How can I create second var automatically?

Question

I parse a text file for the following line.

stp 11441 0 0 0 0

there are always 2 such line occurrences in the txt file. I am looking for the second value in the line (11441 here) and save it as a variable.

I have figured it out how to do such manipulations with only one variable. Here is the code I am using

import re with open('cpu.txt', 'r') as file: for line in file: match = re.search('stp \d{2,100}', line) if match: stp_queue1 = match.group().split( )[1]

However, I can't get my head around how I specify a variable (stp_queue2 in that case) for the second time match occurrence.

In other words: if the file contains 2 following lines:

stp 11441 0 0 0 0 stp 20000 0 0 0 0

then stp_queue1 should be 11441 and stp_queue2 should be 20000 respectively.

Could you please help me with that?

Do you absolutely need to use regex? There may be simpler approaches if all you need is to fetch that first numeric value. — bla
– bla, Commented Dec 29, 2017 at 14:17

Aaditya Ura · Accepted Answer · 2017-12-29 14:20:26Z

There are many patterns you can use for this problem :

i am showing you three pattern, you can choose which you want :

first pattern :

import re pattern=r'stp\s(\d+)' output=[] with open('file.txt','r') as f: for line in f: match=re.search(pattern,line) output.append(match.group(1)) print(output)

output:

['11441', '20000']

Pattern 2:

r'[0-9]{5}'

pattern 3:

Positive Lookbehind (?<=stp\s) pattern=r'(?<=stp\s)\d+'

Zack Tarr · Accepted Answer · 2017-12-29 14:04:34Z

You could add your values to a dictionary rather than each to its own variable. See the code below for adding each match to a dictionary with the key being stp_queue# with the number starting at 1.

import re dictionary={} with open('cpu.txt', 'r') as file: counter=1 for line in file: match = re.search('stp \d{2,100}', line) if match: dictionary["stp_queue"+str(counter)] = match.group().split( )[1] counter++ print dictionary

Then to extract the data dictionary["stp_queue1"] will return the value stored for the first match found.

More on dictionaries here: https://docs.python.org/2/tutorial/datastructures.html#dictionaries

You should go ahead and accept that answer when you get a chance :)

corn3lius · Accepted Answer · 2017-12-29 14:07:52Z

If you put them in a list the order is preserved and look up is as easy as stp_queue[0]

import re stp_queue = [] with open('cpu.txt', 'r') as file: for line in file: match = re.search('stp \d{2,100}', line) if match: stp_queue.append(match.group().split( )[1])

bla · Accepted Answer · 2017-12-29 14:32:51Z

If all you need is a list containing the that first number following stp this may be enough:

with open('cpu.txt', 'r') as f: stp_queue = [line.split()[1] for line in f] print(stp_queue)

If you need to check whether the line begins with stp just add that validation to the comprehension:

with open('cpu.txt', 'r') as f: stp_queue = [line.split()[1] for line in f if line.startswith('stp')] print(stp_queue)

ems · Accepted Answer · 2017-12-29 14:09:01Z

-1

Use groups in your regular expression. Please read python regex documentation

answered Dec 29, 2017 at 14:09

ems

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