[Tutor] path string

Alan Gauld alan.gauld at yahoo.co.uk
Mon Jan 2 19:08:41 EST 2017

On 02/01/17 17:01, anatta anatta wrote:

> I am trying to create unsuccessfully source path as 
> a string 'str7' in part_1 of the code below,

When you say unsuccessfully what do you mean?
What do you expect? What do you get?

> to be used in part_2 of the code.

For that you need to expose it outside the
function, the best way to do that is to return
it as a value, which you comment suggests you
want to do. But the only return is commented out,
so you need to tidy that up.

But personally I think your function is trying to do
too much. You should simplify it to only return the
files and have another function that returns the path.
Functions that try to do too many things (ie more
than one) are notoriously difficult to debug.

> When I define the source path explicitly in part_2 
> of the code (#sourcePath = r'H://TCVFLDAT'), the
> code works right.

I'll take your word for it.

> How else could I find the path in part-1 and use it in part 2?

Return the value (assuming it is the right value)
and in part two assign the return from the
function to a variable.

For my analysis below I've removed all the flag
nonsense which is just cluttering things up for
now and gone with the ENDS_WITH option as

> def fetchFiles(pathToFolder, flag, keyWord):
> 	_pathToFiles = []
> 	_fileNames = []
> 	for dirPath, dirNames, fileNames in os.walk(pathToFolder):
> 			selectedPath = [os.path.join(dirPath,item) for item in fileNames if item.endswith(keyWord)]
> 			_pathToFiles.extend(selectedPath)
> 			selectedFile = [item for item in fileNames if item.endswith(keyWord)]
> 			_fileNames.extend(selectedFile)

You could simplify that by putting the selectedFiles
line before the selectedPath line and use selectedFiles
inside the comprehension.

> 		# Try to remove empty entries if none of the required files are in directory
> 		try:
> 			_pathToFiles.remove('')
> 			_imageFiles.remove('')

It would probably be better to check if they were empty
before putting them in. Since you use the endswith() test
I'm thinking there should never be any empty ones in
this scenario anyway?

> 		except ValueError:
> 			pass

>         #return _pathToFiles, _fileNames

Here is the missing return statement but it's not returning
what you said you wanted, ie str7

>         #print _pathToFiles, _fileNames
>         print 'path to first tuple file is:', _pathToFiles [0]
>         str1 = ' '.join(_pathToFiles [0]) #convert tuple element 0 to string
>         print 'length of str1 is: ', len (str1)

It might be wise to print the string itself to check
you have what you want, I'm not sure you do... But
I'm not really sure what you want since your code
logic is confusing me a bit here.

>         str2 = str1.replace(" ", "") #remove white spaces

So why did you add it above? Why not just use an empty
string in the join?

However, more seriously, what affect does this have
on any paths/filenames that you found with spaces
in them? Is that really what you want?

>         print 'str2 is', str2
>         str3 = str2[13:16] #extract rgeistration
>         print 'str3 is registration:', str3

I'll assume this is right since I've no idea what
format you think your filenames have. However, in general,
relying on fixed positions within a string is not a good
idea. This might be a valid case for using a regex which
can more flexibly match your pattern. But for now just
stick with the simple fixed values...

>         str4 = 'FLDAT'
>         print 'str4 is: ', str4
>         str5 = str3.__add__(str4)

You shouldn't really call the dunder methods directly you
should use the + operator:

str5 = str3 + str4

Or, in this case, save a variable and use the literal:

str5 = str3 + 'FLDAT'

>         print 'str 5 is: ',str5
>         str6 = 'H://'
>         print 'str6 is: ', str5

Did you really mean that? You've already printed str5.
And do you really need a double slash after the
drive letter? That's usually only needed if using
backslashes ('H:\\').

>         str7 = str6.__add__(str5)

Again you could just use the literals:

str7 = 'H://' + str3 = 'FLDAT'

>         print 'str7 is: ', str7  
> fetchFiles('H://','ENDS_WITH','.FLD')

No assignment of any return value here

> #### part_2 #### copying files from sourcePath to destPath

> sourcePath = r'str7'

This assigns the literal string 'str7' is that what you want?
You cannot access the variable str7 that was inside the function.
It was a local variable and will have been destroyed by now.

> print 'Source path is: ', sourcePath
> destPath = r'c://test_o/'
> print 'Destination path is: ', destPath
> for root, dirs, files in os.walk(sourcePath):
>     #figure out where we're going
>     dest = destPath + root.replace(sourcePath, '')
>     #if we're in a directory that doesn't exist in the destination folder
>     #then create a new folder
>     if not os.path.isdir(dest):
>         os.mkdir(dest)
>         print 'Directory created at: ' + dest
>     else:
>         print 'Directory already exists:' + dest
> for root, dirs, files in os.walk(sourcePath):
>         #figure out where we're going
>     dest = destPath + root.replace(sourcePath, '')
>     filetype = '.FLD'# name the file ext to be copied    
>     print 'All files of this type will be copied', filetype    
>     #loop through all files in the directory
>     for f in files:
>         #compute current (old) & new file locations
>         oldLoc = root + '\\' + f
>         newLoc = dest + '\\' + f
>         #print 'Old location is:', oldLoc
>         #print 'New location is:', newLoc
>         if not os.path.isfile(newLoc):
>             try:
>                 filename, file_ext = os.path.splitext(oldLoc)
>                 print 'filename is:', filename
>                 print 'file ext is', file_ext
>                 if file_ext == filetype:
>                     shutil.copy2(oldLoc, newLoc)
>                     print 'File ' + f + ' copied.'

You can save some work by using comma separation:

    print 'File', f,'copied.'

No spaces needed and no string additions going on.

>                 else:
>                     print 'File ' + f + ' not copied'
>             except IOError:
>                 print 'file "' + f + '" already exists'
> ####################
> Below is the output:
> ####################
> Python 2.7.11 |Anaconda 4.0.0 (64-bit)| (default, Feb 16 2016, 09:58:36) [MSC v.1500 64 bit (AMD64)] on win32
> Type "help", "copyright", "credits" or "license" for more information.
> Anaconda is brought to you by Continuum Analytics.
> Please check out: http://continuum.io/thanks and https://anaconda.org
>>>> runfile('D:/university_2/my_code_2/2017_01_02/get_reg_&_copy_file.py', wdir='D:/university_2/my_code_2/2017_01_02')
> path to first tuple file is: H://TCVFLDAT\TCV00000.FLD

Note you have a mix of // and \ as separators. Probably
better to just use one.

> length of str1 is:  49
> str2 is H://TCVFLDAT\TCV00000.FLD
> str3 is registration: TCV
> str4 is:  FLDAT
> str 5 is:  TCVFLDAT
> str6 is:  TCVFLDAT
> str7 is:  H://TCVFLDAT
> Number of files found:  21
> Source path is:  str7
> Destination path is:  c://test_o/

Alan G
Author of the Learn to Program web site
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