Sort a las file keep the header as it is


 
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# 1  
Old 09-11-2015
Question Sort a las file keep the header as it is

I have several las files with a header and each file start Version and text and before the data starts end up with ~Ascii, then the numbers starts:
-------------------------------------------------------------------------
Code:
~Version
.....text....
~Ascii
2  abc  230   1 name
1  abc  400   1 name2
3  abb  350   2  name

----------------------------------------------------------------------------
What I want to do is to keep the header as it is and sort with the first column
the result to look something as following :
Code:
~Version
.....text....
~Ascii
1  abc  400   1 name2
2  abc  230   1 name
3  abb  350   2  name

and if it is possible to run for all files in one go, since all the files are *.las file?

thank
T

Last edited by vbe; 09-11-2015 at 09:58 AM.. Reason: code tags
# 2  
Old 09-11-2015
Operating inline of the file? Hard.

If you can tolerate a stream of output from processing the file twice, then ok. I'm assuming only one ~Ascii in the file.

Code:
sed -n '1,/^~Ascii/p' <test.las; sed  '1,/^~Ascii/d' <test.las | sort -k1,1n

# 3  
Old 09-13-2015
If you are saying that you want to sort all of the *.las files together into a single output file keeping the header from the 1st file and throwing away the other headers, you could try:
Code:
sed -n '1,/~Ascii/p;/~Ascii/q' *.las;sed '/~Version/,/~Ascii/d' *.las|sort -k1,1n

# 4  
Old 09-14-2015
keep each file sorted

Thanks Don/cjcox

I want to sort each file and the output should be new file for each file.
When I tried the one you post. It doesnot give me output file. And
each new file should keep the header and the data sorted.
# 5  
Old 09-14-2015
Assuming you want to sort the files in place and you don't have any hard links or non-default file modes or group-IDs you need to preserve in your .las files, a simple loop of the cjcox's suggestion should do what you want:
Code:
for file in *.las
do	sed -n '1,/~Ascii/p;/~Ascii/q' "$file" > "$file.$$" && \
	sed '1,/~Ascii/d' "$file" | sort -k1,1n >> "$file.$$" && \
	mv "$file.$$" "$file" || rm -f "$file.$$"
done

If you want new files created instead of replacing the original files, try:
Code:
for file in *.las
do	sed -n '1,/~Ascii/p;/~Ascii/q' "$file" > "$file.new"
	sed '1,/~Ascii/d' "$file" | sort -k1,1n >> "$file.new"
done


Last edited by Don Cragun; 09-14-2015 at 01:44 AM.. Reason: Simplify second sed script in both lists.
These 2 Users Gave Thanks to Don Cragun For This Post:
# 6  
Old 09-14-2015
More efficient with awk:
Code:
for file in *.las
do
  awk '(sort==1) {print | "sort -k1,1n"; next} ($1~/^~Version/) {sort=1} {print}' "$file" > "$file.$$" &&
  cp "$file.$$" "$file" &&
  rm -f "$file.$$"
done

Would be even shorter with perl, thanks to its -i option and internal sort.

Last edited by MadeInGermany; 09-14-2015 at 09:41 AM.. Reason: use the precise sort option
These 2 Users Gave Thanks to MadeInGermany For This Post:
# 7  
Old 09-14-2015
Quote:
Originally Posted by MadeInGermany
More efficient with awk:
Code:
for file in *.las
do
  awk '(sort==1) {print | "sort -k1,1n"; next} ($1~/^~Version/) {sort=1} {print}' "$file" > "$file.$$" &&
  cp "$file.$$" "$file" &&
  rm -f "$file.$$"
done

Would be even shorter with perl, thanks to its -i option and internal sort.
I'd have to do some benchmarking to see if sed, sed | sort is slower than using awk to invoke sh to invoke sort; since there are three fork() and exec family (or spawn()) calls either way; but I like the concept.

But, shouldn't:
Code:
($1~/^~Version/) {sort=1}

be:
Code:
($1~/^~Ascii/) {sort=1}

in your script?
This User Gave Thanks to Don Cragun For This Post:
 
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