I would like to know how I can use Python to concatenate multiple Javascript files into just one file.
I am building a component based engine in Javascript, and I want to distribute it using just one file, for example, engine.js.
Alternatively, I'd like the users to get the whole source, which has a hierarchy of files and directories, and with the whole source they should get a build.py Python script, that can be edited to include various systems and components in it, which are basically .js files in components/ and systems/ directories.
How can I load files which are described in a list (paths) and combine them into one file?
For example:
toLoad =
[
"core/base.js",
"components/Position.js",
"systems/Rendering.jd"
]
The script should concatenate these in order.
Also, this is a Git project. Is there a way for the script to read the version of the program from Git and then write it as a comment at the beginning?
This will concatenate your files:
def read_entirely(file):
with open(file, 'r') as handle:
return handle.read()
result = '\n'.join(read_entirely(file) for file in toLoad)
You may then output them as necessary, or write them using code similar to the following:
with open(file, 'w') as handle:
handle.write(result)
How about something like this?
final_script = ''
for script_name in toLoad:
with open(script_name, 'r') as f:
final_script += ('\n' + f.read())
with open('engine.js', 'w') as f:
f.write(final_script)
You can do it yourself, but this is a real problem that real tools are solving more sophisticatedly. Consider "JavaScript Minification", e.g. using http://developer.yahoo.com/yui/compressor/
Related
I have used the following code to read multiple files simultaneously
from contextlib import ExitStack
files_to_parse = [file1, file2, file3]
with ExitStack() as stack:
files = [stack.enter_context(open(i, "r")) for i in files_to_parse]
for rows in zip(*files):
for r in rows:
#do stuff
However I have noticed that since all my files don't have the same number of lines, whenever the shortest file reaches the end, all the files will close.
I used the code above (which I found here on stackoverflow) because I need to parse several files at the same time (to save time). Doing so, divide the computing time by 4. However all files aren't parsed entirely because of the problem I have mentioned above.
Is there any way to solve this problem?
open might be used as context manager, but does not have to. You might use it in ancient way, where you take responsibility to close it, that is
try:
files = [open(fname) for fname in filenames]
# here do what you need to with files
finally:
for file in files:
file.close()
I need to read paths in a csv and go to these file paths in unix. I am wondering if there is any way to do that using unix or python commands. I am unsure of how to proceed and I am not able to find much resources in the net either.
The number of rows in the excel.csv is a lot and similar to below. I need to open the excel.csv and then read the first line and go to this file path. Once this file is opened using the file path, I need to be able to read the file and extract out certain information. I tried using python for this but I am unable to find much information and I am wondering if I can use unix commands to solve this. I am clueless on how to proceed for this one so I would appreciate any reference or help using either python or unix commands. Thank you!
/folder/file1
/folder/file2
/folder/file3
It shouldn't be very difficult to do this in Python, as reading csv files is part of the standard library. Your code could look something like this:
with open('data.csv', newline='') as fh:
# depending if the first row describes the header or not,
# you can also use the simple csv.reader here
for row in csv.DictReader(fh, strict=True):
# again, if you use the simple csv.reader, you'll have to access
# the column via index instead
file_path = row['path']
with open(file_path, 'r') as fh2:
data = fh2.read()
# do something with data
I am new at programming and I have written a script to extract text from a vcf file. I am using a Linux virtual machine and running Ubuntu. I have run this script through the command line by changing my directory to the file with the vcf file in and then entering python script.py.
My script knows which file to process because the beginning of my script is:
my_file = open("inputfile1.vcf", "r+")
outputfile = open("outputfile.txt", "w")
The script puts the information I need into a list and then I write it to outputfile. However, I have many input files (all .vcf) and want to write them to different output files with a similar name to the input (such as input_processed.txt).
Do I need to run a shell script to iterate over the files in the folder? If so how would I change the python script to accommodate this? I.e writing the list to an outputfile?
I would integrate it within the Python script, which will allow you to easily run it on other platforms too and doesn't add much code anyway.
import glob
import os
# Find all files ending in 'vcf'
for vcf_filename in glob.glob('*.vcf'):
vcf_file = open(vcf_filename, 'r+')
# Similar name with a different extension
output_filename = os.path.splitext(vcf_filename)[0] + '.txt'
outputfile = open(output_filename, 'w')
# Process the data
...
To output the resulting files in a separate directory I would:
import glob
import os
output_dir = 'processed'
os.makedirs(output_dir, exist_ok=True)
# Find all files ending in 'vcf'
for vcf_filename in glob.glob('*.vcf'):
vcf_file = open(vcf_filename, 'r+')
# Similar name with a different extension
output_filename = os.path.splitext(vcf_filename)[0] + '.txt'
outputfile = open(os.path.join(output_dir, output_filename), 'w')
# Process the data
...
You don't need write shell script,
maybe this question will help you?
How to list all files of a directory?
It depends on how you implement the iteration logic.
If you want to implement it in python, just do it;
If you want to implement it in a shell script, just change your python script to accept parameters, and then use shell script to call the python script with your suitable parameters.
I have a script I frequently use which includes using PyQt5 to pop up a window that prompts the user to select a file... then it walks the directory to find all of the files in the directory:
pathname = first_fname[:(first_fname.rfind('/') + 1)] #figures out the pathname by finding the last '/'
new_pathname = pathname + 'for release/' #makes a new pathname to be added to the names of new files so that they're put in another directory...but their names will be altered
file_list = [f for f in os.listdir(pathname) if f.lower().endswith('.xls') and not 'map' in f.lower() and not 'check' in f.lower()] #makes a list of the files in the directory that end in .xls and don't have key words in the names that would indicate they're not the kind of file I want
You need to import os to use the os.listdir command.
You can use listdir(you need to write condition to filter the particular extension) or glob. I generally prefer glob. For example
import os
import glob
for file in glob.glob('*.py'):
data = open(file, 'r+')
output_name = os.path.splitext(file)[0]
output = open(output_name+'.txt', 'w')
output.write(data.read())
This code will read the content from input and store it in outputfile.
I am using
for file in fileList:
f.write(open(file).read())
I am combining files if a folder to one csv. However I dont need X amount of headers in the one file.
Is there a way to use this and have it write everything but the first row (header) coming from the files in the files?
Use python csv module
Or something like that:
for file_name in file_list:
file_obj = open(file_name)
file_obj.read()
f.write(file_obj.read())
This solution doesn't load whole file into memory, so when you use file_obj.readlines(), whole file content is load into memory
Note, that it isn't good practice to name variables with builtin names
for file in fileList:
mylines = open(file).readlines()
f.write("".join(mylines[1:]))
This should point you in the right direction. Please don't do your homework on stackoverflow.
If it's a cvs file, look into python csv lib.
I have some data in simplejson format in txt files, which I read using:
with open("my_file.txt") as f: any_variable = simplejson.load(f)
It works fine, no problems. However, I now have 100's of such text files (some of which, I dont know the names for!) to read from and I was wondering, if there was a pythonic way to read all these files and assign them to say: any_variable1 to any_variableN. I dont really care in what order they are read in.
Obviously, a simple way would be to loop and store results, yet, I was wondering if there was a pythonic way here.
If the files are inside a directory, you can use:
variables = []
path = "/your/path"
for filename in os.listdir(path):
variables.append(simplejson.load(open(os.path.join(path, filename))))