grep a line with single number - unix

I have a file with a few lines like this:
1 ab
11 ad
41 ac
1 af
1 ag
and I want the lines where the number is 1:
1 ab
1 af
1 ag
How can I achieve this?
If I write this:
grep "1" file.txt
then I get all the lines that contain 1, even if that's not the entire number:
1 ab
11 ad
41 ac
1 af
1 ag

The -w option tells grep to search for a pattern as a single word:
grep -w 1 file.txt

You can write:
grep '^1 ' file.txt
to get all lines that start with a 1 followed by a space. (The ^ means "start-of-line".)

grep -w ^1 file.txt
To get lines starting with a one.
This is probably very useful for regex with grep.

grep "^1[ \t]" file.txt
^ -> beginning of line
[ \t] -> whitespace after "1"

Related

How to convert multiple lines into fixed column lengths

To convert rows into tab-delimited, it's easy
cat input.txt | tr "\n" " "
But I have a long file with 84046468 lines. I wish to convert this into a file with 1910147 rows and 44 tab-delimited columns. The first column is a text string such as chrXX_12345_+ and the other 43 columns are numerical strings. Is there a way to perform this transformation?
There are NAs present, so I guess sed and substituting "\n" for "\t" if the string preceding is a number doesn't work.
sample input.txt
chr10_1000103_+
0.932203
0.956522
1
0.972973
1
0.941176
1
0.923077
1
1
0.909091
0.9
1
0.916667
0.8
1
1
0.941176
0.904762
1
1
1
0.979592
0.93617
0.934783
1
0.941176
1
1
0.928571
NA
1
1
1
0.941176
1
0.875
0.972973
1
1
NA
0.823529
0.51366
chr10_1000104_-
0.952381
1
1
0.973684
sample output.txt
chr10_1000103_+ 0.932203 (numbers all tab-delimited)
chr10_1000104_- etc
(sorry alot of numbers to type manually)
sed '
# use a delimiter
s/^/M/
:Next
# put a counter
s/^/i/
# test counter
/^\(i\)\{44\}/ !{
$ !{
# not 44 line or end of file, add the next line
N
# loop
b Next
}
}
# remove marker and counter
s/^i*M//
# replace new line by tab
s/\n/ /g' YourFile
some limite if more than 255 tab on sed (so 44 is ok)
Here's the right approach using 4 columns instead of 44:
$ cat file
chr10_1000103_+
0.932203
0.956522
1
chr10_1000104_-
0.952381
1
1
$ awk '{printf "%s%s", $0, (NR%4?"\t":"\n")}' file
chr10_1000103_+ 0.932203 0.956522 1
chr10_1000104_- 0.952381 1 1
Just change 4 to 44 for your real input.
If you are seeing control-Ms in your output it's because they are present in your input so use dos2unix or similar to remove them before running the tool or with GNU awk you could just set -v RS='\n\r'.
When posting questions it's important to make it as clear, simple, and brief as possible so that as many people as possible will be interested in helping you.
BTW, cat input.txt | tr "\n" " " is a UUOC and should just be tr "\n" " " < input.txt
Not the best solution, but should work:
line="nonempty"; while [ ! -z "$line" ]; do for i in $(seq 44); do read line; echo -n "$line "; done; echo; done < input.txt
If there is an empty line in the file, it will terminate. For a more permanent solution I'd try perl.
edit:
If you are concerned with efficiency, just use awk.
awk '{ printf "%s\t", $1 } NR%44==0{ print "" }' < input.txt
You may want to strip the trailing tab character with | sed 's/\t$//' or make the awk script more complicated.
This might work for you (GNU sed):
sed '/^chr/!{H;$!d};x;s/\n/\t/gp;d' file
If a line does not begin with chr append it to the hold space and then delete it unless it is the last. If the line does start chr or it is the last line, then swap to the hold space and replace all newlines by tabs and print out the result.
N.B. the start of the next line will be left untouched in the pattern space which becomes the new hold space.

