How do I remove or address a specific occurrence of a character in sed?
I'm editing a CSV file and I want to remove all text between the third and the fifth occurrence of the comma (that is, dropping fields four and five) . Is there any way to achieve this using sed?
E.g:
% cat myfile
one,two,three,dropthis,dropthat,six,...
% sed -i 's/someregex//' myfile
% cat myfile
one,two,three,,six,...
If it is okay to consider cut command then:
$ cut -d, -f1-3,6- file
awk or any other tools that are able to split strings on delimiters are better for the job than sed
$ cat file
1,2,3,4,5,6,7,8,9,10
Ruby(1.9+)
$ ruby -ne 's=$_.split(","); s[2,3]=nil ;puts s.compact.join(",") ' file
1,2,6,7,8,9,10
using awk
$ awk 'BEGIN{FS=OFS=","}{$3=$4=$5="";}{gsub(/,,*/,",")}1' file
1,2,6,7,8,9,10
A real parser in action
#!/usr/bin/python
import csv
import sys
cr = csv.reader(open('my-data.csv', 'rb'))
cw = csv.writer(open('stripped-data.csv', 'wb'))
for row in cr:
cw.writerow(row[0:3] + row[5:])
But do note the preface to the csv module:
The so-called CSV (Comma Separated
Values) format is the most common
import and export format for
spreadsheets and databases. There is
no “CSV standard”, so the format is
operationally defined by the many
applications which read and write it.
The lack of a standard means that
subtle differences often exist in the
data produced and consumed by
different applications. These
differences can make it annoying to
process CSV files from multiple
sources. Still, while the delimiters
and quoting characters vary, the
overall format is similar enough that
it is possible to write a single
module which can efficiently
manipulate such data, hiding the
details of reading and writing the
data from the programmer.
$ cat my-data.csv
1
1,2
1,2,3
1,2,3,4,
1,2,3,4,5
1,2,3,4,5,6
1,2,3,4,5,6,
1,2,,4,5,6
1,2,"3,3",4,5,6
1,"2,2",3,4,5,6
,,3,4,5
,,,4,5
,,,,5
$ python csvdrop.py
$ cat stripped-data.csv
1
1,2
1,2,3
1,2,3
1,2,3
1,2,3,6
1,2,3,6,
1,2,,6
1,2,"3,3",6
1,"2,2",3,6
,,3
,,
,,
Related
I have a txt file that has only information about location (location.txt)
Another large txt file (all.txt) has a lot of information like id , and a location.txt is subset of all.txt ( some records common in both )
I want to search the location.txt in another file with grep (all.txt)
and print all common records ( but all information like all.txt )
I try to grep by :
grep -f location.txt all.txt
the problem grep just give me the last location not all locations
how can I print all location?
I'm assuming you mean to use one of the files as a set of patterns for grep. If this is the case, you seem to be looking for a way to print all lines in one file not found in the other and this is what you want:
grep -vFf file_with_patterns other_file
Explanation
-F means to interpret the pattern(s) literally, giving no particular meaning to regex metacharacters (like * and +, for example)
-f means read regex patterns from the file named as argument (file_with_patterns in this case).
Am given a list if ID which I need to trace back a name in a file
file: ID contains
1
2
3
4
5
6
The ID are contained in a Large 2 GB file called result.txt
ABC=John,dhds,72828,73737,3939,92929
CDE=John,uubad,32424,ajdaio,343533
FG1=Peter,iasisaio,097282,iosoido
WER=Ann,97391279,89719379,7391739
result,**id=1**,iuhdihdio,ihwoihdoih,iuqhwiuh,ABC
result2,**id=2**,9729179,hdqihi,hidqi,82828,CDE
result3,**id=3**,biasi,8u9829,90u209w,jswjso,FG1
So I cat the ID file into a variable
I then use this variable in a loop to grep out the values to link back to the name using grep and cut -d from results.txt and output to a variable
so variable contains ABS CDE FG1
In the same loop I pass the output of the grep to perform another grep on results.txt, to get the name
ie regrets file for ABC CDE FG1
I do get the answer but takes a long time is their a more efficient way?
Thanks
Making some assumptions about your requirement... ID's that are not found in the big file will not be shown in the output; the desired output is in the format shown below.
Here are mock input files - f1 for the id's and f2 for the large file:
[mathguy#localhost test]$ cat f1
1
2
3
4
5
6
[mathguy#localhost test]$ cat f2
ABC=John,dhds,72828,73737,3939,92929
CDE=John,uubad,32424,ajdaio,343533
FG1=Peter,iasisaio,097282,iosoido
WER=Ann,97391279,89719379,7391739
result,**id=1**,iuhdihdio,ihwoihdoih,iuqhwiuh,ABC
result2,**id=2**,9729179,hdqihi,hidqi,82828,CDE
result3,**id=3**,biasi,8u9829,90u209w,jswjso,FG1
Proposed solution and output:
[mathguy#localhost test]$ sed 's/.*/\*\*id=&\*\*/' f1 | grep -Ff - f2 | \
> sed -E 's/^.*\*\*id=([[:digit:]]*)\*\*.*,([^,]*)$/\1 \2/'
1 ABC
2 CDE
3 FG1
The hard work here is done by grep -F which might be just fast enough for your needs. There is some prep work and some clean-up work done by sed, but those are both on small datasets.
