Not able to read rpt file using Python 3 - python

I am trying to read a .rpt file using the python code:
>>> with open(r'C:\Users\lenovo-pc\Desktop\training2.rpt','r',encoding = 'utf-8', errors = 'replace') as d:
... count = 0
... for i in d.readlines():
... count = count + 1
... print(i+"\n")
...
...
u
i
d
|
e
x
p
i
d
|
n
a
m
e
|
d
o
m
a
i
n
And I am getting the following result as mentioned above.
Kindly, let me know how I can read the .rpt file using python3.

This is, indeed, strange behavior. While I can not easily reproduce the error without knowing the format of the .rpt file here are some hints what might go wrong. I assume it looks something like this:
uid|expid|name|domain
...
Which can be read and printed with the following code:
with open(r'C:\Users\lenovo-pc\Desktop\training2.rpt','r',encoding = 'utf-8', errors = 'replace') as rfile:
count = 0
for line in rfile:
count += 1
print(line.strip()) # this removes white spaces, line breaks etc.
However, the problem seems to be that you iterate over the string of the first line in your file instead of the lines in the file. That would produce the patter of you see, as the print() function adds a line break (in addition to the one you add manually). This leaves you with on character per line (followed by two line breaks).
>>> for i in "foo":
... print(i+"\n")
f
o
o
Make sure you did not reuse variable names from earlier in the session and do not overwrite the file object.

Related

Extract a string between other two in Python

I am trying to extract the comments from a fdf (PDF comment file). In practice, this is to extract a string between other two. I did the following:
I open the fdf file with the following command:
import re
import os
os.chdir("currentworkingdirectory")
archcom =open("comentarios.fdf", "r")
cadena = archcom.read()
With the opened file, I create a string called cadena with all the info I need. For example:
cadena = "\n215 0 obj\n<</D[2.0 2.0]/S/D>>\nendobj\n216 0 obj\n<</D[2.0 2.0]/S/D>>\nendobj\n217 0 obj\n<</D[2.0 2.0]/S/D>>\nendobj\n218 0 obj\n<</W 3.0>>\nendobj\n219 0 obj\n<</W 3.0>>\nendobj\ntrailer\n<</Root 1 0 R>>\n%%EOF\n"
I try to extract the needed info with the following line:
a = re.findall(r"nendobj(.*?)W 3\.0",cadena)
Trying to get:
a = "n216 0 obj\n<</D[2.0 2.0]/S/D>>\nendobj\n217 0 obj\n<</D[2.0 2.0]/S/D>>\nendobj\n218 0 obj\n<<"
But I got:
a = []
The problem is in the line a = re.findall(r"nendobj(.*?)W 3\.0",cadena) but I don't realize where. I have tried many combinations with no success.
I appreciate any comment.
Regards
It seems to me that there are 2 problems:
a) you are looking for nendobj, but the N is actually part of the line break \n. Thus you'll also not get a leading N in the output, because there is no N.
b) Since the text you're looking for crosses some newlines, you need the re.DOTALL flag
Final code:
a = re.findall("endobj(.*?)W 3\.0",cadena, re.DOTALL)
Also note, that there will be a second result, confirmed by Regex101.

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.

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 :)

Bash or Python to go backwards?

