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++;
}
Related
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.
I have this sequences (over 9000) like this:
>TsM_000224500
MTTKWPQTTVTVATLSWGMLRLSMPKVQTTYKVTQSRGPLLAPGICDSWSRCLVLRVYVDRRRPGGDGSLGRVAVTVVETGCFGSAASFSMWVFGLAFVVTIEEQLL
>TsM_000534500
MHSHIVTVFVALLLTTAVVYAHIGMHGEGCTTLQCQRHAFMMKEREKLNEMQLELMEMLMDIQTMNEQEAYYAGLHGAGMQQPLPMPIQ
>TsM_000355900
MESGEENEYPMSCNIEEEEDIKFEPENGKVAEHESGEKKESIFVKHDDAKWVGIGFAIGTAVAPAVLSGISSAAVQGIRQPIQAGRNNGETTEDLENLINSVEDDL
The lines containing the ">" are the ID's and the lines with the letters are the amino acid (aa) sequences. I need to delete (or move to another files) the sequences below 40 aa and over 4000 aa.
Then, the resulting file, should contain only the sequences within this range (>= 40 aa and <= 4K aa).
I've tried writing the following script:
def read_seq(file_name):
with open(file_name) as file:
return file.read().split('\n')[0:]
ts = read_seq("/home/tiago/t_solium/ts_phtm0less.txt")
tsf = open("/home/tiago/t_solium/ts_secp-404k", 'w')
for x in range(len(ts)):
if ([x][0:1] != '>'):
if (len([x]) > 40 or len([x]) < 4000):
tsf.write('%s\n'%(x))
tsf.close()
print "OK!"
I've done some modifications, but all I'm getting are empty files or with all the +9000 sequences.
In your for loop, x is an iterating integer due to using range() (i.e, 0,1,2,3,4...). Try this instead:
for x in ts:
This will give you each element in ts as x
Also, you don't need the brackets around x; Python can iterate over the characters in strings on its own. When you put brackets around a string, you put it into a list, and thus if you tried, for example, to get the second character in x: [x][1], Python will try to get the second element in the list that you put x in, and will run into problems.
EDIT: To include IDs, try this:
NOTE: I also changed if (len(x) > 40 or len(x) < 4000) to if (len(x) > 40 and len(x) < 4000) -- using and instead of or will give you the result you're looking for.
for i, x in enumerate(ts): #NEW: enumerate ts to get the index of every iteration (stored as i)
if (x[0] != '>'):
if (len(x) > 40 and len(x) < 4000):
tsf.write('%s\n'%(ts[i-1])) #NEW: write the ID number found on preceding line
tsf.write('%s\n'%(x))
Try this, simple and easy to understand. It does not load the entire file into memory, instead iterates over the file line by line.
tsf=open('output.txt','w') # open the output file
with open("yourfile",'r') as ts: # open the input file
for line in ts: # iterate over each line of input file
line=line.strip() # removes all whitespace at the start and end, including spaces, tabs, newlines and carriage returns.
if line[0]=='>': # if line is an ID
continue # move to the next line
else: # otherwise
if (len(line)>40) or (len(line)<4000): # if line is in required length
tsf.write('%s\n'%line) # write to output file
tsf.close() # done
print "OK!"
FYI, you could also use awk for a one line solution if working in unix environment:
cat yourinputfile.txt | grep -v '>' | awk 'length($0)>=40' | awk 'length($0)<=4000' > youroutputfile.txt
I am attempting to create a regular expression pattern for strings similar to the below which are stored in a file. The aim is to get any column for any row, the rows need not be on a single line. So for example, consider the following file:
"column1a","column2a","column
3a,", #entity 1
"column\"this is, a test\"4a"
"column1b","colu
mn2b,","column3b", #entity 2
"column\"this is, a test\"4b"
"column1c,","column2c","column3c", #entity 3
"column\"this is, a test\"4c"
Each entity consists of four columns, column 4 for entity 2 would be "column\"this is, a test\"4b", column 2 for entity 3 would be "column2c". Each column begins with a quote and closes with a quote, however you must be careful because some columns have escaped quotes. Thanks in advance!
