I am trying to print out all lines with at least one character that is NOT numeric.
My grep code looks like this: grep '[^[:digit:]]' GTEST
Where GTEST is this:
TEST
55 55 Pink
123
sss
aaa
ss aaa ss
a 1 b 2 a b a
Doop Dap
12 13
77a
59360
And the output is exactly what is in GTEST, except with the matching parts of lines (AKA all of the alpha characters) in red. Instead of displaying the matching characters in red, I /only/ want to print out the lines that contain matching characters.
I've been looking around the grep tags (-o, -w, etc), but none of them seem to do it for me.
Am I missing something?
EDITED:
Expected output would be:
TEST
55 55 Pink
sss
aaa
ss aaa ss
a 1 b 2 a b a
Doop Dap
77a
From your data, I get this output:
grep '[^[:digit:]]' file
TEST
55 55 Pink
sss
aaa
ss aaa ss
a 1 b 2 a b a
Doop Dap
12 13
77a
You get the 12 13, since space between 12 and 13 is a non digit character.
This will also give output if you have space before or after digits, like: 123<space>
To overcome this, you can do like this:
grep '[^[:digit:] ]' file
TEST
55 55 Pink
sss
aaa
ss aaa ss
a 1 b 2 a b a
Doop Dap
77a
Or even better:
grep '[^[:digit:][:blank:]]' file
Related
I have a string in the following way =
"lmn abc 4.0mg 3.50 mg over 12 days. Standing nebs."
I want to convert it into :
"lmn abc 40mg 350 mg over 12 days. Standing nebs."
that is I only convert a.b -> ab where a and b are integer
waiting for help
Assuming you are using Python. You can use captured groups in regex. Either numbered captured group or named captured group. Then use the groups in the replacement while leaving out the ..
import re
text = "lmn abc 4.0mg 3.50 mg over 12 days. Standing nebs."
Numbered: You reference the pattern group (content in brackets) by their index.
text = re.sub("(\d+)\.(\d+)", "\\1\\2", text)
Named: You reference the pattern group by a name you specified.
text = re.sub("(?P<before>\d+)\.(?P<after>\d+)", "\g<before>\g<after>", text)
Which each returns:
print(text)
> lmn abc 40mg 350 mg over 12 days. Standing nebs.
However you should be aware that leaving out the . in decimal numbers will change their value. So you should be careful with whatever you are doing with these numbers afterwards.
Using any sed in any shell on every Unix box:
$ sed 's/\([0-9]\)\.\([0-9]\)/\1\2/g' file
"lmn abc 40mg 350 mg over 12 days. Standing nebs."
Using sed
$ cat input_file
"lmn abc 4.0mg 3.50 mg over 12 days. Standing nebs. a.b.c."
$ sed 's/\([a-z0-9]*\)\.\([a-z0-9]\)/\1\2/g' input_file
"lmn abc 40mg 350 mg over 12 days. Standing nebs. abc."
echo '1.2 1.23 12.34 1. .2' |
ruby -p -e '$_.gsub!(/\d+\K\.(?=\d+)/, "")'
Output
12 123 1234 1. .2
If performance matters:
echo '1.2 1.23 12.34 1. .2' |
ruby -p -e 'BEGIN{$regex = /\d+\K\.(?=\d+)/; $empty_string = ""}; $_.gsub!($regex, $empty_string)'
I need help with one grep command
-single digit occurs one time in line
my solution doesn't work
egrep "^(\s*[1]\s*)(\s*[^1]+\s*)+$|^(\s*[^1]\s*)(\s*[1]+\s*)+$|^(\s*[2]\s*)(\s*[^2]+\s*)+$|^(\s*[^2]\s*)(\s*[2]+\s*)+$|^(\s*[3]\s*)(\s*[^3]+\s*)+$|^(\s*[^3]\s*)(\s*[3]+\s*)+$|^(\s*[4]\s*)(\s*[^4]+\s*)+$|^(\s*[^4]\s*)(\s*[4]+\s*)+$|^(\s*[5]\s*)(\s*[^5]+\s*)+$|^(\s*[^5]\s*)(\s*[5]+\s*)+$|^(\s*[6]\s*)(\s*[^6]+\s*)+$|^(\s*[^6]\s*)(\s*[6]+\s*)+$|^(\s*[7]\s*)(\s*[^7]+\s*)+$|^(\s*[^7]\s*)(\s*[7]+\s*)+$|^(\s*[8]\s*)(\s*[^8]+\s*)+$|^(\s*[^8]\s*)(\s*[8]+\s*)+$|^(\s*[9]\s*)(\s*[^9]+\s*)+$|^(\s*[^9]\s*)(\s*[9]+\s*)+$"
example
for example in this text
012 210 5
6343 232 5 3423
345 689 7 986 543012 210 5
grep color only second line.
