I have a rather big file with 255 coma separated columns and I need to print out every third column only.
I was trying something like this
awk '{ for (i=0;i<=NF;i+=3) print $i }' file
but that doesn't seem to be the solution, since it prints to only one long column. Anybody can help? Thanks
Here is one way to do this.
The script prog.awk:
BEGIN {FS = ","} # field separator
{for (i = 1; i <= NF; i += 3) printf ("%s%c", $i, i + 3 <= NF ? "," : "\n");}
Invocation:
awk -f prog.awk <input.csv >output.csv
Example input.csv:
1,2,3,4,5,6,7,8,9,10
11,12,13,14,15,16,17,18,19,20
Example output.csv:
1,4,7,10
11,14,17,20
It behaves like that because by default awk splits fields in spaces. You have to tell it to split them with commas, and it's done using the FS variable or the -F switch. Besides that, first field is number one. The zero is the whole line, so also change the initial value of the for loop:
awk -F',' '{ for (i=1;i<=NF;i+=3) print $i }' file
Related
I have a file with a column named (effect) which has rows separated by blank lines,
(effect)
1
1
1
(effect)
1
1
1
1
(effect)
1
1
I know how to print the sum of column like
awk '{sum+=$1;} END{print sum;}' file.txt
Using awk how can I print the sum of each (effect) in for loop? such that I have three lines or multiple lines in other cases like below
sum=3
sum=4
sum=2
You can check if there is an (effect) part, and print the sum when encountering either the (effect) part or when in the END block.
awk '
$1 == "(effect)" { if(seen) print "sum="sum; seen = 1; sum = 0 }
/[0-9]/ { sum += $1 }
END { if (seen) print "sum="sum }
' file
Output
sum=3
sum=4
sum=2
With your shown samples, please try following awk code. Written and tested in GNU awk.
awk -v RS='(^|\n)?\\(effect\\)[^(]*' '
RT{
gsub(/\(effect\)\n|\n+[[:space:]]*$/,"",RT)
num=split(RT,arr,ORS)
print "sum="num
}
' Input_file
Explanation: Simple explanation would be, using GNU awk. In awk program set RS as (^|\n)?\\(effect\\)[^(]* regex for whole Input_file. In main program checking condition if RT is NOT NULL then using gsub(Global substitution) function to substitute (effect)\n and \n+[[:space:]]*$(new lines followed by spaces at end of value) with NULL in RT. Then splitting value of RT into array named arr with delimiter of ORS and saving its(total contents value OR array length value) into variable named num, then printing sum= along with value of num here to get required results.
With shown samples, output will be as follows:
sum=3
sum=4
sum=2
This should work in any version of awk:
awk '{sum += $1} $0=="(effect)" && NR>1 {print "sum=" sum; sum=0}
END{print "sum=" sum}' file
sum=3
sum=4
sum=2
Similar to #Ravinder's answer, but does not depend on the name of the header:
awk -v RS='' -v FS='\n' '{
sum = 0
for (i=2; i<=NF; i++) sum += $i
printf "sum=%d\n", sum
}' file
RS='' means that sequences of 2 or more newlines separate records.
The Field Separator is newline.
The for loop omits field #1, the header.
However that means that empty lines truly need to be empty: no spaces or tabs allowed. If your data might have blank lines that contain whitespace, you can set
-v RS='\n[[:space:]]*\n'
$ awk -v RS='(effect)' 'NR>1{sum=0; for(i=1;i<=NF;i++) sum+=$i; print "sum="sum}' file
sum=3
sum=4
sum=2
I will like to duplicate each line 2 times and print values of column 5 and 6 separated.( transpose values of column 5 and 6 from column to row ) for each line
I mean value on column 5 (first line) value in column 6 ( second line)
Input File
08,1218864123180000,3201338573,VV,22,27
08,1218864264864000,3243738789,VV,15,23
08,1218864278580000,3244738513,VV,3,13
08,1218864310380000,3243938789,VV,15,23
08,1218864324180000,3244538513,VV,3,13
08,1218864334380000,3200538561,VV,22,27
Desired Output
08,1218864123180000,3201338573,VV,22
08,1218864123180000,3201338573,VV,27
08,1218864264864000,3243738789,VV,15
08,1218864264864000,3243738789,VV,23
08,1218864278580000,3244738513,VV,3
08,1218864278580000,3244738513,VV,13
08,1218864310380000,3243938789,VV,15
08,1218864310380000,3243938789,VV,23
08,1218864324180000,3244538513,VV,3
08,1218864324180000,3244538513,VV,13
08,1218864334380000,3200538561,VV,22
