
I wonder how to convert a dm3 file into .jpg/jpeg images? there is test annotation and scale bar on the image. I setup a script but it always show that "the format cannot contain the data to be saved". This can be done via file/batch convert function. So how to realize the same function in script？ Thanks
image test:=IntegerImage("test",2,1,100,100)
test.ShowImage()
image frontimage:=GetFrontImage()
string filename=getname(frontimage)
imagedisplay disp = frontImage.ImageGetImageDisplay(0)
disp.applydatabar()
ImageDocument frontDoc = GetFrontImageDocument()
string directoryname, pathname
number length
if(!SaveAsDialog("","Do Not Change Me",directoryname)) exit(0)
length=len(directoryname)-16
directoryname=mid(directoryname,0,length)
pathname=directoryname+filename
frontDoc.ImageDocumentSaveToFile( "JPG Format", pathname )
To convert to jpg you have to use "JPEG/JFIF Format" as the handler (=format).
It has to be exactly this string in the ImageDocument.ImageDocumentSaveToFile() function. Other formats are mentioned in the help (F1 > Scripting > Objects > Document Object Model > ImageDocument Object > ImageDocumentSaveToFile() function). Those are (for example):
'Gatan Format'
'Gatan 3 Format'
'GIF Format'
'BMP Format'
'JPEG/JFIF Format'
'Enhanced Metafile Format'
In your code you are using the SaveAsDialog() to get a directory. This is not necessary. You can use GetDirectoryDialog() to get a directory. This saves you the name operation for the directoryname and avoids problems when users do change your filename.
Also for concatinating paths I prefer using PathConcatenate(). On the first hand this makes your code a lot more readable since its name tells what you are doing. On the other hand this also takes care of the directory ending with \ or not and other path related things.
The following code is what I think you need:
Image test := IntegerImage("test", 2, 1, 100, 100);
test.ShowImage();
Image frontimage := GetFrontImage();
ImageDisplay disp = frontImage.ImageGetImageDisplay(0);
disp.applydatabar();
ImageDocument frontDoc = GetFrontImageDocument();
string directoryname;
if(!GetDirectoryDialog("Select directory", "C:\\\\", directoryname)){
// ↑
// You can of course use something else as the start point for selection here
exit(0);
}
string filename = GetName(frontimage);
string pathname = directoryname.PathConcatenate(filename);
frontDoc.ImageDocumentSaveToFile("JPEG/JFIF Format", pathname);
This answer is correct and should be accepted. Your problem is the wrong file-type string. You want to use "JPEG/JFIF Format"
A bit more general information on image file saving in DigitalMicrograph.
One doesn't save images but always imageDocuments that can contain one, more, or even zero image objects in them. Script-commands that save an image like SaveAsGatan() really just call things like: ImageGetOrCreateImageDocument().ImageDocumentSaveToFile()
The difference doesn't really matter for simple one-image-in-document type images, but it can make a difference when there are multiple images in a document, or when a single image is displayed multiple times simultaneously (which can be done.) So it is always good to know what "really" goes on.
ImageDocuments contain some properties relating to saving:
A save format (“Gatan Format”, “TIFF Format”, …)
Default value: What it was opened with, or last used save-format in case of creation
Script commands: ImageDocumentGetCurrentFileSaveFormat() ImageDocumentSetCurrentFileSaveFormat()
A current file path:
Default value: What it was opened from, or empty
Script commands: ImageDocumentGetCurrentFile() ImageDocumentSetCurrentFile()
A dirty-state:
Default value: clean when opened, dirty when created
Script commands: ImageDocumentIsDirty() ImageDocumentClean()
A linked-to-file state:
Default value: true when opened, false when created
Script commands: ImageDocumentIsLinkedToFile()
There are two ways of saving an imageDocument:
Saving the current document itself to disc:
void ImageDocumentSave( ImageDocument imgDoc, Number save_style ) This utilizes the current properties of the imageDocument to save it to current path in current format, marking it clean in the process. The save_style parameter determines how the program deals with missing info:
0 = never ask for path
1 = ask if not linked (or empty path)
2 = always ask
Saving a copy of the current document to disc:
void ImageDocumentSaveToFile( ImageDocument imgDoc, String handler, String fileName ) This makes a copy and save the file under provided path in the provided format. The imageDocument in memory does not change its properties. Most noticeable: It does not become clean, and it is not linked to the provided file on disc. The filename parameter specifies the saving location including the filename. If a file extension is provided, it has to match the file-format, but it can be left out. The handler parameter specified the file-format and can be anything GMS currently supports, such as:
Gatan Format
Gatan 3 Format
GIF Format
BMP Format
JPEG/JFIF Format
Enhanced Metafile Format
In short:
To save the currently opened imageDocument with a different format, you would want to do:
imageDocument doc = GetFrontImageDocument()
doc.ImageDocumentSetCurrentFileSaveFormat("TIFF Format")
doc.ImageDocumentSave(0)
While to just save a copy of the current state you would use:
imageDocument doc = GetFrontImageDocument()
string path = doc.ImageDocumentGetCurrentFile() // full path including extension!