How to delete words which start with some specific pattern in a file in unix

I want to delete all words in my file that start with 3: and 4:
For Example -
Input is
13 1:12 2:14 3:11
10 1:9 2:7 4:10 5:2
16 3:7 8:24
7 4:7 6:54
Output should be
13 1:12 2:14
10 1:9 2:7 5:2
14 8:24
7 6:54
Can someone tell me if it is possible using sed command or awk command.
This might work for you (GNU sed):
sed 's/\b[34]:\S*\s*//g' file
Looks for a word boundary and then either 3 or 4 followed by : and zero or more non-spaces followed by zero or more spaces and deletes them throughout the line.
With sed
sed -r 's/ 3:[0-9]*| 4:[0-9]*//g'
$ cat input.txt
13 1:12 2:14 3:11 10 1:9 2:7 4:10 5:2 16 3:7 8:24 7 4:7 6:54
$ cat input.txt | sed -r 's/ 3:[0-9]*| 4:[0-9]*//g'
13 1:12 2:14 10 1:9 2:7 5:2 16 8:24 7 6:54
Explanation:
-r = Regex search
3:[0-9]*: Search for a space, then 3, then :, then [0-9] or a number between 0 and 9, the * means that he will search for zero or more hits in the pervious regex search, which is [0-9], so * means on this case that will search for zero or more numbers behind the first number after :
| : Means OR
4:[0-9]*: Same as above except that instead of 3 it will search for 4
//: The substitution strings, if you put POTATOE behind bars it will type it, on this case, sed will simply don't type anything.
/g: Search in all the input passed to sed.
With awk:
awk '{for (i=1; i<=NF; i++)
{if (! sub("^[34]:", "", $i)) d=d$i" "}
print d; d=""
}' file
It loops through the fields and just store in the variable d those that do not start with 3: or 4:. This is done by checking if sub() function returns true or not. When the loop through the line is done, the d variable is printed.
For your given file:
$ awk '{for (i=1; i<=NF; i++) {if (! sub("^[34]:", "", $i)) d=d$i" "} print d; d=""}' file
13 1:12 2:14
10 1:9 2:7 5:2
16 8:24
7 6:54
sed 's/[[:blank:]][34]:[^[:blank:]]\{1,\}[[:blank:]]*/ /g' YourFile
Posix compliant and assuming there is no (as in sample) first word stating with 3: or 4:.
Assuming all words contains : and has at least one digit after the :
sed "s/ \([34]:[^\b]+\)//g" inputfile
This matches SPACE, 3 or 4, colon and then at least one non word boundary. It replaces it forth nothing and does so for the whole line.

Using grep -f and -w together

I have two files like this:
abc.txt
a
b
z
1
10
and abcd.txt
a
b
c
d
1
10
100
1000
I would like:
a
b
1
10
I would like to use grep -fw abc.txt abcd.txt to search through every line of abc.txt and print lines which match the entire word. If I just use grep -f, I get lines 100 since the pattern '10' matches '100'. But grep -f -w abc.txt abcd.txt produces:
a
b
1
and doesn't print out the 10. So, I guess, what is the best way to match every line in abc.txt with the entire line of abcd.txt ?

