topic Re: Parsing list object file paths for file name in Python Questions

Parsing list object file paths for file name

JohnLay — Fri, 18 Jan 2013 14:18:19 GMT

I am trying to automate a naming process that takes the file name from a list of file paths to name other folders, geodatabases, zipfiles, etc.

The problem I'm running into is that, I assume, I am trying to run string functions on list object and I can't figure out how to get the script to do what I need it to do.

For example, I have the file path C:\TEMP\Durham.shp. I want to name a geodatabase, folder, and zipfile "Durham". Instead I get a folder and geodatabase named "C".

Can someone take a look at my script and tel me what I'm doing wrong?

The following script is that last incarnation of several attempts.

import arcpy, os import zipfile from arcpy import env from os.path import basename  BASELAYERS = arcpy.GetParameterAsText(0) CLIPLAYERS = arcpy.GetParameterAsText(1) OUTFOLDERMASTER = arcpy.GetParameterAsText(2) FILETYPES = ["*.shp", "*.dbf", "*.shx", "*.prj", "*.sbn", "*.sbx", "*.xml"]  env.workspace = OUTFOLDERMASTER for LAYERS in CLIPLAYERS:     LAYERSSTR = ''.join(LAYERS)     BASENAME = LAYERSSTR.split("\\")[-1]     FOLDER = os.path.join(OUTFOLDERMASTER, BASENAME)     os.makedirs(FOLDER)     PGDBNAME = "NCFlood_Effective_" + BASENAME + "_PGDB.mdb"     arcpy.CreatePersonalGDB_management(FOLDER, PGDBNAME, "9.3")     PGDB = os.path.join(FOLDER,PGDBNAME)     for FILES in BASELAYERS:         FEATURENAMESTR = ''.join(FILES)         FEATURENAME = os.path.basename(FEATURENAMESTR)         SHAPEFILENAME = FEATURENAME + ".shp"         SHAPEFILE = os.path.join(FOLDER, SHAPEFILENAME)         arcpy.Clip_analysis(FILES, LAYERS, FEATURENAME)         arcpy.Clip_analysis(FILES, LAYERS, SHAPEFILE)     ZIPFILELIST = []             for TYPE in FILETYPES:         ZIPFILELIST.extend(glob.glob(FOLDER))     ZIPFILENAME = os.path.join(FOLDER, "NCFlood_Effective_" + FOLDERNAME + "_SHP.zip")     ARCHIVE = zipfile.ZipFile(ZIPFILENAME, "w")     for ZIP in ZIPFILELIST:         ARCHIVE.write(ZIP, os.path.basename(ZIP), zipfile.ZIP_DEFLATED)     ARCHIVE.close()

I have also tried

    BASENAME = os.path.basename(LAYERS)     FOLDERNAME = os.path.splitext(BASENAME)[0]

Re: Parsing list object file paths for file name

T__WayneWhitley — Fri, 18 Jan 2013 15:44:51 GMT

Well you could use simple variations of the split python functions: 'rsplit', 'split', or, say 'splitext'

So using the string you provided, you want to extract the text 'Durham':
C:\TEMP\Durham.shp

Then, could do it as such:

import os
text = r'C:\TEMP\Durham.shp'
text = text.rsplit(os.sep, 1)[1]
text = os.path.splitext(text)[0]

# or in a single line:
text = os.path.splitext(text.rsplit(os.sep, 1)[1])[0]

EDIT:
Then, for 'assembling' new pathnames, you may find this useful:

newText = r'\\someServer\aFolder\someGDB.gdb'
newFCpath = os.path.join(newText, text)

...which results in the newFCpath string:
\\someServer\aFolder\someGDB.gdb\Durham

Hope that helps.
Wayne

Re: Parsing list object file paths for file name

PhilMorefield — Sat, 11 Dec 2021 21:35:42 GMT

Python is pretty good about recognizing and handling path names. Here's how I handle tasks like yours:

>>>from os import path
>>>text = r'C:\TEMP\Durham.shp'
>>>path.dirname(text)
'C:\\TEMP'
>>>path.basename(text)
'Durham.shp'
>>>newfolderName = path.basename(text)[-4:]
>>>newfolderName
'Durham'
>>>newPath = path.join('C:\\newLocation', newfolderName)
>>>newPath
'C:\\newLocation\\Durham'

Re: Parsing list object file paths for file name

JohnLay — Wed, 23 Jan 2013 10:52:15 GMT

Thanks guys! I found a solution to my issue, but I'm going to have to rewrite the script to handle SDE database pathnames, so I will give these options a shot.