08,1218864334380000,3200538561,VV,27
I use this code to duplicate the lines 2 times, but i cant'n figer out the condition with values of column 5 and 6
awk '{print;print}' file
Thanks in advance
To repeatedly print the start of a line for each of the last N fields where N is 2 in this case:
$ awk -v n=2 '
BEGIN { FS=OFS="," }
{
base = $0
sub("("FS"[^"FS"]+){"n"}$","",base)
for (i=NF-n+1; i<=NF; i++) {
print base, $i
}
}
' file
08,1218864123180000,3201338573,VV,22
08,1218864123180000,3201338573,VV,27
08,1218864264864000,3243738789,VV,15
08,1218864264864000,3243738789,VV,23
08,1218864278580000,3244738513,VV,3
08,1218864278580000,3244738513,VV,13
08,1218864310380000,3243938789,VV,15
08,1218864310380000,3243938789,VV,23
08,1218864324180000,3244538513,VV,3
08,1218864324180000,3244538513,VV,13
08,1218864334380000,3200538561,VV,22
08,1218864334380000,3200538561,VV,27
In this simple case where the last field has to be removed and placed on the last line, you can do
awk -F , -v OFS=, '{ x = $6; NF = 5; print; $5 = x; print }'
Here -F , and -v OFS=, will set the input and output field separators to a comma, respectively, and the code does
{
x = $6 # remember sixth field
NF = 5 # Set field number to 5, so the last one won't be printed
print # print those first five fields
$5 = x # replace value of fifth field with remembered value of sixth
print # print modified line
}
This approach can be extended to handle fields in the middle with a function like the one in the accepted answer of this question.
EDIT: As Ed notes in the comments, writing to NF is not explicitly defined to trigger a rebuild of $0 (the whole-line record that print prints) in the POSIX standard. The above code works with GNU awk and mawk, but with BSD awk (as found on *BSD and probably Mac OS X) it fails to do anything.
So to be standards-compliant, we have to be a little more explicit and force awk to rebuild $0 from the modified field state. This can be done by assigning to any of the field variables $1...$NF, and it's common to use $1=$1 when this problem pops up in other contexts (for example: when only the field separator needs to be changed but not any of the data):
awk -F , -v OFS=, '{ x = $6; NF = 5; $1 = $1; print; $5 = x; print }'
I've tested this with GNU awk, mawk and BSD awk (which are all the awks I can lay my hands on), and I believe this to be covered by the awk bit in POSIX where it says "setting any other field causes the re-evaluation of $0" right at the top. Mind you, the spec could be more explicit on this point, and I'd be interested to test if more exotic awks behave the same way.
Could you please try following(considering that your Input_file always is same as shown and you need to print every time 1st four fields and then rest of the fields(one by one printing along with 1st four)).
awk 'BEGIN{FS=OFS=","}{for(i=5;i<=NF;i++){print $1,$2,$3,$4,$i}}' Input_file
This might work for you (GNU awk):
awk '{print gensub(/((.*,).*),/,"\\1\n\\2",1)}' file
Replace the last comma by a newline and the previous fields less the penultimate.
The Problem
I just need to combine a whole bunch of files and strip out the header (line 1) from the 1st file.
The Data
Here are the last three lines (with line 1: header) from three of these files:
"START_DATE","END_DATE","UNITS","COST","COST_CURRENCY","AMOUNT"
"20170101","20170131","1","5.49","EUR","5.49"
"20170101","20170131","1","4.27","EUR","4.27"
"","","","","9.76",""
"START_DATE","END_DATE","UNITS","COST","COST_CURRENCY","AMOUNT"
"20170201","20170228","1","5.49","EUR","5.49"
"20170201","20170228","1","4.88","EUR","4.88"
"20170201","20170228","1","0.61","EUR","0.61"
"20170201","20170228","1","0.61","EUR","0.61"
"","","","","11.59",""
START_DATE","END_DATE","UNITS","COST","COST_CURRENCY","AMOUNT"
"20170301","20170331","1","4.88","EUR","4.88"
"20170301","20170331","1","4.27","EUR","4.27"
"","","","","9.15",""
Problem (Continued)
As you can see, the last line has a number (it's a column total) in column 5. Of course, I don't want that last line. But it's (obviously) on a different line number in each file.
(G)awk is clearly the solution, but I don't know (g)awk.