path = PathExtractDirectory(path,0) + PathExtractBaseName(path,0) // path without file extension
doc.ImageDocumentSaveToFile("TIFF Format", path )
I have a list of filenames that I want to make (they don't exist yet). I want to loop through the list and create each file. Next I want to write to each file a path (along with other text not shown here) that includes the name of the file. I have written something similar to below so far but cannot see how to get the index i to take the file name values. Please help.
import os
biglist=['sleep','heard','shed']
for i in biglist:
myfile=open('C:\autumn\winter\spring\i.txt','w')
myfile.write('DATA = c:\autumn\winter\spring\i.dat')
myfile.close
Maybe you can try this below python function.
import sys
biglist=['sleep','heard','shed']
def create_file():
for i in biglist:
try:
file_name_with_ext = "C:\autumn\winter\spring\"+ i + ".txt"
file = open(file_name_with_ext, 'a')
file.close()
except:
print("caught error!")
sys.exit(0)
create_file() #invoking the function
I want to read a file having size 4 MB using python xlrd in GAE.
i am getting the file from Blobstore. Code used is given below.
book = xlrd.open_workbook(file_contents=temp_file)
sh = book.sheet_by_index(0)
for col_no in range(sh.ncols):
its gives me DeadlineExceededError.
book = xlrd.open_workbook(file_contents=file_data)
File "/base/data/home/apps/s~appid/app-version.369475363369053908/xlrd/__init__.py", line 416, in open_workbook
ragged_rows=ragged_rows,
File "/base/data/home/apps/s~appid/app-version.369475363369053908/xlrd/xlsx.py", line 756, in open_workbook_2007_xml
x12sheet.process_stream(zflo, heading)
File "/base/data/home/apps/s~appid/app-version.369475363369053908/xlrd/xlsx.py", line 520, in own_process_stream
for event, elem in ET.iterparse(stream):
DeadlineExceededError
But i am able to read files with smaller size.
Actually i need to get only first few rows(30 to 50) of the file. Is there any other method, other than adding it as a task and getting the details using channel API to get the details with out causing deadline error ?
What i can do to handle this....?
I read a file about 1000 rows excel and it works okay the library.
I leave a link that might be useful https://github.com/cjhendrix/HXLator-SpaceAppsVersion/blob/master/gae/main.py
the code I see that this crossing of columns and rows must be at lists for each row
example:
wb = xlrd.open_workbook(file_contents=inputfile.read())
sh = wb.sheet_by_index(0)
for rownum in range(sh.nrows):
val_row = sh.row_values(rownum)
#here print element of list
self.response.write(val_row[1]) #depending for number for columns
regards!!!
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.
I have WPF DataGrid (VS2010 C#). I copied the data from DataGrid to Clipboard and write it to an Excel file. Below is my code.
dataGrid1.SelectAllCells();
dataGrid1.ClipboardCopyMode = DataGridClipboardCopyMode.IncludeHeader;
ApplicationCommands.Copy.Execute(null, dataGrid1);
dataGrid1.UnselectAllCells();
string path1 = "C:\\test.xls";
string result1 = (string)Clipboard.GetData(DataFormats.CommaSeparatedValue);
Clipboard.Clear();
System.IO.StreamWriter file1 = new System.IO.StreamWriter(path1);
file1.WriteLine(result1);
file1.Close();
Everything works out OK except when I open the excel file it gives me two warning:
"The file you are trying to open
'test.xls' is in a different format
than specified by the file extension.
Verify that the file is not corrupted
and is from a trusted source before
opening the file. Do you want to open
the file now?"
"Excel has detected that 'test.xls' is
a SYLK file, but cannot load it."
But after I click through it, it still open the excel file OK and data are formated as it supposed to be. But I can't find how to get rid of the two warnings before the excel file is open.
You need to use csv as extension. Xls is the Excel file extension.
So
string path1 = "C:\\test.csv";
should work.
A problem like yours has already been described here : generating/opening CSV from console - file is in wrong format error.
Does it helps to solve yours ?
Edit : Here is the Microsoft KB related => http://support.microsoft.com/kb/323626