This sed command is described as follows
Delete the cars that are $10,000 or more. Pipe the output of the sort into a sed to do this, by quitting as soon as we match a regular expression representing 5 (or more) digits at the end of a record (DO NOT use repetition for this):
So far the command is:
$ grep -iv chevy cars | sort -nk 5
I have to add another pipe at the end of that command I think which "quits as soon as we match a regular expression representing 5 or more digits at the end of a record"
I tried things like
$ grep -iv chevy cars | sort -nk 5 | sed "/[0-9][0-9][0-9][0-9][0-9]/ q"
and other variations within the // but nothing works! What is the command which matches a regular expression representing 5 or more digits and quits according to this question?
Nominally, you should add a $ before the second / to match 5 digits at the end of the record. If you omit the $, then any sequence of 5 digits will cause sed to quit, so if there is another number (a VIN, perhaps) before the price, it might match when you didn't intend it to.
grep -iv chevy cars | sort -nk 5 | sed '/[0-9][0-9][0-9][0-9][0-9]$/q'
On the whole, it's safer to use single quotes around the regex, unless you need to substitute a shell variable into it (or unless the regex contains single quotes itself). You can also specify the repetition:
grep -iv chevy cars | sort -nk 5 | sed '/[0-9]\{5,\}$/q'
The \{5,\} part matches 5 or more digits. If for any reason that doesn't work, you might find you're using GNU sed and you need to do something like sed --posix to get it working in the normal mode. Or you might be able to just remove the backslashes. There certainly are options to GNU sed to change the regex mechanism it uses (as there are with GNU grep too).
Another way.
As you don't post a file sample, a did it as a guess.
Here I'm looking for lines with the word "chevy" where the field 5 is less than 10000.
awk '/chevy/ {if ( $5 < 10000 ) print $0} ' cars
I forgot the flag -i from grep ... so the correct is:
awk 'BEGIN{IGNORECASE=1} /chevy/ {if ( $5 < 10000 ) print $0} ' cars
$ cat > cars
Chevy 2 3 4 10000
Chevy 2 3 4 5000
chEvy 2 3 4 1000
CHEVY 2 3 4 10000
CHEVY 2 3 4 2000
Prevy 2 3 4 1000
Prevy 2 3 4 10000
$ awk 'BEGIN{IGNORECASE=1} /chevy/ {if ( $5 < 10000 ) print $0} ' cars
Chevy 2 3 4 5000
chEvy 2 3 4 1000
CHEVY 2 3 4 2000
grep -iv chevy cars | sort -nk 5 | sed '/[0-9][0-9][0-9][0-9][0-9]$/d'
Related
I want to select from a file random lines/units but where the units are consisted of 2 lines.
For example a file looks like this
Adam
Apple
Mindy
Candy
Steve
Chips
David
Meat
Carol
Carrots
And I want to randomly subselect lets say 2 units group
For example
Adam
Apple
David
Meat
or
Steve
Chips
Carol
Carrots
I've tried using shuf and sort -R but they only shuffle 1 lines. Could someone help me please?
Thank you.
You could do it with shuf by joining the lines before shuffling (that might not be a bad idea for a file format in general, if the lines describe a single item):
$ < file sed -e 'N;s/\n/:/' | shuf | head -1 | tr ':' '\n'
Carol
Carrots
The sed loads two lines at a time, and joins them with a colon.
Pick a random number in the correct range, ensure that it is odd (if desired), then use sed to print the 2 lines:
$ a=$(expr $RANDOM % \( $(wc -l < input) / 2 \) \* 2 + 1)
$ sed -n -e ${a}p -e $((a+1))p input
Rather than selecting lines to print, you could walk the file and print each "unit" with a particular probability. For example, to print (roughly) 10% of the "units" in the file, you could do:
awk 'BEGIN{srand()} NR%2 && (rand() < .1) {print; getline; print}' input
Hello I have a file containing these lines:
apple
12
orange
4
rice
16
how to use bash to sort it by numbers ?
Suppose each number is the price for the above object.