What I've Tried
I've tried a number of combinations of things, but I guess the one that I'm most surprised does not work is:
gawk '
{ if (!$1 ) nextfile }
NR == 1 {$0 = "Filename" "StartDate" OFS $0; print}
FNR > 1 {$0 = FILENAME StartDate OFS $0; print}
' OFS=',' */*.csv > ../path/file.csv
Expected Output (by request)
"START_DATE","END_DATE","UNITS","COST","COST_CURRENCY","AMOUNT
20170101","20170131","1","5.49","EUR","5.49
20170101","20170131","1","4.27","EUR","4.27
20170201","20170228","1","5.49","EUR","5.49
20170201","20170228","1","4.88","EUR","4.88
20170201","20170228","1","0.61","EUR","0.61
20170201","20170228","1","0.61","EUR","0.61
20170301","20170331","1","4.88","EUR","4.88
20170301","20170331","1","4.27","EUR","4.27"
And, of course, I've tried searching both Google and SO. Most of the answers I see require much more awk knowledge than I have, just to understand them. (I'm not a data wrangler, but I have a data wrangling task.)
Thanks for any help!
this should do...
awk 'NR==1; FNR==1{next} FNR>2{print p} {p=$0}' file{1..3}
print first header, skip other headers and last lines.
Another awk approach:-
awk -F, '
NR == 1 {
header = $0
print
next
}
FNR > 1 && $1 != "\"\""
' *.csv
Something like the following should do the trick:
awk -F"," 'NR==1{header=$0; print $0} $0!=header && $1!=""{print $0}' */*.csv > ../path/file.csv\
Here awk will:
Split the records by comma -F","
If this is the first record awk encounters, it sets variable header to the entire contents of the line and then prints the header NR==1{header=$0; print $0}
If the contents of the current line are not a header and the first field isn't empty (indicating a "total" line), then print the line $0!=header && $1!=""{print $0}'
As mentioned in my comment below, if the first field of your records always begin with an 8 digit date, then you could simplify (this is less generic than the code above):
awk -F"," 'NR == 1 || $1 ~ /"[0-9]{8}"/ {print $0} /*.csv > outfile.csv
Essentially that says if this is the first record to process then print it (it's a header) OR || if the first field is an 8 digit number surrounded by double quotes then print it.
I have a text file with 3 columns as below.
$ cat test.txt
1,A,300
1,B,300
1,C,300
Till now i have tried as, awk -F, '{$3=$3+1;print}' OFS=, test.txt
But output is coming as:
1,A,301
1,B,301
1,C,301
& below is my desired output
Now i want to increment the third column only, the output should be like below
1,A,300
1,B,301
1,C,302
How can I achieve the desired output?
could be, assuming line are sequential like your sample)
awk -F ',' '{sub($3"$",$3+NR-1)}7' YourFile
it use the line numer as increment value, changing the line end and not the field value (different from an awk POV, that don't need to rebuild the line with separator)
Alternative if empty or other line between modifiable lines (i arbitrary use NF as filter but it depend of your criteria if any)
awk -F ',' 'NF{sub($3"$",$3+i++)}7' YourFile
awk 'BEGIN{x=0;FS=OFS=","} NF>1{$3=$3+x;x++}1' inputfile
1,A,300
1,B,301
1,C,302
Explanation:
BEGIN Block : It contains x which is a counter initially set to zero, FS and OFS . /./ is used to ignore blank lines(Remove this part if there are no blank lines). $3=$3+x : This will add the value of counter to $3. x++ : To increment the current value of the counter.
try this NR starts at 1 so NR -1 should give you the correct number
awk -F, '{$3=$3+NR-1;print}' OFS=, test.txt
Yet another:
awk 'BEGIN{ FS=OFS="," } ($3+=i++)||1 ' file
awk 'BEGIN{i=0;FS=OFS=","} NF>1{$3=$3+i;i++}1' filename
It contains x which is a counter initially set to zero, FS and OFS . /./ is used to ignore blank lines(Remove this part if there are no blank lines).
$3=$3+i : This will add the value of counter to $3. i++ : To increment the value of counter. Must and should give space betwen awk and begin as well as filename and end of the file
I was wondering if there's a built in command in awk to get the field number of the phrase that you just matched.
Banana is yellow.
awk {
/yellow/{ for (i=1;i<=NF;i++) if($i ~/yellow/) print $i}'
Is there a way to avoid writing the loop?
Your command doesn't work when I test it. Here's my version:
echo "banana is yellow" | awk '{for (i=1;i<=NF;i++) if($i ~/yellow/) print i}'
The output is :
3
As far as I know, there's no such built-in feature, to improve your command, the pattern match /yellow/ at the beginning is not necessary, and also $i will print the matching field other than the field number that you need.
Alternatively, you can use an array to store each field and its corresponding index number, and then print field by arr["yellow"]
If the input string is a oneline string you can set the record delimiter to the field delimiter. Doing so you can use NR to print the position:
awk 'BEGIN{RS=FS}/yellow/{print NR}' <<< 'banana is yellow'
3