Develop Reference

Python Regex - Reference first line on every match, until the start of a new group

Sample text:
This is HeaderA
Line 1
Line 2
Line 3
Line 4
Line 5
This is HeaderB
Line 1
Line 2
Intended result:
HeaderA1 HeaderA2 HeaderA3 HeaderA4 HeaderA5
HeaderB1, HeaderB2
Regex Attempts:
(?:^This is (?P<H>HeaderB)\s) (Line (?P<L>\d)\s)*?
Matches only the Header 'H' and 1st 'L' Line
(?:^This is (?P<H>HeaderB)\s)? (Line (?P<L>\d)\s)*?
manage to match multiple 'L' Lines however, only first 2 line are of the same match, not the subsequent L lines does not reference the Header capture group.
I tried other attempts to adjust the regex but ended up screwing up the expression. I have limited experience with regex, so I am not entirely sure if it is possible to get the desired output.

Mix of regex and substitutions with format.
It is assumed that below a Header you always have a Line i
import re
text = """This is HeaderA
Line 1
Line 2
Line 3
Line 4
Line 5
This is HeaderB
Line 1
Line 2"""
ordered_matches = [] # global
def custom_match(m, all_matches=ordered_matches):
p = m.group(0)
if p.isdigit():
all_matches[-1] += [p]
else:
all_matches += [[p]]
return '' # doesn't matter
r = re.sub(r'([A-Z0-9]+)$', custom_match, text, flags=re.M)
for m in ordered_matches:
print(('Header{}{{}} '.format(m[0]) * (len(m)-1)).format(*m[1:]))
Output
HeaderA1 HeaderA2 HeaderA3 HeaderA4 HeaderA5
HeaderB1 HeaderB2

IIUC you're trying to combine the Header(A|B) with the integers in the following lines. With the given output, it's probably easier to work with simple split() operations instead of re.
for group in text.split('This is ')[1:]:
header, *lines = group.splitlines()
print(*[header+line.split()[-1] for line in lines])
Output:
HeaderA1 HeaderA2 HeaderA3 HeaderA4 HeaderA5
HeaderB1 HeaderB2

Formatting output of CSV data?

I'm fairly new to python and made something that had this output:
(The text is in a csv file so so:
1,A
2,B
3,C etc)
Number Letter
1 A
2 B
3 C
26 Z
Unfortunately, I spent a good amount of time making it using a complicated method in which I manually made spaces like this:
Updated Code rn
fx = int(input('Number?\n'))
f=open('nums.txt','r')
lines=f.readlines()
line = lines[fx - 1]
with open('nums.txt','r') as f:
for i, line in enumerate(f):
if i >= 5:
break
NUM, LTR, SMB = line.rsplit(',', 1)
print(NUM.ljust(13) + LTR.ljust(13) + SMB)
How do I get it to make 3 columns? Right now it comes up with a
ValueError: not enough values to unpack (expected 3, got 2)
So is there a simpler method of achieving this that doesn't move the strings around like this:
Number Letter
1 A
2 B
3 C
26 Z #< string moves with spaces.

For simple alignment, you can use ljust or rjust. There is also no need to read the entire file for each line you want to process:
with open('numberletter','r') as f:
for i, line in enumerate(f):
if i >= 5:
break
number, letter = line.rsplit(',', 1)
print(number.ljust(13) + letter)
For more complex output formatting, look at str.format() and the formatting syntax

You can use sys module for that.
import sys
a=[1,"A"]
sys.stdout.write("%-6s %-50s " % (a[0],a[1]))

Regex remove certain characters from a file

I'd like to write a python script that reads a text file containing this:
FRAME
1 J=1,8 SEC=CL1 NSEG=2 ANG=0
2 J=8,15 SEC=CL2 NSEG=2 ANG=0
3 J=15,22 SEC=CL3 NSEG=2 ANG=0
And output a text file that looks like this:
1 1 8
2 8 15
3 15 22
I essentially don't need the commas or the SEC, NSEG and ANG data. Could someone help me use regex to do this?
So far I have this:
import re
r = re.compile(r"\s*(\d+)\s+J=(\S+)\s+SEC=(\S+)\s+NSEG=(\S+)+ANG=(\S+)\s")
with open('RawDataFile_445.txt') as a:
# open all 4 files with a meaningful name
file=[open(outputfile.txt","w")
for line in a:

Without regex:
for line in file:
keep = []
line = line.strip()
if line.startswith('FRAME'):
continue
first, second, *_ = line.split()
keep.append(first)
first, second = second.split('=')
keep.extend(second.split(','))
print(' '.join(keep))