I want they are formatted like this:
12 apple
4 orange
16 rice
or
apple 12
orange 4
rice 16
Thanks
A solution using paste + sort to get each product sorted by its price:
$ paste - - < file|sort -k 2nr
rice 16
apple 12
orange 4
Explanation
From paste man:
Write lines consisting of the sequentially corresponding lines from
each FILE, separated by TABs, to standard output. With no FILE, or
when FILE is -, read standard input.
paste gets the stream coming from the stdin (your <file) and figures that each line belongs to the fictional archive represented by - , so we get two columns using - -
sort use the flag -k 2nr to get paste output sorted by second column in reverse numerical order.
you can use awk:
awk '!(NR%2){printf "%s %s\n" ,$0 ,p}{p=$0}' inputfile
(slightly adapted from this answer)
If you want to sort the output afterwards, you can use sort (quite logically):
awk '!(NR%2){printf "%s %s\n" ,$0 ,p}{p=$0}' inputfile | sort -n
this would give:
4 orange
12 apple
16 rice
Another solution using awk
$ awk '/[0-9]+/{print prev, $0; next} {prev=$0}' input
apple 12
orange 4
rice 16
while read -r line1 && read -r line2;do
printf '%s %s\n' "$line1" "$line2"
done < input_file
If you want lines to be sorted by price, pipe the result to sort -k2:
while read -r line1 && read -r line2;do
printf '%s %s\n' "$line1" "$line2"
done < input_file | sort -k2
You can do this using paste and awk
$ paste - - <lines.txt | awk '{printf("%s %s\n",$2,$1)}'
12 apple
4 orange
16 rice
an awk-based solution without needing external paste / sort, using regex, calculating modulo % of anything, or awk/bash loops
{m,g}awk '(_*=--_) ? (__ = $!_)<__ : ($++NF = __)_' FS='\n'
12 apple
4 orange
16 rice
I have a file with many lines of tab separated data in the following format:
1 1 2 2
3 3 4 4
5 5 6 6
...
and I would like to change the format to:
1 1
2 2
3 3
4 4
5 5
6 6
Is there a not too complicated way to do this? I don't have any experience with using awk, sed, etc.
Thanks
If you just want to group your file in blocks of X columns, you can make use of xargs -nX:
$ xargs -n2 < file
1 1
2 2
3 3
4 4
5 5
6 6
To have more control and print an empty line after 4th field, you can also use this awk:
$ awk 'BEGIN{FS=OFS="\t"} {for (i=1;i<=NF;i++) printf "%s%s", $i, (i%2?OFS:RS); print ""}' file
1 1
2 2
3 3
4 4
5 5
6 6
# <-- note there is an empty line here
Explanation
On odd fields, it print FS after it.
On even fields, print RS.
Note FS stands for field separator, which defaults to space, and RS stands for record separator, which defaults to new line. As you have tab as field separator, we redefine it in the BEGIN block.
This is probably the simplest way which allows for customisation
awk '{print $1,$2"\n"$3,$4}' file
For a line between
awk '{print $1,$2"\n"$3,$4"\n"}' file
although fedorquis answer with xargs is probably the simplest if this isn't needed
As Ed pointed out this wouldn't work if there were blanks in the fields, this could be resolved using
awk 'BEGIN{FS=OFS="\t"} {print $1,$2 ORS $3,$4 ORS}' file
Through perl,
perl -pe 's/\t(\d\t\d)$/\n$1\n/g' file
Fed the above command's output to the sed command to delete the last blank line.
perl -pe 's/\t(\d\t\d)$/\n$1\n/g' file | sed '$d'
Acttualy this is my assignment.I have three-four file,related by student record.Every file have two-three student record.like this
Course Name:Opreating System
Credit: 4
123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25
243567 0 1 1 0 1 1 0 1 0 0 0 7 9 12 15 17 15
Every file have different coursename.I did every coursename and studentid move
in one file but now i don't know how to add all marks and move to another file on same place where is id? Can you please tell me how to do it?
It looks like this:
Student# Operating Systems JAVA C++ Web Programming GPA
123456 76 63 50 82 67.75
243567 80 - 34 63 59
I did like this:
#!/bin/sh
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
echo "STUDENT ID" > rsh2
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
echo "GPA" >> rsh2
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
Some comments and a few pointers :
It would help to add 'comments' for each line of code that is not self evident ; i.e. code like mv f f.bak doesn't need to be commented, but I'm not sure what the intent of your many lines of code are.
You insert a comment with the '#' char, like
# concatenate all files that contain the word CREDITS into a file called rsh1
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
Also note that you consistently use all uppercase for your search targets, i.e. CREDITS, when your sample files shows mixed case. Either used correct case for your search targets, i.e.
`grep -l 'Credits'`
OR tell grep to -i(gnore case), i.e.
`grep -il 'Credits'
Your line
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
can be reduced to 1 call to sed (and you have the same case confusion thing going on), try
sed -n '/COURSE/i{;s/COURSE NAME: //gip;}' rsh1 >> rsh2
This means (-n don't print every line by default),
`gip` = global substitute,
= ignore case in matching
print only lines where substituion was made
So you're editing out the string COURSE NAME for any line that has COURSE in it, and only printing those lines' (you probably don't need the 'g' (global) specifier given that you expect only 1 instance per line)
Your line
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
Actually looks pretty good, very advanced, you're trying to 'fold' each 2 lines together into 1 line, right?