I want to grep color every line because in each line any digit occurs one time.In first line this is 5 in second line this is 5 in third line this is 7
A pattern that detects if a digit is unique on a line (if I'm understanding the question correctly):
For the digit 5:
^[^5]*(5)[^5]*$
^ // start of line
[^5]* // any char not 5, 0-or-more
(5) // 5
[^5]* // any char not 5, 0-or-more
$ // end of line
To test all digits, it becomes:
^(?:[^0]*(0)[^0]*|[^1]*(1)[^1]*)$ etc for all digits. The digit is captured in the first group.
Demo
Steps: 509 steps
Flags: g, m
I'm really unsure what the expected output should be (PLEASE UPDATE IT PROPERLY TO THE QUESTION), but here using GNU awk. First test data:
$ cat foo
012 210 5
6343 232 5 3423
345 689 7 986 543012 210 5
234 12 43
Then:
$ awk -F '' '{
delete a
for(i=1;i<=NF;i++)
if($i~/[0-9]/)
a[$i]++
for(i in a)
if(a[i]==1 && match($0, "[^" i "]*" i "[^" i "]*")) {
print $0
next # second data line has 2 matches
}
}' foo
012 210 5
6343 232 5 3423
345 689 7 986 543012 210 5
234 12 43
Then again, its shorter just to:
$ awk '{for(i=0;i<=9;i++)if(gsub(i,i,$0)==1){print;next}}' foo
I'm not absolutely sure what you're after, but if it's matching lines that only contain one instance of a digit, try this:
[^0]*0[^0]*|[^1]*1[^1]*|[^2]*2[^2]*|[^3]*3[^3]*|[^4]*4[^4]*|[^5]*5[^5]*|[^6]*6[^6]*|[^7]*7[^7]*|[^8]*8[^8]*|[^9]*9[^9]*
or grepified
grep -x "[^0]*0[^0]*\|[^1]*1[^1]*\|[^2]*2[^2]*\|[^3]*3[^3]*\|[^4]*4[^4]*\|[^5]*5[^5]*\|[^6]*6[^6]*\|[^7]*7[^7]*\|[^8]*8[^8]*\|[^9]*9[^9]*"
(-x makes grep match the full line.)
The regex uses 10 identical alternations, one for each digit. Each of the alternations
make sure zero or more of anything but the digit starts the line.
match the one allowed digit
make sure zero or more of anything but the digit ends the line.
See it here at regex101.
My file is:
ab 12ab 1cd uu 88 ab 33 33 1 1
ab cd uu 88 88 33 33 33 cw ab
And I need to extract the words and numbers that appears 2-4 times.- {2,4}
I've tried many regex lines and even regex101.
I cant really put my finger on what's not working.
this is the closest I've got so far:
egrep -o '[\w]{2,4}' A1
Native grep doesn't supoort \w and {} notations. You have to use extended regular expressions.
Use
-E option as,
-E, --extended-regexp
Interpret pattern as an extended regular expression (i.e. force grep to behave as egrep).
Also use
-w to match words, so that it matches the entire words instead of partial.
-w, --word-regexp
The expression is searched for as a word (as if surrounded by [[:<:]]' and[[:>:]]'; see re_format(7)).
Example
$ grep -Ewo "\w{2,4}" file
ab
12ab
1cd
uu
88
ab
33
33
ab
cd
uu
88
88
33
33
33
cw
Note
You can eliminated use of an un-necessary cat by providing file as input to grep instead.
