Python 像 Linux grep 一样搜索并打印出整行？

Question

Let's use this as example.

>>> t = '''Line 1
... Line 2
... Line 3'''
>>> 

re.findall only print out the specific pattern which is similar to Linux grep -o

>>> re.findall('2', t)
['2']
>>> 

Linux grep

wolf@linux:~$ echo 'Line 2' | grep 2
Line 2
wolf@linux:~$ 

Linux grep -o

wolf@linux:~$ echo 'Line 2' | grep 2 -o
2
wolf@linux:~$ 

I know it's possible to print out the whole output, I just can't think the logic at the moment.

Expected Output in Python

Line 2

If there's better way to do this, please let me know.

Answer 1

print([l for l in t.splitlines() if "2" in l])

Or, if you want it separated as in grep ,

print('\n'.join([l for l in t.splitlines() if "2" in l]))

Answer 2

t = '''Line 1
Line 2
Line 3'''

for line in t.split('\n'):
    if(line.find("2")!=-1):
        print(line)

This should work for your usecase. find() is used to check if a pattern is present in a string

Answer 3

Put.* around what you want to find:

re.findall(r'.*2.*', t)