But,
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
I'm really confused by this, is this where you're trying to total a students score? (with a sort embedded I guess not). Why do you think you need a sort,
While it is possible to perform arithmetic in sed, it is super-crazy hard, so you can either use bash variables to calculate the values OR use a unix tool that is designed to process text AND perform logical and mathematical operations of the data presented, awk or perl come to mind here
Anyway, one solution to total each score is to use awk
echo "123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25" |\
awk '{for (i=2;i<=NF;i++) { tot+=$i }; print $1 "\t" tot }'
Will give you a clue on how to proceed for that.
Awk has predefined variables that it populates for each file, and each line of text that it reads, i.e.
$0 = complete line of text (as defined by the internal variables RS (RecordSeparator)
which defaults to '\n' new-line char, the unix end-of-line char
$1 = first field in text (as defined by the internal variables FS (FieldSeparator)
which defaults to (possibly multiple) space chars OR tab char
a line with 2 connected spaces chars and 1 tab char has 3 fields)
NF = Number(of)Fields in current line of data (again fields defined by value of FS as
described above)
(there are many others, besides, $0, $n, $NF, $FS, $RS).
you can programatically increment for values like $1, $2, $3, by using a variable as in the example code, like $i (i is a variable that has a number between 2 and NF. The leading '$'
says give me the value of field i (i.e. $2, $3, $4 ...)
Incidentally, your problem could be easily solved with a single awk script, but apparently, you're supposed to learn about cat, cut, grep, etc, which is a very worthwhile goal.
I hope this helps.
Is there a Linux utility or a Bash command I can use to sort a space delimited string of numbers?
Here's a simple example to get you going:
echo "81 4 6 12 3 0" | tr " " "\n" | sort -g
tr translates the spaces delimiting the numbers, into carriage returns, because sort uses carriage returns as delimiters (ie it is for sorting lines of text). The -g option tells sort to sort by "general numerical value".
man sort for further details about sort.
This is a variation from #JamesMorris answer:
echo "81 4 6 12 3 0" | xargs -n1 | sort -g | xargs
Instead of tr, I use xargs -n1 to convert to new lines. The final xargs is to convert back, to a space separated sequence of numbers.
This is a variation on ghostdog74's answer that's too big to fit in a comment. It shows digits instead of names of numbers and both the original string and the result are in space-delimited strings (instead of an array which becomes a newline-delimited string).
$ s="3 2 11 15 8"
$ sorted=$(echo $(printf "%s\n" $s | sort -n))
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2 3 8 11 15
If you didn't use the echo when setting the value of sorted, then the string has newlines in it. In that case echoing it without quotes puts it all on one line, but, as echoing it with quotes would show, each number would appear on its own line. This is the case whether the original is an array or a string.
# demo
$ s="3 2 11 15 8"
$ sorted=$(printf "%s\n" $s | sort -n)
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2
3
8
11
15
$ s=(one two three four)
$ sorted=$(printf "%s\n" ${s[#]}|sort)
$ echo $sorted
four one three two
Using Bash parameter expansion (to replace spaces with newlines) we can do:
str="3 2 11 15 8"
sort -n <<< "${str// /$'\n'}"
# alternative
NL=$'\n'
str="3 2 11 15 8"
sort -n <<< "${str// /${NL}}"
If you actually have a space-delimited string of numbers, then one of the other answers provided would work fine. If your list is a bash array, then:
oldIFS="$IFS"
IFS=$'\n'
array=($(sort -g <<< "${array[*]}"))
IFS="$oldIFS"
might be a better solution. The newline delimiter would help if you want to generalize to sorting an array of strings instead of numbers.
Improving on Evan Krall's nice Bash "array sort" by limiting the scope of IFS to a single command:
printf "%q\n" "${IFS}"
array=(3 2 11 15 8)
array=($(IFS=$'\n' sort -n <<< "${array[*]}"))
echo "${array[#]}"
printf "%q\n" "${IFS}"
$ awk 'BEGIN{split(ARGV[1], numbers);for(i in numbers) {print numbers[i]} }' \
"6 7 4 1 2 3" | sort -n
I added this to my .zshrc (or .bashrc) file:
#sort a space-separated list of words (e.g. a list of HTML classes)
sortwords() {
echo $1 | xargs -n1 | sort -g | xargs
}
Call it from the terminal like this:
sortwords "banana date apple cherry"
# apple banana cherry date
Thanks to #FranMowinckel and others for inspiration.