print the first character after the same element in every line-CodePudding

Im trying to print the first character after an element that occurs in each line.

w.txt:

USER TTY FROM LOGIN@ IDLE JCPU PCPU WHAT

osboxes tty7 :0 22:23 42:15 2.27s 0.23s cinnamon-session --session cinnamon

osboxes pts/1 192.168.1.17 22:25 1:26 0.06s 0.00s nano urandom.py

osboxes pts/3 192.168.1.17 22:25 1:26 0.06s 0.00s nano urandom.py

osboxes pts/0 192.168.56.117 22:25 1:26 0.06s 0.00s nano urandom.py

f = open("w.txt", "r") w = f.read()

tty = (w.split('/')[1])print(tty[0])

The above returns the id of the first pts

1

but I want it to do this for every line, and return all the integers seperated by a comma eg:

1,3,0

Ive tried the following but just returns every character in the line after 1 a bunch of times.

tty = w
for line in w:
  print(tty.split('/')[1])

thanks:)

CodePudding user response：

You could use re.findall() to scan for any digit after a /. This will give you strings, but you can convert those to integers fairly easily.

import re

with open(path, 'r') as file:
    found = re.findall(r'/(\d)', file.read())
    # ['1', '3', '0']

[int(i) for i in found]
# [1, 3, 0]

CodePudding user response：

You can use list commprehension to check '/' in each character of each line then print unpacked int splitted index 0 from list with comma separator

with open("w.txt") as f:
    print(*[int(line.split('/')[1][0]) for line in f.readlines() if '/' in line], sep=',')

# 1,3,0