I want to write all mp3 files in a file that are in a certain directory.
I used this code
import os
path = 'P:\dn\test55'
wrname = r'P:\dn\path\test55.txt'
test_files = [f for f in os.listdir(path) if f.endswith('.mp3')]
f = open(wrname, "w")
f.write(str(test_files))
f.close()
the file is also written, but it looks like this
['001-file.mp3', '002-file.mp3', '003-file.mp3']
but i want the file to look like this :
001-file.mp3
002-file.mp3
003-file.mp3
How can I change this?
Thanks a lot
the write method writes its input string in the file. You then need pass to write the actual string that you want in your file: the mp3 names separated by the \n character that means "go to line"
f.write("\n".join(test_files))
the join method of strings takes a list as input and then join the elements of the list separated by the string from which you call the method.
Related
What is the appropriate way to take in files that have a filename with a timestamp in it and read properly?
One way I'm thinking of so far is to take these filenames into one single text file to read all at once.
For example, filenames such as
1573449076_1570501819_file1.txt
1573449076_1570501819_file2.txt
1573449076_1570501819_file3.txt
Go into a file named filenames.txt
Then something like
with open('/Documents/filenames.txt', 'r') as f:
for item in f:
if item.is_file():
file_stat = os.stat(item)
item = item.replace('\n', '')
print("Fetching {}".format(convert_times(file_stat)))
My question is how would I go about this where I can properly read the names in the text file given that they have timestamps in the actual names? Once figuring that out I can convert them.
If you just want to get the timestamps from the file names, assuming that they all use the same naming convention, you can do so like this:
import glob
import os
from datetime import datetime
# Grab all .txt files in the specified directory
files = glob.glob("<path_to_dir>/*.txt")
for file in files:
file = os.path.basename(file)
# Check that it contains an underscore
if not '_' in file:
continue
# Split the file name using the underscore as the delimiter
stamps = file.split('_')
# Convert the epoch to a legible string
start = datetime.fromtimestamp(int(stamps[0])).strftime("%c")
end = datetime.fromtimestamp(int(stamps[1])).strftime("%c")
# Consume the data
print(f"{start} - {end}")
...
You'll want to add some error checking and handling; for instance, if the first or second index in the stamps array isn't a parsable int, this will fail.
I am a little bit confused in how to read all lines in many files where the file names have format from "datalog.txt.98" to "datalog.txt.120".
This is my code:
import json
file = "datalog.txt."
i = 97
for line in file:
i+=1
f = open (line + str (i),'r')
for row in f:
print (row)
Here, you will find an example of one line in one of those files:
I need really to your help
I suggest using a loop for opening multiple files with different formats.
To better understand this project I would recommend researching the following topics
for loops,
String manipulation,
Opening a file and reading its content,
List manipulation,
String parsing.
This is one of my favourite beginner guides.
To set the parameters of the integers at the end of the file name I would look into python for loops.
I think this is what you are trying to do
# create a list to store all your file content
files_content = []
# the prefix is of type string
filename_prefix = "datalog.txt."
# loop from 0 to 13
for i in range(0,14):
# make the filename variable with the prefix and
# the integer i which you need to convert to a string type
filename = filename_prefix + str(i)
# open the file read all the lines to a variable
with open(filename) as f:
content = f.readlines()
# append the file content to the files_content list
files_content.append(content)
To get rid of white space from file parsing add the missing line
content = [x.strip() for x in content]
files_content.append(content)
Here's an example of printing out files_content
for file in files_content:
print(file)
I am trying to see if I can extract the file names from a os.listdir() output by omitting the '.csv' part in one single line for loop.
for example my list of file names look like this :
files = ['OPS020.csv','OPS340.csv',OPS230.csv','OPS349.csv']
Then all i could do was this
file_names = [f.split('.') for f in files]
file_names = [f[0] for f in file_names]
Is there a more elegant and shorter way to do this ?
the output i'm expecting is
file_names : ['OPS020','OPS340','OPS230','OPS349']
I guess, something like this would work.
from os import path
files = ['OPS020.csv','OPS340.csv','OPS230.csv','OPS349.csv']
filenames = [path.splitext(x)[0] for x in files]
Docs
I write a list of dics into a csv file. But the output is in one line. How could witer each value in new lines?
f = open(os.getcwd() + '/friend1.csv','w+',newline='')
for Member in MemberList:
f.write(str(Member))
f.close()
Take a look at the writing example in the csv module of the standard library and this question. Either that, or simply append a newline ("\n") after each write: f.write(str(Member)) + "\n").
I have 4 lists called I_list, Itiso, ItHDKR and Itperez and I would like to receive .txt output files with the data of these lists. I am trying to make Python rename automatically the name of the .txt output files in terms of some of my input data. In this way, the .txt output files will always have different names.
Now I am programming the following commands:
Horizontal_radiation = []
Isotropic_radiation = []
HDKR_radiation = []
Perez_radiation = []
Horizontal = open("outputHorizontal.txt", 'w')
Isotropic = open("outputIsotropic.txt", 'w')
HDKR = open("outputHDKR.txt", 'w')
Perez = open("outputPerez.txt", 'w')
for i in I_list:
Horizontal_radiation.append(i)
for x in Itiso:
Isotropic_radiation.append(x)
for y in ItHDKR:
HDKR_radiation.append(y)
for z in Itperez:
Perez_radiation.append(z)
Horizontal.write(str(Horizontal_radiation))
Isotropic.write(str(Isotropic_radiation))
HDKR.write(str(HDKR_radiation))
Perez.write(str(Perez_radiation))
Horizontal.close()
Isotropic.close()
HDKR.close()
Perez.close()
As you can see, the name of the .txt output file is fixed as "outputHorizontal.txt" (the first one). Is there any way to change this name and put it according to a input? For example, one of my inputs is the latitude, as 'lat'. I am trying to make the output file name be expressed in terms of 'lat', in this way everytime I run the program the name would be different, because now I always get the same name and the file is overwritten.
Thank you very much people, kind regards.
You can pass a string variable as the output file name. For example you could move the file declarations after you add elements to the lists (and before you write them) and use
Horizontal = open(str(Horizontal_radiation[0]), 'w')
Or just add a timestamp to the file name if it's all about don't overwriting files
Horizontal = open("horizontal-%s".format(datetime.today()), 'w')