You were very close; within character class notation [], the special notation \w is being treated literally, put it out of []:
egrep -o '\w{2,4}'
Also egrep is deprecated in favor of grep -E, and you don't need the cat as grep takes file(s) as argument(s):
grep -Eo '\w{2,4}' file.txt
I would use awk for it:
awk '{for(i=1;i<=NF;i++)a[$i]++}
END{for(x in a)if(a[x]>1&&a[x]<5)print x}' file
It will scan the whole file, find out the words with occurrence (in the file) in this range [2,4]
Output is:
uu
ab
88
1
Using AWK, this solution counts the word instances per line not per file:
awk '{delete array; for(i = 1; i <= NF; i++) array[$i]+=1; for(i in array) if(array[i] >= 2 && array[i] <= 4) printf "%s ", i; printf "\n" }' input.txt
Delete to clear the array for each new line. Use fields as hash for array indexes and increment it's value by one. Print the index (field) with values between 2 and 4 inclusive.
Output:
ab 1 33
ab 88 33
Perl implementation for a file small enough to process its content as a single string:
$/ = undef;
$_ = <>;
#_ = /(\b\w+\b)/gs;
my %h; $h{$_}++ for #_;
for (keys %h) {
print "$_\n" if $h{$_} >= 2 and $h{$_} <= 4;
}
Save it into a script.pl and run:
perl script.pl < file
Of course, you can pass the code via -e option as well: perl -e 'the code' < file.
Input
ab 12ab 1cd uu 88 ab 33 33 1 1
ab cd uu 88 88 33 33 33 cw ab
Output
88
uu
ab
1
There is no 33 in the output, since it occurs 5 times in the input.
The code reads the file in slurp mode into the default variable ($_), then collects all the words (\w with word boundaries around) into #_ array. Then it counts the number of times each word occurred in the file and stores the result into %h hash. The final block prints only the items that occurred 2, 3, or 4 times, no more and no less.
Note, in Perl you should always use strict; and use warnings; in order to detect issues at early phase.
I try to use "uniq -c" to count 2nd string in line
My file A has around 500,000 lines, and looks like this
File_A
30-Nov 20714 GHI 235
30-Nov 10005 ABC 101
30-Nov 10355 DEF 111
30-Nov 10005 ABC 101
30-Nov 10005 ABC 101
30-Nov 10355 DEF 111
30-Nov 10005 ABC 101
30-Nov 20714 GHI 235
...
The command I used
sort -k 2 File_A | uniq -c
I find out the result i get doesn't match the lines.
How can i fix this problem? or Does there has other way to count unique string in line?
The result i get will similar like this (i just random made out the number)
70 30-Nov 10005 ABC 101
5 30-Nov 10355 DEF 111
55 30-Nov 20714 GHI 235
You need to also tell uniq to consider only that field, the same way you did with sort. Perhaps you can use -f of --skip-fields for that. The problem you then have is that uniq doesn't take a "number of fields to check".
Otherwise, if you don't need to keep the original string you can just:
cut -d' ' -f2 | sort ...
Here are a couple, or three, other ways to do it. These solutions have the benefit that the file is not sorted - rather they rely on hashes (associative arrays) to keep track of unique occurrences.
Method 1:
perl -ane 'END{print scalar keys %h,"\n"}$h{$F[1]}++' File_A
The "-ane" makes Perl loop through the lines in File_A, and sets elements of the array F[] equal to the fields of each line as it goes. So your unique numbers end up in F[1]. %h is a hash. The hash element indexed by $F[1] is incremented as each line is processed. At the end, the END{} block is run, and it simply prints the number of elements in the hash.
Method 2:
perl -ane 'END{print "$u\n"}$u++ if $h{$F[1]}++==1' File_A
Similar to the method above, but this time a variable $u is incremented each time incrementing the hash results in it becoming 1 - i.e. the first time we see that number.
I am sure #mpapec or #fedorqui could do it in half the code, but you get the idea!
