Python search string from output variable and print next two lines

Question

How do I Search string from command output and print next two lines from the output.

Below is code:

a = """
Some lines I do not want 
----- -------- --
I need this line
I need this line also
Again few lines i do not want
"""
for line in a.split("\n"):
    if line.startswith("----"):
        print "I need this line"
        print "I need this line also"

What I am doing in above code is I am checking if line starts with "----" and This works fine. Now How do i print exactly two lines after line starts with "----". In this example code print, " I need this line and I need this line also"

Jean-François Fabre · Accepted Answer · 2018-04-01 16:40:21Z

3

you can create an iterator out of the list (no need with a file handle BTW). Then let for iterate, but allow to use next manually within the loop:

a = """
Some lines I do not want
----- -------- --
I need this line
I need this line also
Again few lines i do not want
"""
my_iter = iter(a.splitlines())
for line in my_iter:
    if line.startswith("----"):
        print(next(my_iter))
        print(next(my_iter))

This code will raise StopIteration if there aren't enough lines after the dashes. One alternative that avoids this issue is (courtesy Jon Clements)

from itertools import islice

my_iter = iter(a.splitlines(True))  # preserves \n (like file handle would do)
for line in my_iter:
    if line.startswith("----"):
        print(''.join(islice(my_iter, 2)))

Another way, without splitting the string:

print(re.search("-----.*\n(.*\n.*)",a).group(1))

this searches for 2 lines after the pattern in the unsplitted, multi-line string. Can crash if re.search returns None because there are no more lines.

In both cases you get:

I need this line
I need this line also

edited Apr 1, 2018 at 16:40

answered Apr 1, 2018 at 15:46

Jean-François Fabre♦

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4 Comments

Jon Clements Over a year ago

Or print ''.join(islice(my_iter, 2)) which scales a bit better if you want more than 2 lines and won't raise a StopIteration...

Jean-François Fabre Over a year ago

'\n'.join(islice(my_iter, 2)) then. nice. Long time no hear, always fan of your code.

Jon Clements Over a year ago

Changed from '\n' to '' as if it's coming from a file (or your example string) - you don't want additional new lines between lines (as they'll end with '\n' anyway... (or most likely will)

Jon Clements Over a year ago

Ahhh... didn't see the splitlines... could just change it to a.splitlines(True) and then you get the same as though it was a file :)

Austin · Accepted Answer · 2018-04-01 15:37:29Z

0

This is one simple way:

a = """
Some lines I do not want 
----- -------- --
I need this line
I need this line also
Again few lines i do not want
"""

lines = a.split("\n")
for i in range(len(lines)):
    if lines[i].startswith("----"):  # if current line starts with ----
        print(lines[i+1])  # print next line.
        print(lines[i+2])  # print line following next line.

# I need this line
# I need this line also

answered Apr 1, 2018 at 15:37

Austin

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Comments

linusg · Accepted Answer · 2018-04-01 15:37:41Z

0

You can store the index of the current line and thus get the next n lines:

a = """
Some lines I do not want 
----- -------- --
I need this line
I need this line also
Again few lines i do not want
"""
lines = a.split("\n")
for index, line in enumerate(lines):
    if line.startswith("----"):
        print lines[index+1]
        print lines[index+2]

You may want to check for IndexErrors though.

answered Apr 1, 2018 at 15:37

linusg

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Comments

mshsayem · Accepted Answer · 2018-04-01 15:42:18Z

0

Plain way (almost C):

>>> a = """
Some lines I do not want
----- -------- --
I need this line
I need this line also
Again few lines i do not want
No nee
---- ------- --
Need this
And this
But Not this
"""

>>> start_printing, lines_printed = False, 0
>>> for line in a.split('\n'):
        if line.startswith('----'):
            start_printing = True
        elif start_printing:
            print line
            lines_printed += 1
        if lines_printed>=2:
            start_printing=False
            lines_printed = 0


I need this line
I need this line also
Need this
And this

answered Apr 1, 2018 at 15:42

mshsayem

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Comments

amanb · Accepted Answer · 2018-04-01 16:36:42Z

Here something with list comprehensions:

    a = """
    Some lines I do not want 
    ----- -------- --
    I need this line
    I need this line also
    -----------------------
    A line after (---)
    A consecutive line after (---)
    """

   lines = a.split("\n")
   test = [print(lines[index+1] + '\n' + lines[index+2]) for index in range(len(lines)) if lines[index].startswith("----")]

 #Output:I need this line
         #I need this line also
         #A line after (---)
         #A consecutive line after (---)

I bumped into IndexError on further testing with more sentences, so I added an exception block:

a = """
Some lines I do not want 
----- -------- --
I need this line
I need this line also
-----------------------
A line after (---)
A consecutive line after (---)
-------------------------
Just for fun
Another one
-------------------------
"""
lines = a.split("\n")
try:
    test = [print(lines[index+1] + '\n' + lines[index+2]) for index in range(len(lines)) if lines[index].startswith("----")]
except:
    pass

Now, the desired output without exceptions:

I need this line
I need this line also
A line after (---)
A consecutive line after (---)
Just for fun
Another one

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