Here is the command I execute in shell. I want to get the same results in Python. Can I do this using the os module and how? I'm using grep -v here as some file names also have that pattern. Please note that i dont want to invoke this from shell.
du -ah 2> >(grep -v "permission denied") |grep [1-9][0-9]G | grep -v [0-9][0-9]K|grep -v [0-9][0-9]M|sort -nr -k 1| head -50
You can use this python program. It does not spawn any child processes in the shell.
#!/usr/bin/env python
from __future__ import absolute_import
from __future__ import print_function
import subprocess
import os
import argparse
def files_larger_than_no_child_process(min_bytes, count):
"""Return the top count files that are larger than the given min_bytes"""
# A list that will have (size, name) tuples.
file_info = []
for root, dirs, files in os.walk("."):
for f in files:
file_path = os.path.abspath(os.path.realpath(os.path.join(root, f)))
try:
size = os.path.getsize(file_path)
# Discard all smaller files than the given threshold
if size > min_bytes:
file_info.append((size,file_path))
except OSError as e:
pass
# Sort the files with respect to their sizes
file_info = sorted(file_info, key=lambda x: x[0], reverse=True)
# Print the top count entries
for l in file_info[:count]:
print(l[0], " ", l[1])
def main():
parser = argparse.ArgumentParser("""Prints the top files that are larger than the
given bytes in the current directory recusrively.""")
parser.add_argument("min_bytes",help="Print files larger than this value",
type=int)
parser.add_argument("count",help="Print at most the given number of files",
type=int)
args = parser.parse_args()
files_larger_than_no_child_process(args.min_bytes, args.count)
if __name__ == "__main__":
main()
