Python Regex re.search列出

Question

I have some code to parse the linux 'df -h', the normal command line output looks like this:

Filesystem      Size  Used Avail Use% Mounted on
udev            987M     0  987M   0% /dev
tmpfs           201M  9.2M  191M   5% /run
/dev/sda1        38G   11G   25G  30% /
tmpfs          1001M  416K 1000M   1% /dev/shm
tmpfs           5.0M     0  5.0M   0% /run/lock
tmpfs          1001M     0 1001M   0% /sys/fs/cgroup
tmpfs           201M   28K  201M   1% /run/user/132
tmpfs           201M   28K  201M   1% /run/user/0

Currently my code achieves the desired output:

['/run', '/run/lock', '/run/user/132', '/run/user/0']

But the 'print ([x.split(" ")[-1] for x in newlist])' line shown below feels like a hack, I'm struggling to get this working as a regex using 'r.search' below, can anyone advise a better way of doing this please ?

import subprocess
import re


cmd = 'df -h'
output = subprocess.check_output(cmd, shell=True).decode('utf8')
ln = output.split('\n')
r = re.compile('/run.*')
newlist = list(filter(r.search, ln))

print ([x.split(" ")[-1] for x in newlist])

Edit * I am using 'df -h' as some random output to practice regex on, so while @romanPerekhrest offers the best real world solution for this problem I was looking for a regex solution.

Answer 1

The fastest approach:

df -h --output=target | grep '/run.*'

The output:

/run
/run/lock
/run/user/132
/run/user/0

---

• --output=target - to output only mount points

Answer 2

how about

re.findall(r'/run.*$', output, re.MULTILINE)

I don't know about better or speed, but it cuts your code down to 3 lines, and you're regexing anyway.