need code in awk unix or use substr

I want to print words between "ctr{words}" and count the same words in a file.
I tried:
sed -n 's/.*ctr{\(.[^}]*\).*/\1/p' file
but it didn't search all words, just one word
The file is:
962796057604|mar0101|0|00000107A20E00000A6C331650B920340C00|0|0|400019FD7DBFBF7F|1001|962796057604|0 |01001|||-1|795971936| 00962795971936|16||-1| 00962795971936|-1|0|2|0|416019000659493|0||||||0|0|2012.12.01 00:07:09|12|30|0|516|16|1|2012.12.01 00:06:39|1|0||202|20001||0B12F1001104697209100300000000000000|1|1|11000|0|0||0881006972091003F000||0 714F610045584E6|000000000000|3|1|0000000000000000|0|140|0|0|0|0|0|0|||0|2|||||||||||||||||||||0|||0| |0|1|143|acf{0}cif{0}fcf{0}con{0}cuf{0}ctr{**Mo7afazat**}cgpa{962796057604}vlr{0096279001300}cff{0}roaf{0}mpty{0}ftksn{JMT}ftksr{0001}ftktp{CallTicketCPOCS} ||
1|34|2012.12.01 00:08:35|12|4|921-*203-0000000000-962796298894|mar0101|0|000001028225AE4AD868A8B750B900980C00|1|0|4000018001000002||962796298894|||||-1|||||-1||-1|0||||0||||||-1|-1|||-1|0|-1|-1|-1|2012.12.01 00:08:35|1|0||-1|0|||||||||||||0|0|||3797|0|12|-2147483648|-2147483648|-2147483648|-2147483648|||||||||||||||||||||||||0|||0||1|6|244|tid{111210532409329884}pfid{20}gob{1}rid{globitel} afid{}uid1{962796298894}aid1{1}ar1{0}uid2{globitel}aid2{-1}pid{1234}pur{!GDRC COMMIT AMOUNT 0}ratinf{}rec{0}rots{0}tda{}mid{}exd{0}reqa{0}ctr{**JaishanaIN**}ftksn{JMT}ftksr{0001}ftktp{PayCallTicket}||
1|34|2012.12.01 00:08:35|12|4|100-50-0-962796605155|mar0101|0|00000102A20400000A6A439D50B920520C00|0|0|400019FD7DBFBF7F|1001|962796605155|1 6||||-1|b116c||16||-1||-1|0|0|0|416017002233360|0||||||0|0|1970.01.01 02:00:00|0|0|0|220|0|1|1970.01.01 02:00:00|1|0||194|0||000000000000000000000000000000000000|0|0||0|0||00000000000000000000||0000000000 000000|000000000000|0|0|0000000000000000|0|370|0|0|0|0|0|0|||0|0|||||||||||||||||||||0|||0||0|1|70|a cf{3}ussd{1}ctr{**ZainElKul**}ftksn{JMT}ftksr{0001}ftktp{CallTicketCPOCS}||
1|34|2012.12.01 00:08:35|12|4|100-10-0
1|34|2012.12.01 00:08:35|12|4|921-*203-0000000000-962797611253|mar0101|0|0000010282B54BD015FF4C4B50B8F96E0C00|1|0|4000018001000002||962797611253|||||-1|||||-1||-1|0||||0||||||-1|-1|||-1|0|-1|-1|-1|2012.12.01 00:08:35|1|0||-1|0|||||||||||||0|0|||885|0|12|-2147483648|-2147483648|-2147483648|-2147483648|||||||||||||||||||||||||0|||0||1|6|243|tid{111220371293561120}pfid{20}gob{1}rid{globitel} afid{}uid1{962797611253}aid1{1}ar1{0}uid2{globitel}aid2{-1}pid{1234}pur{!GDRC COMMIT AMOUNT 0}ratinf{}rec{0}rots{0}tda{}mid{}exd{0}reqa{0}ctr{**ZainElKul**}ftksn{JMT}ftksr{0001}ftktp{PayCallTicket}||
