Feel like a dunce. I'm trying to interact with a zip file and can't seem to use the zipfile library. Fairly new to python
from zipfile import *
#set filename
fpath = '{}_{}_{}.zip'.format(strDate, day, week)
#use zipfile to get info about ftp file
zip = zipfile.Zipfile(fpath, mode='r')
# doesn't matter if I use
#zip = zipfile.Zipfile(fpath, mode='w')
#or zip = zipfile.Zipfile(fpath, 'wb')
I'm getting this error
zip = zipfile.Zipfile(fpath, mode='r')
NameError: name 'zipfile' is not defined
if I just use import zipfile I get this error:
TypeError: 'module' object is not callable
Two ways to fix it:
1) use from, and in that case drop the zipfile namespace:
from zipfile import *
#set filename
fpath = '{}_{}_{}.zip'.format(strDate, day, week)
#use zipfile to get info about ftp file
zip = ZipFile(fpath, mode='r')
2) use direct import, and in that case use full path like you did:
import zipfile
#set filename
fpath = '{}_{}_{}.zip'.format(strDate, day, week)
#use zipfile to get info about ftp file
zip = zipfile.ZipFile(fpath, mode='r')
and there's a sneaky typo in your code: Zipfile should be ZipFile (capital F, so I feel slightly bad for answering...
So the lesson learnt is:
avoid from x import y because editors have a harder time to complete words
with a proper import zipfile and an editor which proposes completion, you would never have had this problem in the first place.
Easiest way to zip a file using Python:
import zipfile
zf = zipfile.ZipFile("targetZipFileName.zip",'w', compression=zipfile.ZIP_DEFLATED)
zf.write("FileTobeZipped.txt")
zf.close()
Related
I'm trying to read all yaml files in a directory, but I am having trouble. First, because I am using Python 2.7 (and I cannot change to 3) and all of my files are utf-8 (and I also need them to keep this way).
import os
import yaml
import codecs
def yaml_reader(filepath):
with codecs.open(filepath, "r", encoding='utf-8') as file_descriptor:
data = yaml.load_all(file_descriptor)
return data
def yaml_dump(filepath, data):
with open(filepath, 'w') as file_descriptor:
yaml.dump(data, file_descriptor)
if __name__ == "__main__":
filepath = os.listdir(os.getcwd())
data = yaml_reader(filepath)
print data
When I run this code, python gives me the message:
TypeError: coercing to Unicode: need string or buffer, list found.
I want this program to show the content of the files. Can anyone help me?
I guess the issue is with filepath.
os.listdir(os.getcwd()) returns the list of all the files in the directory. so you are passing the list to codecs.open() instead of filename
There are multiple problems with your code, apart from that it is invalide Python, in the way you formatted this.
def yaml_reader(filepath):
with codecs.open(filepath, "r", encoding='utf-8') as file_descriptor:
data = yaml.load_all(file_descriptor)
return data
however it is not necessary to do the decoding, PyYAML is perfectly capable of processing UTF-8:
def yaml_reader(filepath):
with open(filepath, "rb") as file_descriptor:
data = yaml.load_all(file_descriptor)
return data
I hope you realise your trying to load multiple documents and always get a list as a result in data even if your file contains one document.
Then the line:
filepath = os.listdir(os.getcwd())
gives you a list of files, so you need to do:
filepath = os.listdir(os.getcwd())[0]
or decide in some other way, which of the files you want to open. If you want to combine all files (assuming they are YAML) in one big YAML file, you need to do:
if __name__ == "__main__":
data = []
for filepath in os.listdir(os.getcwd()):
data.extend(yaml_reader(filepath))
print data
And your dump routine would need to change to:
def yaml_dump(filepath, data):
with open(filepath, 'wb') as file_descriptor:
yaml.dump(data, file_descriptor, allow_unicode=True, encoding='utf-8')
However this all brings you to the biggest problem: that you are using PyYAML, that will mangle your YAML, dropping flow-style, comment, anchor names, special int/float, quotes around scalars etc. Apart from that PyYAML has not been updated to support YAML 1.2 documents (which has been the standard since 2009). I recommend you switch to using ruamel.yaml (disclaimer: I am the author of that package), which supports YAML 1.2 and leaves comments etc in place.
And even if you are bound to use Python 2, you should use the Python 3 like syntax e.g. for print that you can get with from __future__ imports.
So I recommend you do:
pip install pathlib2 ruamel.yaml
and then use:
from __future__ import absolute_import, unicode_literals, print_function
from pathlib import Path
from ruamel.yaml import YAML
if __name__ == "__main__":
data = []
yaml = YAML()
yaml.preserve_quotes = True
for filepath in Path('.').glob('*.yaml'):
data.extend(yaml.load_all(filepath))
print(data)
yaml.dump(data, Path('your_output.yaml'))
I'm new to this, and trying to download a snappy.parquet file from Amazon s3 I can later convert to CSV file.