You could do like this, ie
Read the whole file.
Split the input according to the newline character which was not preceded by a comma.
Iterate over the spitted elements and again do splitting on the comma (and also the following optional newline character) which was preceded and followed by double quotes.
Code:
import re
with open(file) as f:
fil = f.read()
m = re.split(r'(?<!,)\n', fil.strip())
for i in m:
print(re.split('(?<="),\n?(?=")', i))
Output:
['"column1a"', '"column2a"', '"column3a,"', '"column\\"this is, a test\\"4a"']
['"column1b"', '"column2b,"', '"column3b"', '"column\\"this is, a test\\"4b"']
['"column1c,"', '"column2c"', '"column3c"', '"column\\"this is, a test\\"4c"']
Here is the check..
$ cat f
"column1a","column2a","column3a,",
"column\"this is, a test\"4a"
"column1b","column2b,","column3b",
"column\"this is, a test\"4b"
"column1c,","column2c","column3c",
"column\"this is, a test\"4c"
$ python3 f.py
['"column1a"', '"column2a"', '"column3a,"', '"column\\"this is, a test\\"4a"']
['"column1b"', '"column2b,"', '"column3b"', '"column\\"this is, a test\\"4b"']
['"column1c,"', '"column2c"', '"column3c"', '"column\\"this is, a test\\"4c"']
f is the input file name and f.py is the file-name which contains the python script.
Your problem is terribly familiar to what I have to deal thrice every month :) Except I'm not using python to solve it, but I can 'translate' what I usually do:
text = r'''"column1a","column2a","column
3a,",
"column\"this is, a test\"4a"
"column1a2","column2a2","column3a2","column4a2"
"column1b","colu
mn2b,","column3b",
"column\"this is, a test\"4b"
"column1c,","column2c","column3c",
"column\"this is, a test\"4c"'''
import re
# Number of columns one line is supposed to have
columns = 4
# Temporary variable to hold partial lines
buffer = ""
# Our regex to check for each column
check = re.compile(r'"(?:[^"\\]*|\\.)*"')
# Read the file line by line
for line in text.split("\n"):
# If there's no stored partial line, this is a new line
if buffer == "":
# Check if we get 4 columns and print, if not, put the line
# into buffer so we store a partial line for later
if len(check.findall(line)) == columns:
print matches
else:
# use line.strip() if you need to trim whitespaces
buffer = line
else:
# Update the variable (containing a partial line) with the
# next line and recheck if we get 4 columns
# use line.strip() if you need to trim whitespaces
buffer = buffer + line
# If we indeed get 4, our line is complete and print
# We must not forget to empty buffer now that we got a whole line
if len(check.findall(buffer)) == columns:
print matches
buffer = ""
# Optional; always good to have a safety backdoor though
# If there is a problem with the csv itself like a weird unescaped
# quote, you send it somewhere else
elif len(check.findall(buffer)) > columns:
print "Error: cannot parse line:\n" + buffer
buffer = ""
ideone demo
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 :)
I have a set of 1000 text files with names in_s1.txt, in_s2.txt and so. Each file contains millions of rows and each row has 7 columns like:
ccc245 1 4 5 5 3 -12.3
For me the most important is the values from the first and seventh columns; the pairs ccc245 , -12.3
What I need to do is to find between all the in_sXXXX.txt files, the 10 cases with the lowest values of the seventh column value, and I also need to get where each value is located, in which file. I need something like:
FILE 1st_col 7th_col
in_s540.txt ccc3456 -9000.5
in_s520.txt ccc488 -723.4
in_s12.txt ccc34 -123.5
in_s344.txt ccc56 -45.6
I was thinking about using python and bash for this purpose but at the moment I did not find a practical approach. All what I know to do is:
concatenate all in_ files in IN.TXT
search the lowest values there using: for i in IN.TXT ; do sort -k6n $i | head -n 10; done
given the 1st_col and 7th_col values of the top ten list, use them to filter the in_s files, using grep -n VALUE in_s*, so I get for each value the name of the file
It works but it is a bit tedious. I wonder about a faster approach only using bash or python or both. Or another better language for this.
