Unzip all zipped files in a folder to that same folder using Python 2.7.5
Question:
I would like to write a simple script to iterate through all the files in a folder and unzip those that are zipped (.zip) to that same folder. For this project, I have a folder with nearly 100 zipped .las files and I’m hoping for an easy way to batch unzip them. I tried with following script
import os, zipfile
folder = 'D:/GISData/LiDAR/SomeFolder'
extension = ".zip"
for item in os.listdir(folder):
if item.endswith(extension):
zipfile.ZipFile.extract(item)
However, when I run the script, I get the following error:
Traceback (most recent call last):
File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 10, in <module>
extract = zipfile.ZipFile.extract(item)
TypeError: unbound method extract() must be called with ZipFile instance as first argument (got str instance instead)
I am using the python 2.7.5 interpreter. I looked at the documentation for the zipfile module (https://docs.python.org/2/library/zipfile.html#module-zipfile) and I would like to understand what I’m doing incorrectly.
I guess in my mind, the process would go something like this:
- Get folder name
- Loop through folder and find zip files
- Extract zip files to folder
Thanks Marcus, however, when implementing the suggestion, I get another error:
Traceback (most recent call last):
File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 12, in <module>
zipfile.ZipFile(item).extract()
File "C:Python27ArcGIS10.2libzipfile.py", line 752, in __init__
self.fp = open(file, modeDict[mode])
IOError: [Errno 2] No such file or directory: 'JeffCity_0752.las.zip'
When I use print statements, I can see that the files are in there. For example:
for item in os.listdir(folder):
if item.endswith(extension):
print os.path.abspath(item)
filename = os.path.basename(item)
print filename
yields:
D:GISDataToolsMO_ToolsJeffCity_0752.las.zip
JeffCity_0752.las.zip
D:GISDataToolsMO_ToolsJeffCity_0753.las.zip
JeffCity_0753.las.zip
As I understand the documentation,
zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])
Open a ZIP file, where file can be either a path to a file (a string) or a file-like object
It appears to me like everything is present and accounted for. I just don’t understand what I’m doing wrong.
Any suggestions?
Thank You
Answers:
You need to construct a ZipFile object with the filename, and then extract it:
zipfile.ZipFile.extract(item)
is wrong.
zipfile.ZipFile(item).extractall()
will extract all files from the zip file with the name contained in item.
I think you should more closely read the documentation to zipfile 🙂 but you’re on the right track!
Below is the code that worked for me:
import os, zipfile
dir_name = 'C:\SomeDirectory'
extension = ".zip"
os.chdir(dir_name) # change directory from working dir to dir with files
for item in os.listdir(dir_name): # loop through items in dir
if item.endswith(extension): # check for ".zip" extension
file_name = os.path.abspath(item) # get full path of files
zip_ref = zipfile.ZipFile(file_name) # create zipfile object
zip_ref.extractall(dir_name) # extract file to dir
zip_ref.close() # close file
os.remove(file_name) # delete zipped file
Looking back at the code I had amended, the directory was getting confused with the directory of the script.
The following also works while not ruining the working directory. First remove the line
os.chdir(dir_name) # change directory from working dir to dir with files
Then assign file_name as
file_name = dir_name + "/" + item
The accepted answer works great!
Just to extend the idea to unzip all the files with .zip extension within all the sub-directories inside a directory the following code seems to work well:
import os
import zipfile
for path, dir_list, file_list in os.walk(dir_path):
for file_name in file_list:
if file_name.endswith(".zip"):
abs_file_path = os.path.join(path, file_name)
# The following three lines of code are only useful if
# a. the zip file is to unzipped in it's parent folder and
# b. inside the folder of the same name as the file
parent_path = os.path.split(abs_file_path)[0]
output_folder_name = os.path.splitext(abs_file_path)[0]
output_path = os.path.join(parent_path, output_folder_name)
zip_obj = zipfile.ZipFile(abs_file_path, 'r')
zip_obj.extractall(output_path)
zip_obj.close()
I think this is shorter and worked fine for me. First import the modules required:
import zipfile, os
Then, I define the working directory:
working_directory = 'my_directory'
os.chdir(working_directory)
After that you can use a combination of the os and zipfile to get where you want:
for file in os.listdir(working_directory): # get the list of files
if zipfile.is_zipfile(file): # if it is a zipfile, extract it
with zipfile.ZipFile(file) as item: # treat the file as a zip
item.extractall() # extract it in the working directory
Recursive version of @tpdance answer.