My advice? Since I don't write many regex's I avoid writing big ones all at once. Since you've already done that I would try to verify it a small chunk at a time, as illustrated in this code.
import re
r = re.compile(r"\s*(\d+)\s+J=(\S+)\s+SEC=(\S+)\s+NSEG=(\S+)+ANG=(\S+)\s")
r = re.compile(r"\s*(\d+)")
r = re.compile(r"\s*(\d+)\s+J=(\d+)")
with open('RawDataFile_445.txt') as a:
a.readline()
for line in a.readlines():
result = r.match(line)
if result:
print (result.groups())
The first regex is your entire brute of an expression. The next line is the first chunk I verified. The next line is the second, bigger chunk that worked. Notice the slight change.
At this point I would go back, make the correction to the original, whole regex and then copy a bigger chunk to try. And re-run.

Let's focus on an example string we want to parse:
1 J=1,8
We have space(s), digit(s), more space(s), some characters, then digit(s), a comma, and more digit(s). If we replace them with regex characters, we get (\d+)\s+J=(\d+),(\d+), where + means we want 1 or more of that type. Note that we surround the digits with parentheses so we can capture them later with .groups() or .group(#), where # is the nth group.

Find and copy a line using regex in Python

I am new to this forum and to programming and apologize in advance if I violate any of the forum rules. I have researched this extensively, but I couldn't find a solution for my problem.
So I have a very long file that has this general structure:
data="""
20.020001 563410 9
20.520001 577410 20
21.022001 591466 9
21.522001 605466 120
23.196001 652338 2
25.278001 710634 7
25.780001 724690 144
26.280001 738690 9
26.782001 752746 40
27.282001 766746 9
27.784001 780802 140
29.372001 825266 2
31.458001 883674 7
31.958002 897674 8
32.458002 911674 9
32.958002 925674 10
"""
I imported the file using
with open("C:\blablabla\text.txt", 'r+') as infile:
data = infile.read()
Now I am trying to use a regular expression to find all lines that end with 140 through 146, so I did this:
items=re.findall('.......................14[0-6]\n',data,re.MULTILINE)
for x in items:
print x
This works, but when I now try to copy those lines that contain the regular expression,
for x in items:
if items in data:
data.write(items)
I get the following error:
if items in data:
TypeError: 'in <string>' requires string as left operand, not list
I understand what the problem is, but I don't know how to solve it. How can I feed the left operand a string when the outcome of my regex is a list?
Any help is much appreciated!

You should simply handle each line separately:
data = infile.readlines()
for line in data:
if re.match('.......................14[0-6]\n', line):
print line[:-1]
The last character of the line is a trailing newline, which would be duplicated by the one the print statement includes.

You can read the file line by line:
data=""
with open("file.txt", 'r+') as infile:
for line in infile:
if (146 >= int(line.split()[-1]) >= 140) :
data = data + line
print data

Your Regex can be simplified further
re.findall('.*?14[0-6]\n')
To overcome your further problems
items = re.findall('.*?14[0-6]\n',data)
result=""""""
for x in items:
result+=str(x)
print result

matching and dispalying specific lines through python

I have 15 lines in a log file and i want to read the 4th and 10 th line for example through python and display them on output saying this string is found :
abc
def
aaa
aaa
aasd
dsfsfs
dssfsd
sdfsds
sfdsf
ssddfs
sdsf
f
dsf
s
d
please suggest through code how to achieve this in python .
just to elaborate more on this example the first (string or line is unique) and can be found easily in logfile the next String B comes within 40 lines of the first one but this one occurs at lots of places in the log file so i need to read this string withing the first 40 lines after reading string A and print the same that these strings were found.
Also I cant use with command of python as this gives me errors like 'with' will become a reserved keyword in Python 2.6. I am using Python 2.5

You can use this:
fp = open("file")
for i, line in enumerate(fp):
if i == 3:
print line
elif i == 9:
print line
break
fp.close()

def bar(start,end,search_term):
with open("foo.txt") as fil:
if search_term in fil.readlines()[start,end]:
print search_term + " has found"
>>>bar(4, 10, "dsfsfs")
"dsfsfs has found"

#list of random characters
from random import randint
a = list(chr(randint(0,100)) for x in xrange(100))
#look for this
lookfor = 'b'
for element in xrange(100):
if lookfor==a[element]:
print a[element],'on',element
#b on 33
#b on 34
is one easy to read and simple way to do it. Can you give part of your log file as an example? There are other ways that may work better :).
after edits by author:
The easiest thing you can do then is:
looking_for = 'findthis' i = 1 for line in open('filename.txt','r'):
if looking_for == line:
print i, line
i+=1
it's efficient and easy :)