I have a text file which a lot of random occurrences of the string #STRING_A, and I would be interested in writing a short script which removes only some of them. Particularly one that scans the file and once it finds a line which starts with this string like
#STRING_A
then checks if 3 lines backwards there is another occurrence of a line starting with the same string, like
#STRING_A
#STRING_A
and if it happens, to delete the occurrence 3 lines backward. I was thinking about bash, but I do not know how to "go backwards" with it. So I am sure that this is not possible with bash. I also thought about python, but then I should store all information in memory in order to go backwards and then, for long files it would be unfeasible.
What do you think? Is it possible to do it in bash or python?
Thanks
Funny that after all these hours nobody's yet given a solution to the problem as actually phrased (as #John Machin points out in a comment) -- remove just the leading marker (if followed by another such marker 3 lines down), not the whole line containing it. It's not hard, of course -- here's a tiny mod as needed of #truppo's fun solution, for example:
from itertools import izip, chain
f = "foo.txt"
for third, line in izip(chain(" ", open(f)), open(f)):
if third.startswith("#STRING_A") and line.startswith("#STRING_A"):
line = line[len("#STRING_A"):]
print line,
Of course, in real life, one would use an iterator.tee instead of reading the file twice, have this code in a function, not repeat the marker constant endlessly, &c;-).
Of course Python will work as well. Simply store the last three lines in an array and check if the first element in the array is the same as the value you are currently reading. Then delete the value and print out the current array. You would then move over your elements to make room for the new value and repeat. Of course when the array is filled you'd have to make sure to continue to move values out of the array and put in the newly read values, stopping to check each time to see if the first value in the array matches the value you are currently reading.
Here is a more fun solution, using two iterators with a three element offset :)
from itertools import izip, chain, tee
f1, f2 = tee(open("foo.txt"))
for third, line in izip(chain(" ", f1), f2):
if not (third.startswith("#STRING_A") and line.startswith("#STRING_A")):
print line,
Why shouldn't it possible in bash? You don't need to keep the whole file in memory, just the last three lines (if I understood correctly), and write what's appropriate to standard-out. Redirect that into a temporary file, check that everything worked as expected, and overwrite the source file with the temporary one.
Same goes for Python.
I'd provide a script of my own, but that wouldn't be tested. ;-)
As AlbertoPL said, store lines in a fifo for later use--don't "go backwards". For this I would definitely use python over bash+sed/awk/whatever.
I took a few moments to code this snippet up:
from collections import deque
line_fifo = deque()
for line in open("test"):
line_fifo.append(line)
if len(line_fifo) == 4:
# "look 3 lines backward"
if line_fifo[0] == line_fifo[-1] == "#STRING_A\n":
# get rid of that match
line_fifo.popleft()
else:
# print out the top of the fifo
print line_fifo.popleft(),
# don't forget to print out the fifo when the file ends
for line in line_fifo: print line,
This code will scan through the file, and remove lines starting with the marker. It only keeps only three lines in memory by default:
from collections import deque
def delete(fp, marker, gap=3):
"""Delete lines from *fp* if they with *marker* and are followed
by another line starting with *marker* *gap* lines after.
"""
buf = deque()
for line in fp:
if len(buf) < gap:
buf.append(line)
else:
old = buf.popleft()
if not (line.startswith(marker) and old.startswith(marker)):
yield old
buf.append(line)
for line in buf:
yield line
I've tested it with:
>>> from StringIO import StringIO
>>> fp = StringIO('''a
... b
... xxx 1
... c
... xxx 2
... d
... e
... xxx 3
... f
... g
... h
... xxx 4
... i''')
>>> print ''.join(delete(fp, 'xxx'))
a
b
xxx 1
c
d
e
xxx 3
f
g
h
xxx 4
i
This "answer" is for lyrae ... I'll amend my previous comment: if the needle is in the first 3 lines of the file, your script will either cause an IndexError or access a line that it shouldn't be accessing, sometimes with interesting side-effects.
Example of your script causing IndexError:
>>> lines = "#string line 0\nblah blah\n".splitlines(True)
>>> needle = "#string "
>>> for i,line in enumerate(lines):
... if line.startswith(needle) and lines[i-3].startswith(needle):
... lines[i-3] = lines[i-3].replace(needle, "")
...
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
IndexError: list index out of range
and this example shows not only that the Earth is round but also why your "fix" to the "don't delete the whole line" problem should have used .replace(needle, "", 1) or [len(needle):] instead of .replace(needle, "")
>>> lines = "NEEDLE x NEEDLE y\nnoddle\nnuddle\n".splitlines(True)
>>> needle = "NEEDLE"
>>> # Expected result: no change to the file
... for i,line in enumerate(lines):
... if line.startswith(needle) and lines[i-3].startswith(needle):
... lines[i-3] = lines[i-3].replace(needle, "")
...
>>> print ''.join(lines)
x y <<<=== whoops!
noddle
nuddle
<<<=== still got unwanted newline in here
>>>
My awk-fu has never been that good... but the following may provide you what you're looking for in a bash-shell/shell-utility form:
sed `awk 'BEGIN{ORS=";"}
/#STRING_A/ {
if(LAST!="" && LAST+3 >= NR) print LAST "d"
LAST = NR
}' test_file` test_file
Basically... awk is producing a command for sed to strip certain lines. I'm sure there's a relatively easy way to make awk do all of the processing, but this does seem to work.
The bad part? It does read the test_file twice.
The good part? It is a bash/shell-utility implementation.
Edit: Alex Martelli points out that the sample file above might have confused me. (my above code deletes the whole line, rather than the #STRING_A flag only)
This is easily remedied by adjusting the command to sed:
sed `awk 'BEGIN{ORS=";"}
/#STRING_A/ {
if(LAST!="" && LAST+3 >= NR) print LAST "s/#STRING_A//"
LAST = NR
}' test_file` test_file
This may be what you're looking for?
lines = open('sample.txt').readlines()
needle = "#string "
for i,line in enumerate(lines):
if line.startswith(needle) and lines[i-3].startswith(needle):
lines[i-3] = lines[i-3].replace(needle, "")
print ''.join(lines)
this outputs:
string 0 extra text
string 1 extra text
string 2 extra text
string 3 extra text
--replaced -- 4 extra text
string 5 extra text
string 6 extra text
#string 7 extra text
string 8 extra text
string 9 extra text
string 10 extra text
In bash you can use sort -r filename and tail -n filename to read the file backwards.
$LINES=`tail -n filename | sort -r`
# now iterate through the lines and do your checking
I would consider using sed. gnu sed supports definition of line ranges. if sed would fail, then there is another beast - awk and I'm sure you can do it with awk.
O.K. I feel I should put my awk POC. I could not figure out to use sed addresses. I have not tried combination of awk+sed, but it seems to me it's overkill.
my awk script works as follows:
It reads lines and stores them into 3 line buffer
once desired pattern is found (/^data.*/ in my case), the 3-line buffer is looked up to check, whether desired pattern has been seen three lines ago
if pattern has been seen, then 3 lines are scratched
to be honest, I would probably go with python also, given that awk is really awkward.
the AWK code follows:
function max(a, b)
{
if (a > b)
return a;
else
return b;
}
BEGIN {
w = 0; #write index
r = 0; #read index
buf[0, 1, 2]; #buffer
}
END {
# flush buffer
# start at read index and print out up to w index
for (k = r % 3; k r - max(r - 3, 0); k--) {
#search in 3 line history buf
if (match(buf[k % 3], /^data.*/) != 0) {
# found -> remove lines from history
# by rewriting them -> adjust write index
w -= max(r, 3);
}
}
buf[w % 3] = $0;
w++;
}
/^.*/ {
# store line into buffer, if the history
# is full, print out the oldest one.
if (w > 2) {
print buf[r % 3];
r++;
buf[w % 3] = $0;
}
else {
buf[w] = $0;
}
w++;
}

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