I am doing some processing on wave audio files using Tensorflow
and saving them using the tf.print with output_stream option.
pcm =contrib_audio.encode_wav(processed_audio,16000)
tf.print(output_stream="file:///tmp/test.wav",summarize=-1)
The problem is that I am not able to change value of /tmp/test.wav dynamically
so that multiple wave files are stored.
Kindly refer to the below code.
# Using a counter
for i in range(1,10):
fname = "test_"+str(i)+".wav" #filename
path = "//content/sample_data/" #path to save
fname = "file://{path}{fname}".format(fname=fname, path = path)
tf.print(output_stream=fname,summarize=-1)
You can create a dynamic text, for it to be a unique filename.
Related
I'm trying to create a python program that allows you to dynamically choose the folder to save the download.
'outtmpl' lets you choose an output as a simple path such that
'e:/python/downloadedsongs/%(title)s.%(ext)s'
is a valid path and will let you save to the downloadedsongs folder
But With the function that defines the choose folder button
def openLocation():
global Folder_Name
Folder_Name = filedialog.askdirectory()
print(Folder_Name)
why does
f'{Folder_name}/%(title)s.%(ext)s'
not return a valid path, the program outright ignores the variable. I thought at first i was having a fstring issue because %s was pythons old format for f'strings but I have also tried these variances with no success
'outtmpl': Folder_Name + '/' + '%(title)s.%(ext)s', (no errors but cannot find output)
'outtmpl': '%(Folder_Name)s%(title)s.%(ext)s',
%()s is the old version of fstrings I was conflating the two.
answer ended up looking like 'outtmpl': Folder + '/%(title)s.%(ext)s'
My goal is to do following:
I am using Win 10 and I have files like so:
folder
2020-04-23_19-30-52_UTC.mp4
2020-04-23_19-30-52_UTC.txt which contains string "This video is me at a wedding"
2020-05-25_19-30-52_UTC.mp4
2020-05-25_19-30-52_UTC.txt which contains string "This video is dogwalk at the sunset"
where .txt contains the name of the mp4 from the same date and I want to do the following:
folder
This video is me at a wedding.mp4
2020-04-23_19-30-52_UTC.txt
This video is dogwalk at the sunset.mp4
2020-05-25_19-30-52_UTC.txt
there is a few ways how to achieve this but I am not that good with coding. My only priority is to have it done and I am for now not limited to use of any tool or programming language.
Thanks
I'd tackle this problem with Python.
import os
dir = ('[path to original folder]')
files = os.listdir(dir)
# Iterate through all the files in the folder
for path in files:
filetype = path[-4:] # Grabs last 4 characters of the filepath
# Checks if it's a textfile
if (filetype == '.txt'):
f = open(os.path.join(dir, path), "r") # open the textfile
new_name = f.read() # grab the description
f.close() # close the textfile
new_name = new_name + '.mp4' # Add proper filetype
path = path[:-4] # Throws away the last 4 characters of the filepath
path = path + '.mp4' # Add proper filetype
os.rename(os.path.join(dir, path), os.path.join(dir, new_name)) # Rename
If any more issues arise please let me know so I can help.
I have saved a pb file with a Tensorflow script but it seems that the output folder, which I specified, is used as a parent folder and Tensorflow created a subfolder with a numeric title (e.g. 1586312775) then saved the pb file within it.
estimator.export_savedmodel('/model_output', serving_input_receiver_fn=serving_input_receiver_fn())
And the output folder structure looks like this:
model_output -- 1586312775 -- variables
-- saved_model.pb
Is there any method to access the automatic folder name (1586312775) which is created by Tensorflow?
Can I change the name of the folder before it saves the pb file under it?
Thanks,
def get_dist_folder_name(estimator):
job_name = estimator.latest_training_job.job_name
base_job_name = estimator.base_job_name
job_timestamp = job_name[job_name.find(base_job_name)+len(base_job_name)+1:][:-4]
target_folder_name = str(int(time.mktime(datetime.datetime.strptime(job_timestamp, "%Y-%m-%d-%H-%M-%S").timetuple())))
return target_folder_name
BTW, I was doing this this on AWS and this will get the timestamp folder name....
I have a folder called "before_manipulation ".
It contains 3 CSV files with names File_A.CSV, File_B.CSV ,File_C.CSV
Current_path : c:/users/before_manipulation [file_A.CSV, File_B.CSV,File_C.CSV]
I have a data manipulation that I need to do in each of the files and after manipulation ,I need to save with the same file names in another directory.
Targeted_path : C:/users/after_manipulation [file_A.CSV, File_B.CSV,File_C.CSV]
I have the logic to do the data manipulation when there is only a single file with Pandas dataframe. When I have multiple files, how to read each file and its name and pass it to my logic ?
Pseudo Code of how I am working if there was one file.
import pandas as pd
df = pd.read_csv('c:/users/before_manipulation/file_A.csv')
... do logic/manipulation
df.to_csv('c:/users/after_manipuplation/file_A.csv')
any help is appreciated.
You can use os.listdir(<path>) to return a list of the files contained within a directory. If you do not pass a variable to <path> it will return the working directory listing.
With the list from os.listdir you can iterate over it, passing the capture filename to the function you already have for data manipulation. Then on the save to you can use the captured filename to save in your desired directory.
In summary the code would look something like this.
import os
import pandas as pd
in_dir = r'c:/users/before_manipulation/'
out_dir = r'c:/users/after_manipulation/'
files_to_run = os.listdir(in_dir)
for file in files_to_run:
print('Running {}'.format(in_dir + file))
df = pd.read_csv(in_dir + file)
...do your logic here to return the changed df you want to save
...
df.to_csv(out_dir + file)
For this to work you would need to have the same shape files for each file you have in the directory, and also you would need to want to do the same logic for each file.
If that is not the case you will need something like a dictionary to save the different manipulations you need to do based on the file name and call those when appropriate.
Assuming you have some logic that works for one file, I'd just put that logic into a function and run it on a for loop.
You'd end up with something like this:
directory = r'c:/users/before_manipulation'
files = ['file_A.CSV', 'File_B.CSV','File_C.CSV']
for file in files:
somefunction(directory + '/' + file)
If you need more info on functions I'd check this out: https://www.w3schools.com/python/python_functions.asp
using pathlib
from pathlib import Path
new_dir = '\\your_path'
files = [file for file in Path(your_dir).glob('*.csv')]
for file in files:
df = pd.read_csv(file)
# .. your logic
df.to_csv(f'{new_dir}\\{file.name}',index=False)
I have about 200 folders with X images in each of them.
I have a master script in the root folder that does some stuff to the images.
Each folder has some variables specific to it and its contents.
I want my master script, when it parses folder Y, load some sort of a config file from within folder Y to get those variables, then when folder Z is to be parsed, load the config file from that one.
I know of #include "config.jsx" that I use at the moment to load it but its at the beginning of the script, I need something dynamic and doesn't need to be a jsx at all.
I store all my parameters in xml format and read that in using the XML objects in extendscript. As long as your parameters file is always named something like 'config.xml' it is easily located.
var file = new File( /c/folder/file.xml );
file.open("r");
var str = file.read();
var xml = new XML(str);