Use this for for subfolders and subfolder. Working on Python 3.8
import os
import zipfile
base_dir = '/Users/john/data' # absolute path to the data folder
extension = ".zip"
os.chdir(base_dir) # change directory from working dir to dir with files
def unpack_all_in_dir(_dir):
for item in os.listdir(_dir): # loop through items in dir
abs_path = os.path.join(_dir, item) # absolute path of dir or file
if item.endswith(extension): # check for ".zip" extension
file_name = os.path.abspath(abs_path) # get full path of file
zip_ref = zipfile.ZipFile(file_name) # create zipfile object
zip_ref.extractall(_dir) # extract file to dir
zip_ref.close() # close file
os.remove(file_name) # delete zipped file
elif os.path.isdir(abs_path):
unpack_all_in_dir(abs_path) # recurse this function with inner folder
unpack_all_in_dir(base_dir)
I would like to write a simple script to iterate through all the files in a folder and unzip those that are zipped (.zip) to that same folder. For this project, I have a folder with nearly 100 zipped .las files and I’m hoping for an easy way to batch unzip them. I tried with following script
import os, zipfile
folder = 'D:/GISData/LiDAR/SomeFolder'
extension = ".zip"
for item in os.listdir(folder):
if item.endswith(extension):
zipfile.ZipFile.extract(item)
However, when I run the script, I get the following error:
Traceback (most recent call last):
File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 10, in <module>
extract = zipfile.ZipFile.extract(item)
TypeError: unbound method extract() must be called with ZipFile instance as first argument (got str instance instead)
I am using the python 2.7.5 interpreter. I looked at the documentation for the zipfile module (https://docs.python.org/2/library/zipfile.html#module-zipfile) and I would like to understand what I’m doing incorrectly.
I guess in my mind, the process would go something like this:
- Get folder name
- Loop through folder and find zip files
- Extract zip files to folder
Thanks Marcus, however, when implementing the suggestion, I get another error:
Traceback (most recent call last):
File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 12, in <module>
zipfile.ZipFile(item).extract()
File "C:Python27ArcGIS10.2libzipfile.py", line 752, in __init__
self.fp = open(file, modeDict[mode])
IOError: [Errno 2] No such file or directory: 'JeffCity_0752.las.zip'
When I use print statements, I can see that the files are in there. For example:
for item in os.listdir(folder):
if item.endswith(extension):
print os.path.abspath(item)
filename = os.path.basename(item)
print filename
yields:
D:GISDataToolsMO_ToolsJeffCity_0752.las.zip
JeffCity_0752.las.zip
D:GISDataToolsMO_ToolsJeffCity_0753.las.zip
JeffCity_0753.las.zip
As I understand the documentation,
zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])
Open a ZIP file, where file can be either a path to a file (a string) or a file-like object
It appears to me like everything is present and accounted for. I just don’t understand what I’m doing wrong.
Any suggestions?
Thank You
You need to construct a ZipFile object with the filename, and then extract it:
zipfile.ZipFile.extract(item)
is wrong.
zipfile.ZipFile(item).extractall()
will extract all files from the zip file with the name contained in item.
I think you should more closely read the documentation to zipfile 🙂 but you’re on the right track!
Below is the code that worked for me:
import os, zipfile
dir_name = 'C:\SomeDirectory'
extension = ".zip"
os.chdir(dir_name) # change directory from working dir to dir with files
for item in os.listdir(dir_name): # loop through items in dir
if item.endswith(extension): # check for ".zip" extension
file_name = os.path.abspath(item) # get full path of files
zip_ref = zipfile.ZipFile(file_name) # create zipfile object
zip_ref.extractall(dir_name) # extract file to dir
zip_ref.close() # close file
os.remove(file_name) # delete zipped file
Looking back at the code I had amended, the directory was getting confused with the directory of the script.
The following also works while not ruining the working directory. First remove the line
os.chdir(dir_name) # change directory from working dir to dir with files
Then assign file_name as
file_name = dir_name + "/" + item
The accepted answer works great!
Just to extend the idea to unzip all the files with .zip extension within all the sub-directories inside a directory the following code seems to work well:
import os
import zipfile
for path, dir_list, file_list in os.walk(dir_path):
for file_name in file_list:
if file_name.endswith(".zip"):
abs_file_path = os.path.join(path, file_name)
# The following three lines of code are only useful if
# a. the zip file is to unzipped in it's parent folder and
# b. inside the folder of the same name as the file
parent_path = os.path.split(abs_file_path)[0]
output_folder_name = os.path.splitext(abs_file_path)[0]
output_path = os.path.join(parent_path, output_folder_name)
zip_obj = zipfile.ZipFile(abs_file_path, 'r')
zip_obj.extractall(output_path)
zip_obj.close()
I think this is shorter and worked fine for me. First import the modules required:
import zipfile, os
Then, I define the working directory:
working_directory = 'my_directory'
os.chdir(working_directory)
After that you can use a combination of the os and zipfile to get where you want:
for file in os.listdir(working_directory): # get the list of files
if zipfile.is_zipfile(file): # if it is a zipfile, extract it
with zipfile.ZipFile(file) as item: # treat the file as a zip
item.extractall() # extract it in the working directory
Recursive version of @tpdance answer.
Use this for for subfolders and subfolder. Working on Python 3.8
import os
import zipfile
base_dir = '/Users/john/data' # absolute path to the data folder
extension = ".zip"
os.chdir(base_dir) # change directory from working dir to dir with files
def unpack_all_in_dir(_dir):
for item in os.listdir(_dir): # loop through items in dir
abs_path = os.path.join(_dir, item) # absolute path of dir or file
if item.endswith(extension): # check for ".zip" extension
file_name = os.path.abspath(abs_path) # get full path of file
zip_ref = zipfile.ZipFile(file_name) # create zipfile object
zip_ref.extractall(_dir) # extract file to dir
zip_ref.close() # close file
os.remove(file_name) # delete zipped file
elif os.path.isdir(abs_path):
unpack_all_in_dir(abs_path) # recurse this function with inner folder
unpack_all_in_dir(base_dir)