I tried working with the following example I've found online, and I get an empty folder. can anyone please help me?
import boto
import sys, os
from boto.s3.key import Key
from boto.exception import S3ResponseError
DOWNLOAD_LOCATION_PATH =""
BUCKET_NAME = ""
AWS_ACCESS_KEY_ID= ""
AWS_ACCESS_SECRET_KEY = ""
conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_ACCESS_SECRET_KEY)
bucket = conn.get_bucket(BUCKET_NAME)
#goto through the list of files
bucket_list = bucket.list()
for l in bucket_list:
key_string = str(l.key)
s3_path = DOWNLOAD_LOCATION_PATH + key_string
try:
print ("Current File is ", s3_path)
l.get_contents_to_filename(s3_path)
except (OSError, S3ResponseError) as e:
pass
# check if the file has been downloaded locally
if not os.path.exists(s3_path):
try:
os.makedirs(s3_path)
except OSError as exc:
# let guard againts race conditions
import errno
if exc.errno != errno.EEXIST:
raise
The script you are using appears to recursively download the contents of the specified S3 bucket (BUCKET_NAME) to the specified local directory (DOWNLOAD_LOCATION_PATH). FWIW, I notice this script looks like it comes from here.
The "Current File is ..." output line should show you the progress of these files being written. One problem you might be having is due to this line:
s3_path = DOWNLOAD_LOCATION_PATH + key_string
If you had specified DOWNLOAD_LOCATION_PATH at the top as a directory without a trailing '/' character, e.g. like this:
DOWNLOAD_LOCATION_PATH = '/tmp/my_dir'
then the files being downloaded would be written not underneath the /tmp/my_dir directory, but directly in /tmp/ with a my_dir prefix on each filename! You can fix this by changing this line to:
s3_path = os.path.join(DOWNLOAD_LOCATION_PATH, key_string)
Other than that, the script appears to work alright. You may want to add this line at the very top:
from __future__ import print_function
if you are still using Python 2.x, otherwise the print output will look a bit odd (print will think you are printing a 2-Tuple).
Your question also makes it sound like you really only want/need to download a single file from the bucket -- if so, this isn't really a great script to be using, since it's downloading everything.
Hello sorry if this question has been asked before.
But I have tried a lot of methods that provided.
Basically, I want to download the file from a website, which is I will show my coding below. The code works perfectly, but the problem is the file was auto download in our download folder path directory.
My concern is to download the file and save it to a specific folder.
I'm aware we can change our browser setting since this was a server that will remote by different users. So, it will automatically download to their temporarily /users/adam_01/download/ folder.
I want it to save in server disk which is, C://ExcelFile/
Below are my script and some of the data have been changing because it is confidential.
import pandas as pd
import html5lib
import time from bs4
import BeautifulSoup
import requests
import csv
from datetime
import datetime
import urllib.request
import os
with requests.Session() as c:
proxies = {"http": "http://:911"}
url = 'https://......./login.jsp'
USERNAME = 'mwirzonw'
PASSWORD = 'Fiqr123'
c.get(url,verify= False)
csrftoken = ''
login_data = dict(proxies,atl_token = csrftoken, os_username=USERNAME, os_password=PASSWORD, next='/')
c.post(url, data=login_data, headers={"referer" : "https://.....com"})
page = c.get('https://........s...../SearchRequest-96010.csv')
location = 'C:/Users/..../Downloads/'
with open('asdsad906010.csv', 'wb') as output:
output.write(page.content )
print("Done!")
Thank you, be pleased to ask if any confusing information was given.
Regards,
Fiqri
It seems that from your script you are writing the file to asdsad906010.csv. You should be able to change the output directory as follows.
# Set the output directory to your desired location
output_directory = 'C:/ExcelFile/'
# Create a file path by joining the directory name with the desired file name
file_path = os.path.join(output_directory, 'asdsad906010.csv')
# Write the file
with open(file_path, 'wb') as output:
output.write(page.content)
right so I'm working on a Python script (Python 2.7) that will extract the metadata from OLE files. I am using OleFileIO_PL and it work perfectly file with OLE files 97 - 2003, but any later then that it just says that it is not an OLE2 file type.
Any way I can modify my code to support both .doc and .docx ? Same with .ppt and .pptx etc.