First we take the id's from the input file and we output strings in the format **id=<number>**. The output is presented as the fixed-character patterns to grep -F via the option -f (take the patterns from file, in this case from stdin, invoked as -; that is, from the output of sed).
After we find the needed lines from the big file, the final sed just extracts the id and the name from each line.
Note: this assumes that each id is only found once in the big file. (Actually the command will work regardless; but if there are duplicate lines for an id, your business users will have to tell you how to handle. What if you get contradictory names for the same id? Etc.)
I'm trying to reduce a .sm file1 - around 10 GB by filtering it using a fair long set of words (around 180.108 items) listed in a text file file2.
File1 is structured as follows:
word <http://internet.address.com> 1
i.e. one word followed by a blank space, an internet address, and a number.
File2 is a simple .txt file, a list of words, one on each line.
My aim is to create a third file File3 containing only those lines in file1 whose first word matches with the word-list of file2, and disregard the rest.
My attempt is the following:
grep -w -F -f file2.txt file1.sm > file3.sm
I've also attempted something along this line:
gawk 'FNR==NR {a[$1]; next } !($2 in a)' file2.txt file1.sm > file3.sm
but with no success. I understand /^ and \b might play a part here, but I don't know how to fit them in the syntax. I've looked around extensively but no solution seems to fit.
My problem is that here grep reads the entire file1's line, and it can happen that the matching word lies in the webpage address, which I'm not interested in finding out.
sed 's/^/^/' file2.txt | grep -f - file1.sm
join is the best tool for this, not grep/awk:
join -t' ' <(sort file1.sm) <(sort file2.txt) >file3.sm
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
how to use grep to find occurrences of words from a dictionary file which have a given set of letters with the restriction that each letter occurs once and only once.
EG if the letters are abc then the expected output is:
cab
EDIT:
Given a dictionary file (that is a file containing one word per line such as /usr/share/dict/words on mac os x operating system) and a set of (unique) characters, I want to print out all of the dictionary file's words that contain each character of the input set once and only once. For example if the set of characters is {a,b,c} then print out all (3-letter) words that contain each character of the set.
I am looking, preferably, for a solution that uses just grep expressions.
Given a series of letters, for example abc, you can convert each one to a lookahead, like this:
^(?=[^a]*a[^a]*)(?=[^b]*b[^b]*)(?=[^c]*c[^c]*)$
You may need to use the "extended regex" flag -E to use this regex with grep.
To create this regex from a string, you could use sed (an exercise for the reader)
grep -E ^[abc]{3}.$ <Dictionary file> | grep -v -e a.*a -e b.*b -e c.*c
i.e. Find all three letter strings matching the input and pipe these through inverse grep to remove strings with double letters.
I'm using the '.' after {3} because my dictionary file is windows based so has an extra carriage return or line feed. So, that's probably not necessary.
Below is a Perl solution. Note, you'll need to add more words to the dictionary, and read input in to the $input variable. An array of valid words will end up in #results.
#!/usr/bin/env perl
use Data::Dumper;
my $input = "abc";
my #dictionary = qw(aaa aac aad aal aam aap aar aas aat aaw aba abc abd abf abg
abh abm abn abo abr abs abv abw aca acc ace aci ack acl acp acs act acv ada adb
adc add adf adh adl adn ado adp adq adr ads adt adw aea aeb aec aed aef aes aev
afb afc afe aff afg afi afk afl afn afp aft afu afv agb agc agl agm agn ago agp
...
PUT A REAL DICTIONARY HERE!
...
zie zif zig zii zij zik zil zim zin zio zip zir zis zit ziu ziv zlm zlo zlx zma
zme zmi zmu zna zoa zob zoe zog zoi zol zom zon zoo zor zos zot zou zov zoy zrn
zsr zub zud zug zui zuk zul zum zun zuo zur zus zut zuz zva zwo zye zzz);
# Generate a lookahead expression for each character in the input word
my $regexp = join("", map { "(?=.*$_)" } split(//, $input));
my #results;
foreach my $word (#dictionary) {
# If the size of the input doesn't match the dictionary word, skip to the
# next word.
if (length($input) != length($word)) {
next;
}
if ($word =~ /$regexp/) {
push(#results, $word);
}
}
print Dumper #results;
The solution I found involves using grep first to extract all n-letter words that contain only letters from the input set - although some letters might appear more than once, some may not appear; (again I am assuming that the input letters are unique). Then it does a series of 1-letter greps to make sure each letter occurs at least once. Because the words are of length n this ensures the word contains each letter once and only once. For example, if the input character set is (a,b,c} then the solution would be:
grep -E '^[abc]{3}$' /usr/share/dict/words | grep a | grep b | grep c
a simple bash script can be written which creates this grep string and executes it against the word file, using $1 as the input letter set. It might not be the most efficient method of generating the string, but as I am not familiar with sed or awk it does seem to solve my problem. The script I created is:
#!/bin/sh
slen=${#1}
g2="'^[$1]{$slen}\$'"
g3=""
ix1=0
while [ $ix1 -lt $slen ]
do
g3="$g3 | grep ${1:$ix1:1}"
ix1=$((ix1+1))
done
eval grep -E $g2 /usr/share/dict/words $g3