I'm trying to generate some .mp3 files, but i don't have control where they are saved! (basically they're saved on the .py's location).
I want to change the save location, it's possible?
# -*- coding: <UTF-8> -*-
from gtts import gTTS as gtts
from datetime import datetime as dt
#.MP3 FUNCTIONS FROM GTTS , THIS IS THE STANDARD
def audio_br(words, mp3name, language="pt"):
teste = gtts(text=words,lang=language)
teste.save("%s.mp3" % mp3name)
#GET THE TIME TO CREATE MY VARIABLE TO BE TURN ON MP3
tempo = dt.now()
begin = "Olá! são " + str(tempo.hour) + " horas e "+ str(tempo.minute) + " minutos"
print begin
#GENERATE MY .MP3
audio_br(begin,"A_BEGIN")
The problem resides in the fact that the read from gtts don't talk about save path location.
EDIT 1: Sorry for forgiving the code
# -*- coding: <UTF-8> -*-
from gtts import gTTS as gtts
from datetime import datetime as dt
import os
#.MP3 FUNCTIONS FROM GTTS , THIS IS THE STANDARD
def audio_br(words, mp3name, language="pt"):
teste = gtts(text=words,lang=language)
teste.save("%s.mp3" % os.path.join(<desired folder>,mp3name))
You need to call teste.save with the complete path to where you want the files to go. By default, if no path is provided, python writes files to the current working directory, which is the folder where the script is called from.
Using os.path.join is best b/c it joins paths using the system path separator, making your code more platform independent.
Related
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 have been experimenting with python by creating some programs .The thing is, I have no idea how to import something OUT of the default python directory.
OK
So I did some heavy research and the conclusion is
if u want to access a file saved at different location
use
f = open('E:/somedir/somefile.txt', 'r')
r = f.read()
NOTE: Dont use '\' that were I went wrong.Our system addresses uses '\' So be careful
If you need to just read in a file and not import a module the documentation covers this extensively.
https://docs.python.org/2/tutorial/inputoutput.html#reading-and-writing-files
Specifically for Windows file systems you will need to do one of the following:
1.) Use forwardslashes vs backslashes. This should work with most OSes.
f = open("c:/somedir/somefile.txt", "r")
2.) Use a raw string.
f = open(r"c:\somedir\somefile.txt", "r")
3.) Escape the backslashes.
f = open("c:\\somedir\\somefile.txt", "r")
If you need to import a module to use in your program from outside your programs directory you can use the below information.
Python looks in the sys.path to see if the module exists there and if so does the import. If the path where you files/modules are located is not in the sys.path, Python will raise an ImportError. You can update the path programmatically by using the sys module.
import sys
dir = "path to mymodule"
if dir not in sys.path:
sys.path.append(dir)
import mymodule
You can check the current sys.path by using:
print(sys.path)
Example:
>>> print(sys.path)
['', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python34.zip', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/plat-darwin', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/lib-dynload', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages']
>>> sys.path.append("/Users/ddrummond/pymodules")
>>> print(sys.path)
['', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python34.zip', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/plat-darwin', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/lib-dynload', '/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages', '/Users/ddrummond/pymodules']
>>>
You can see that sys.path now contains '/Users/ddrummond/pymodules'.
I can't make output in Russian language only output of Unicode=(
I use Pythonv.2.7.9
Microsoft 8
How I can do that with list?
#! /usr/bin/env python
# -*- coding: utf-8 -*-
import requests
from bs4 import BeautifulSoup
r = requests.get("http://fs.to/video/films/group/film_genre/")
response = r.content.decode('utf-8')
page = BeautifulSoup(response)
for tag in page.findAll('li'):
a = tag.find('a')
for b in a.contents:
print (u'{0}'.format(u'○'),unicode(b.string))
Example of output must be like:
Аниме
Биография
...
Фэнтези
Эротика
Change the last line to:
print (u'{0}'.format(u'○'),b.string
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)