Thank you in advance
Source Code:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import OleFileIO_PL
import StringIO
import optparse
import sys
import os
def printMetadata(fileName):
data = open(fileName, 'rb').read()
f = StringIO.StringIO(data)
OLEFile = OleFileIO_PL.OleFileIO(f)
meta = OLEFile.get_metadata()
print('Author:', meta.author)
print('Title:', meta.title)
print('Creation date:', meta.create_time)
meta.dump()
OLEFile.close()
def main():
parser = optparse.OptionParser('usage = -F + Name of the OLE file with the extention For example: python Ms Office Metadata Extraction Script.py -F myfile.docx ')
parser.add_option('-F', dest='fileName', type='string',\
help='specify OLE (MS Office) file name')
(options, args) = parser.parse_args()
fileName = options.fileName
if fileName == None:
print parser.usage
exit(0)
else:
printMetadata(fileName)
if __name__ == '__main__':
main()
To answer your question, this is because the newer MS Office 2007+ files (docx, xlsx, xlsb, pptx, etc) have a completely different structure from the legacy MS Office 97-2003 formats.
It is mainly a collection of XML files within a Zip archive. So with a little bit of work, you can extract everything you need using zipfile and ElementTree from the standard library.
If openxmllib does not work for you, you may try other solutions:
officedissector: https://www.officedissector.com/
python-opc: https://pypi.python.org/pypi/python-opc
openpack: https://pypi.python.org/pypi/openpack
paradocx: https://pypi.python.org/pypi/paradocx
BTW, OleFileIO_PL has been renamed to olefile, and the new project page is https://github.com/decalage2/olefile
I'm trying to get data from a zipped csv file. Is there a way to do this without unzipping the whole files? If not, how can I unzip the files and read them efficiently?
I used the zipfile module to import the ZIP directly to pandas dataframe.
Let's say the file name is "intfile" and it's in .zip named "THEZIPFILE":
import pandas as pd
import zipfile
zf = zipfile.ZipFile('C:/Users/Desktop/THEZIPFILE.zip')
df = pd.read_csv(zf.open('intfile.csv'))
If you aren't using Pandas it can be done entirely with the standard lib. Here is Python 3.7 code:
import csv
from io import TextIOWrapper
from zipfile import ZipFile
with ZipFile('yourfile.zip') as zf:
with zf.open('your_csv_inside_zip.csv', 'r') as infile:
reader = csv.reader(TextIOWrapper(infile, 'utf-8'))
for row in reader:
# process the CSV here
print(row)
A quick solution can be using below code!
import pandas as pd
#pandas support zip file reads
df = pd.read_csv("/path/to/file.csv.zip")
zipfile also supports the with statement.
So adding onto yaron's answer of using pandas:
with zipfile.ZipFile('file.zip') as zip:
with zip.open('file.csv') as myZip:
df = pd.read_csv(myZip)
Thought Yaron had the best answer but thought I would add a code that iterated through multiple files inside a zip folder. It will then append the results:
import os
import pandas as pd
import zipfile
curDir = os.getcwd()
zf = zipfile.ZipFile(curDir + '/targetfolder.zip')
text_files = zf.infolist()
list_ = []
print ("Uncompressing and reading data... ")
for text_file in text_files:
print(text_file.filename)
df = pd.read_csv(zf.open(text_file.filename)
# do df manipulations
list_.append(df)
df = pd.concat(list_)
Yes. You want the module 'zipfile'
You open the zip file itself with zipfile.ZipInfo([filename[, date_time]])
You can then use ZipFile.infolist() to enumerate each file within the zip, and extract it with ZipFile.open(name[, mode[, pwd]])
this is the simplest thing I always use.
import pandas as pd
df = pd.read_csv("Train.zip",compression='zip')
Supposing you are downloading a zip file that contains a CSV and you don't want to use temporary storage. Here is what a sample implementation looks like:
#!/usr/bin/env python3
from csv import DictReader
from io import TextIOWrapper, BytesIO
from zipfile import ZipFile
import requests
def all_tickers():
url = "https://simfin.com/api/bulk/bulk.php?dataset=industries&variant=null"
r = requests.get(url)
zip_ref = ZipFile(BytesIO(r.content))
for name in zip_ref.namelist():
print(name)
with zip_ref.open(name) as file_contents:
reader = DictReader(TextIOWrapper(file_contents, 'utf-8'), delimiter=';')
for item in reader:
print(item)
This takes care of all python3 bytes/str issues.
Modern Pandas since version 0.18.1 natively supports compressed csv files: its read_csv method has compression parameter : {'infer', 'gzip', 'bz2', 'zip', 'xz', None}, default 'infer'.
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html
If you have a file name: my_big_file.csv and you zip it with the same name my_big_file.zip
you may simply do this:
df = pd.read_csv("my_big_file.zip")
Note: check your pandas version first (not applicable for older versions)