I am using python to create a program that takes attendance. I want it to create a new text document each day that is saved as the date of that day. I know how to write into a file that already exists, but do not know how to create a new one through python. I also do not know how to make it create a new one each new day. Thanks guys!
This can be done easily, you just need to provide the date as filename which can be calculated like
import time
timestr = time.strftime("%Y%m%d-%H%M%S")
with open(timestr, 'w+') as f:
f.write('XXXXX')
# Write what further needed
A file will be created with the proper timestamp.
Related
I frequent a real estate website that shows recent transactions, from which I will download data to parse within a Pandas dataframe. Everything about this dataset remains identical every time I download it (regarding the column names, that is).
The name of the Excel output may change, though. For example, if I already have download a few of these in my Downloads folder, the file that's exported may read "Generic_File_(3)" or "Generic_File_(21)" if I already have a few older "Generic_File" exports in that folder from a previous export.
Ideally, I'd like my workflow to look like this: export this Excel file of real estate sales, then run a Python script to read in the most recent export as a Pandas dataframe. The catch is, I don't want to have to go in and change the filename in the script to match the appending number of the Excel export everytime. I want the pd.read_excel method to simply read the "Generic_File" that is appended with the largest number (which will obviously correspond to the most rent export).
I suppose I could always just delete old exports out of my Downloads folder so the newest, freshest export is always named the same ("Generic_File", in this case), but I'm looking for a way to ensure I don't have to do this. Are wildcards the best path forward, or is there some other method to always read in the most recently downloaded Excel file from my Downloads folder?
I would use the OS package and create a method to read to file names in the downloads folder. Parsing string filenames you could then find the file following your specified format with the highest copy number. Something like the following might help you get started.
import os
downloads = os.listdir('C:/Users/[username here]/Downloads/')
is_file = [True if '.' in item else False for item in downloads]
files = [item for keep, item in zip(is_file, downloads) if keep]
** INSERT CODE HERE TO IDENTIFY THE FILE OF INTEREST **
Regex might be the best way to find matches if you have a diverse listing of files in your downloads folder.
So, I am looking to have a Lambda function (python) that will pick up the
file name
date & time of creation
of the last created file in that directory (having certain suffix)
My use-case requires something like this:
import boto3
def lambda_handler(event, context):
S3Path="s3://my_bucket/my_dir/"
file_name = get_file_name()
created_datetime = get_datetime()
if (current_datetime-created_datetime < 48 hours)
#do something with "file_name"
Please help with the above required functions
It is not possible to "get the last created file".
You will need to list the objects with that Prefix, then programmatically determine the latest object by sorting the list. You can also add some code to check the Suffix of the Key.
Give it a try and let us know if you have any specific issues.
I have trouble trying to keep the sharing properties of an excel. I tried this :Python and openpyxl is saving my shared workbook as unshared but the part of vout just cancels all the modification I made with the script
To explain the problem :
There's an excel file that is shared in which people can do some modification
Python reads and writes on it
When I save the workbook in the excel file, it automatically either drops the sharing property or when I try to keep it, it just doesn't do any modification
Can someone help me please ?
I'll get a little more precise, as requested.
The sharing mode is the one Microsoft provides. You can see the button below:
Share button Excel
The excel is stored on a server. Several users can write on it at the same time but when I launch my script, it stops automatically the sharing property, so everyone that is writing on it just can't do modification anymore and every modif they did is lost.
First I treated my Excel normally :
DLT=openpyxl.load_workbook(myPath)
ws=DLT['DLT']
...my modifications on ws...
DLT.save()
DLT.close()
But then I tried this (Python and openpyxl is saving my shared workbook as unshared)
DLT=openpyxl.load_workbook(myPath)
ws=DLT['DLT']
zin = zipfile.ZipFile(myPath, 'r')
buffers = []
for item in zin.infolist():
buffers.append((item, zin.read(item.filename)))
zin.close()
...my modif on ws...
DLT.save()
zout = zipfile.ZipFile(myPath, 'w')
for item, buffer in buffers:
zout.writestr(item, buffer)
zout.close()
DLT.close()
The second one just doesn't save my modification on ws.
The thing I would like to do, is not to get rid of the sharing property. I would need to keep it while I write on it. Not sure if it is possible. I have one alternative solution that is to use another file, and just copy/paste by hand the new data from this file to the DLT one.
well... after playing with it back and forth, for some weird reason zipfile.infolist() does contains the sheet data as well, so here's my way to fine tune it, using the shared_pyxl_save example the previous gentleman provided
basically instead of letting the old file overriding the sheet's data, use the old one
def shared_pyxl_save(file_path, workbook):
"""
`file_path`: path to the shared file you want to save
`workbook`: the object returned by openpyxl.load_workbook()
"""
zin = zipfile.ZipFile(file_path, 'r')
buffers = []
for item in zin.infolist():
if "sheet1.xml" not in item.filename:
buffers.append((item, zin.read(item.filename)))
zin.close()
workbook.save(file_path)
""" loop through again to find the sheet1.xmls and put it into buffer, else will show up error"""
zin2 = zipfile.ZipFile(file_path, 'r')
for item in zin2.infolist():
if "sheet1.xml" in item.filename:
buffers.append((item, zin2.read(item.filename)))
zin2.close()
#finally saves the file
zout = zipfile.ZipFile(file_path, 'w')
for item, buffer in buffers:
zout.writestr(item, buffer)
zout.close()
workbook.close()
I have a list of dataframes (df_cleaned) created from multiple csv files chosen by the user.
My objective is to save each dataframe within the df_cleaned list as a separate csv file locally.
I have the following code done which saves the file with its original title. But I see that it overwrites and manages to save a copy of only the last dataframe.
How can I fix it? According to my very basic knowledge perhaps I could use a break-continue statement in the loop? But I do not know how to implement it correctly.
for i in range(len(df_cleaned)):
outputFile = df_cleaned[i].to_csv(r'C:\...\Data Docs\TrainData\{}.csv'.format(name))
print('Saving of files as csv is complete.')
You can create a different name for each file, as an example in the following I attach the index to name:
for i in range(len(df_cleaned)):
outputFile = df_cleaned[i].to_csv(r'C:\...\Data Docs\TrainData\{0}_{1}.csv'.format(name,i))
print('Saving of files as csv is complete.')
this will create a list of files named <name>_N.csv with N = 0, ..., len(df_cleaned)-1.
A very easy way of solving. Just figured out the answer myself. Posting to help someone else.
fileNames is a list I created at the start of the code to save the
names of the files chosen by the user.
for i in range(len(df_cleaned)):
outputFile = df_cleaned[i].to_csv(r'C:\...\TrainData\{}.csv'.format(fileNames[i]))
print('Saving of files as csv is complete.')
Saves a separate copy for each file in the defined directory.
I have a python code and I am saving the results in a destination with specific file name, this file name will change every time and it is a recurring event.
Here is my code:
import csv
outfile=open('path.macrovariable.csv','w',newline='')
writer=csv.writer(outfile)
writer.writerow(["Jobname","employement","company","descrption","location"])
writer.writerows([job_name])
writer.writerows([emplomnt_type])
writer.writerows([organisation])
writer.writerows([job_descrption])
writer.writerows([job_location])
In the outfile the file name here as "macrovariable" will change every time. I want to create a macrovariable at the top of the program which will be called later in the program in the place of "macrovariable" instead of hardcoding.
Thanks & regards,
Sanjay.