Thanks
In python, use the nsmallest function in the heapq module -- it's designed for exactly this kind of task.
Example (tested) for Python 2.5 and 2.6:
import heapq, glob
def my_iterable():
for fname in glob.glob("in_s*.txt"):
f = open(fname, "r")
for line in f:
items = line.split()
yield fname, items[0], float(items[6])
f.close()
result = heapq.nsmallest(10, my_iterable(), lambda x: x[2])
print result
Update after above answer accepted
Looking at the source code for Python 2.6, it appears that there's a possibility that it does list(iterable) and works on that ... if so, that's not going to work with a thousand files each with millions of lines. If the first answer gives you MemoryError etc, here's an alternative which limits the size of the list to n (n == 10 in your case).
Note: 2.6 only; if you need it for 2.5 use a conditional heapreplace() as explained in the docs. Uses heappush() and heappushpop() which don't have the key arg :-( so we have to fake it.
import glob
from heapq import heappush, heappushpop
from pprint import pprint as pp
def my_iterable():
for fname in glob.glob("in_s*.txt"):
f = open(fname, "r")
for line in f:
items = line.split()
yield -float(items[6]), fname, items[0]
f.close()
def homegrown_nlargest(n, iterable):
"""Ensures heap never has more than n entries"""
heap = []
for item in iterable:
if len(heap) < n:
heappush(heap, item)
else:
heappushpop(heap, item)
return heap
result = homegrown_nlargest(10, my_iterable())
result = sorted(result, reverse=True)
result = [(fname, fld0, -negfld6) for negfld6, fname, fld0 in result]
pp(result)
I would:
take first 10 items,
sort them and then
for every line read from files insert the element into those top10:
in case its value is lower than highest one from current top10,
(keeping the sorting for performance)
I wouldn't post the complete program here as it looks like homework.
Yes, if it wasn't ten, this would be not optimal
Try something like this in python:
min_values = []
def add_to_min(file_name, one, seven):
# checks to see if 7th column is a lower value than exiting values
if len(min_values) == 0 or seven < max(min_values)[0]:
# let's remove the biggest value
min_values.sort()
if len(min_values) != 0:
min_values.pop()
# and add the new value tuple
min_values.append((seven, file_name, one))
# loop through all the files
for file_name in os.listdir(<dir>):
f = open(file_name)
for line in file_name.readlines():
columns = line.split()
add_to_min(file_name, columns[0], float(columns[6]))
# print answers
for (seven, file_name, one) in min_values:
print file_name, one, seven
Haven't tested it, but it should get you started.
Version 2, just runs the sort a single time (after a prod by S. Lott):
values = []
# loop through all the files and make a long list of all the rows
for file_name in os.listdir(<dir>):
f = open(file_name)
for line in file_name.readlines():
columns = line.split()
values.append((file_name, columns[0], float(columns[6]))
# sort values, print the 10 smallest
values.sort()
for (seven, file_name, one) in values[:10]
print file_name, one, seven
Just re-read you question, with millions of rows, you might run out of RAM....
A small improvement of your shell solution:
$ cat in.txt
in_s1.txt
in_s2.txt
...
$ cat in.txt | while read i
do
cat $i | sed -e "s/^/$i /" # add filename as first column
done |
sort -n -k8 | head -10 | cut -d" " -f1,2,8
This might be close to what you're looking for:
for file in *; do sort -k6n "$file" | head -n 10 | cut -f1,7 -d " " | sed "s/^/$file /" > "${file}.out"; done
cat *.out | sort -k3n | head -n 10 > final_result.out
If your files are million lines, you might want to consider using "buffering". the below script goes through those million lines, each time comparing field 7 with those in the buffer. If a value is smaller than those in the buffer, one of them in buffer is replaced by the new lower value.
for file in in_*.txt
do
awk -vt=$t 'NR<=10{
c=c+1
val[c]=$7
tag[c]=$1
}
NR>10{
for(o=1;o<=c;o++){
if ( $7 <= val[o] ){
val[o]=$7
tag[o]=$1
break
}
}
}
END{
for(i=1;i<=c;i++){
print val[i], tag[i] | "sort"
}
}' $file
done