Method 3:
awk 'FNR==NR{a[$2]++;next}{print a[$2],$0}END{for(i in a)u++;print u}' File_A File_A
Result:
2 30-Nov 20714 GHI 235
4 30-Nov 10005 ABC 101
2 30-Nov 10355 DEF 111
4 30-Nov 10005 ABC 101
4 30-Nov 10005 ABC 101
2 30-Nov 10355 DEF 111
4 30-Nov 10005 ABC 101
2 30-Nov 20714 GHI 235
3
This uses awk and runs through your file twice - that is why it appears twice at the end of the command. On the first pass, the code in curly braces after "FNR==NR" is run and it increments the element of associative array a[] as indexed by field 2 ($2) so it is essentially counting the number of times each id in field 2 is seen. Then, on the second pass, the part in the second set of curly braces is run and it prints the total number of times the id was seen on the first pass, plus the current line. At the end, the END{} block is run and it counts the elements in associative array a[] and prints that out.
If your intention is to count the unique values in the second column, the one that has 20714, 10005, ... in it, then you need to extract it first using cut.
cut -d' ' -f 2 File_A | sort | uniq -c
I have a whitespace delimited file with a variable number of entries on each line. I want to replace the first two whitespaces with commas to create a comma delimited file with three columns.
Here's my input:
a b 1 2 3 3 2 1
c d 44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z y 2 3 33
And here's my desired output:
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33
I'm trying to use perl regular expressions in a sed command but I can't quite get it to work. First I try capturing a word, followed by a space, then another word, but that only works for lines 1, 2, and 5:
$ cat test | sed -r 's/(\w)\s+(\w)\s+/\1,\2,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z,y,2 3 33
I also try capturing whitespace, a word, and then more whitespace, but that gives me the same result:
$ cat test | sed -r 's/\s+(\w)\s+/,\1,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z,y,2 3 33
I also try doing this with the .? wildcard, but that does something funny to line 4.
$ cat test | sed -r 's/\s+(.?)\s+/,\1,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line http://google.com 100 200 300
ef jh,,77 88 99
z,y,2 3 33
Any help is much appreciated!
How about this:
sed -e 's/\s\+/,/' | sed -e 's/\s\+/,/'
It's probably possible with a single sed command, but this is sure an easy way :)
My output:
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33
Try this:
sed -r 's/\s+(\S+)\s+/,\1,/'
Just replaced \w (one "word" char) with \S+ (one or more non-space chars) in one of your attempts.
You can provide multiple commands to a single instance of sed by just providing multiple -e arguments.
To do the first two, just use:
sed -e 's/\s\+/,/' -e 's/\s\+/,/'
This basically runs both commands on the line in sequence, the first doing the first block of whitespace, the second doing the next.
The following transcript shows this in action:
pax$ echo 'a b 1 2 3 3 2 1
c d 44 55 66 2355
line http://google.com 100 200 300
ef jh 77 88 99
z y 2 3 33
' | sed -e 's/\s\+/,/' -e 's/\s\+/,/'
a,b,1 2 3 3 2 1
c,d,44 55 66 2355
line,http://google.com,100 200 300
ef,jh,77 88 99
z,y,2 3 33
Sed s/// supports a way to say which occurrence of a pattern to replace: just add the n to the end of the command to replace only the nth occurrence. So, to replace the first and second occurrences of whitespace, just use it this way:
$ sed 's/ */,/1;s/ */,/2' input
a,b ,1 2 3 3 2 1
c,d ,44 55 66 2355
line,http://google.com 100,200 300
ef,jh ,77 88 99
z,y 2,3 33
EDIT: reading another proposed solutions, I noted that the 1 and 2 after s/ */,/ is not only unnecessary but plainly wrong. By default, s/// just replaces the first occurrence of the pattern. So, if we have two identical s/// in sequence, they will replace the first and the second occurrence. What you need is just
$ sed 's/ */,/;s/ */,/' input
(Note that you can put two sed commands in one expression if you separate them by a semicolon. Some sed implementations do not accept the semicolon after the s/// command; use a newline to separate the commands, in this case.)
A Perl solution is:
perl -pe '$_=join ",", split /\s+/, $_, 3' some.file
Not sure about sed/perl, but here's an (ugly) awk solution. It just prints fields 1-2, separated by commas, then the remaining fields separated by space:
awk '{
printf("%s,", $1)
printf("%s,", $2)
for (i=3; i<=NF; i++)
printf("%s ", $i)
printf("\n")
}' myfile.txt