-962795292027|mar0101|0|00000101A20200000A6A96B750B920300C00|0|0|400019FD7DBFBF7F|1001|962795292027|0 |01004|||-1|797196452| 00962797196452|16||-1| 00962797196452|-1|0|2|0|416018002276781|0||||||0|0|2012.12.01 00:07:09|12|12|23|516|16|1|2012.12.01 00:06:34|1|0||202|1||0B12F1001104697209100300000000000000|1|1|11000|0|0||0881006972091003F000||0714F 6100455AD67|000000000000|3|1|0000000000000000|0|30|0|0|0|0|0|0|||0|0|||||||||||||||||||||0|||0||0|1| 171|acf{0}cif{0}fcf{0}con{0}cuf{0}ctr{ZainUnlimited}cgpa{962795292027}vlr{0096279001300}cff{0}roaf{0}mpty{0}cacc{1;0;30}cquo{1;230;}ftksn{JMT}ftksr{000 1}ftktp{CallTicketCPOCS}||
1|34|2012.12.01 00:08:35|12|4|921-*203-0000000000-962796012818|mar0101|0|0000010882218115085D5F9150B920520C00|0|0|4000018001000002||962796012818|||||-1|||||-1||-1|0||||0||||||-1|-1|||-1|0|-1|-1|-1|2012.12.01 00:08:35|1|0||-1|1|||||||||||||0|0|||70|0|0|-2147483648|-2147483648|-2147483648|-2147483648|||||||||||||||||||||||||0|||0||1|6|258|tid{111221366974701289}pfid{17}gob{1}rid{globitel} afid{}uid1{962796012818}aid1{1}ar1{-2147483648}uid2{}aid2{-1}pid{DEFAULT_DECISION}pur{!GDRC Balance Check}ratinf{}rec{0}rots{0}tda{}mid{}exd{0}reqa{0}ctr{**AlBarakehNew**}ftksn{JMT}ftksr{0001}ftktp{PayCallTicket}||
1|34|2012.12.01 00:08:35|12|4|921-*203-0000000000-962797251349|mar0101|0|0000010282A451483EDFCFD350B920400C00|1|0|4000018001000002||962797251349|||||-1|||||-1||-1|0||||0||||||-1|-1|||-1|0|-1|-1|-1|2012.12.01 00:08:35|1|0||-1|0|||||||||||||0|0|||440|0|12|-2147483648|-2147483648|-2147483648|-2147483648|||||||||||||||||||||||||0|||0||1|6|245|tid{111211342745325133}pfid{20}gob{1}rid{globitel} afid{}uid1{962797251349}aid1{1}ar1{0}uid2{globitel}aid2{-1}pid{1234}pur{!GDRC COMMIT AMOUNT 0}ratinf{}rec{0}rots{0}tda{}mid{}exd{0}reqa{0}ctr{**ZainElKulSN**}ftksn{JMT}ftksr{0001}ftktp{PayCallTicket}||
1|34|2012.12.01 00:08:35|12|4|921-*203-0000000000-
It looks like you are missing the counts. The easiest way to do this is to pipe your output through uniq -c:
$ sed -n 's/.*ctr{\(.[^}]*\).*/\1/p' file | sort | uniq -c
1 **Mo7afazat**
1 **JaishanaIN**
2 **ZainElKul**
1 ZainUnlimited
1 **AlBarakehNew**
1 **ZainElKulSN**
Another way, only using awk:
$ awk 'match($0,".*ctr{([^}]*)}.*",m){a[m[1]]++}END{for(i in a) print i,a[i]}' file
ZainUnlimited 1
**ZainElKulSN** 1
**Mo7afazat** 1
**ZainElKul** 2
**JaishanaIN** 1
**AlBarakehNew** 1
When searching for matches in files grep is the best choice more often than not.
Using grep with postive lookahead and uniq -c:
$ grep -Po "(?<=ctr{)[^}]+" file | uniq -c
1 Mo7afazat
1 JaishanaIN
2 ZainElKul
1 ZainUnlimited
1 AlBarakehNew
1 ZainElKulSN
From man uniq:
Note: 'uniq' does not detect repeated lines unless they are adjacent.
For files where the duplicates are not adjacent pipe to sort first however the order in which each match is found in the orignal file will be lost:
grep -Po "(?<=ctr{)[^}]+" file | sort | uniq -c
1 AlBarakehNew
1 JaishanaIN
1 Mo7afazat
2 ZainElKul
1 ZainElKulSN
1 ZainUnlimited

sed/awk or other: one-liner to increment a number by 1 keeping spacing characters

EDIT: I don't know in advance at which "column" my digits are going to be and I'd like to have a one-liner. Apparently sed doesn't do arithmetic, so maybe a one-liner solution based on awk?
I've got a string: (notice the spacing)
eh oh 37
and I want it to become:
eh oh 36
(so I want to keep the spacing)
Using awk I don't find how to do it, so far I have:
echo "eh oh 37" | awk '$3>=0&&$3<=99 {$3--} {print}'
But this gives:
eh oh 36
(the spacing characters where lost, because the field separator is ' ')
Is there a way to ask awk something like "print the output using the exact same field separators as the input had"?
Then I tried yet something else, using awk's sub(..,..) method:
' sub(/[0-9][0-9]/, ...) {print}'
but no cigar yet: I don't know how to reference the regexp and do arithmetic on it in the second argument (which I left with '...' for now).
Then I tried with sed, but got stuck after this:
echo "eh oh 37" | sed -e 's/\([0-9][0-9]\)/.../'
Can I do arithmetic from sed using a reference to the matching digits and have the output not modify the number of spacing characters?
Note that it's related to my question concerning Emacs and how to apply this to some (big) Emacs region (using a replace region with Emacs's shell-command-on-region) but it's not an identical question: this one is specifically about how to "keep spaces" when working with awk/sed/etc.
Here is a variation on ghostdog74's answer that does not require the number to be anchored at the end of the string. This is accomplished using match instead of relying on the number to be in a particular position.
This will replace the first number with its value minus one:
$ echo "eh oh 37 aaa 22 bb" | awk '{n = substr($0, match($0, /[0-9]+/), RLENGTH) - 1; sub(/[0-9]+/, n); print }'
eh oh 36 aaa 22 bb
Using gsub there instead of sub would replace both the "37" and the "22" with "36". If there's only one number on the line, it doesn't matter which you use. By doing it this way, though, it will handle numbers with trailing whitespace plus other non-numeric characters that may be there (after some whitespace).
If you have gawk, you can use gensub like this to pick out an arbitrary number within the string (just set the value of which):
$ echo "eh oh 37 aaa 22 bb 19" |
awk -v which=2 'BEGIN { regex = "([0-9]+)\\>[^0-9]*";
for (i = 1; i < which; i++) {regex = regex"([0-9]+)\\>[^0-9]*"}}
{ match($0, regex, a);
n = a[which] - 1; # do the math
print gensub(/[0-9]+/, n, which) }'
eh oh 37 aaa 21 bb 19
The second (which=2) number went from 22 to 21. And the embedded spaces are preserved.
It's broken out on multiple lines to make it easier to read, but it's copy/pastable.
$ echo "eh oh 37" | awk '{n=$NF+1; gsub(/[0-9]+$/,n) }1'
eh oh 38
or
$ echo "eh oh 37" | awk '{n=$NF+1; gsub(/..$/,n) }1'
eh oh 38
something like
number=`echo "eh oh 37" | grep -o '[0-9]*'`
sed 's/$number/`expr $number + 1`/'
How about:
$ echo "eh oh 37" | awk -F'[ \t]' '{$NF = $NF - 1;} 1'
eh oh 36
The solution will not preserve the number of decimals, so if the number is 10, then the result is 9, even if one would like to have 09.
I did not write the shortest possible code, it should stay readable
Here I construct the printf pattern using RLENGTH so it becomes %02d (2 being the length of the matched pattern)
$ echo "eh oh 10 aaa 22 bb" |
awk '{n = substr($0, match($0, /[0-9]+/), RLENGTH)-1 ;
nn=sprintf("%0" RLENGTH "d", n)
sub(/[0-9]+/, nn);
print
}'
eh oh 09 aaa 22 bb

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