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copying columns with headers' specific pattern

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Top Forums Shell Programming and Scripting copying columns with headers' specific pattern
# 8  
Old 06-01-2010
Quote:
Originally Posted by jacks
it keeps selecting all columns.
Hi, could you indicate where your input file differs from my test sample?

Code:
$ cat infile
*sml1   *sml3   *smln7  sm*fk9  smllf56 *tr     yty
1       2       3       4       5       6       7
a       b       c       d       e       f       g

Code:
$ awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^*/) A[i]; if (i in A) printf $i"\t"} print ""}' infile
*sml1   *sml3   *smln7  *tr
1       2       3       6
a       b       c       f



---------- Post updated at 13:15 ---------- Previous update was at 13:13 ----------

I tried it with your newly supplied input and I get this:
Code:
$ awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^*/) A[i]; if (i in A) printf $i"\t"} print ""}' infile2
*smt_0018       *smlf_0042      *sugk_0053
11.2902 10.7395 7.9672
8.98292 9.69812 7.7105
7.71334 8.05211 7.57295
9.07925 8.96038 8.1141
10.1289 9.17119 7.52616
11.5821 8.76748 8.19344
9.18122 10.1511 13.4099

# 9  
Old 06-01-2010
This is my output with your file:
Code:
$ cat file
*smt_0018 smf_0031 *smlf_0042 *sugk_0053
11.2902 12.4807 10.7395 7.9672
8.98292 9.03125 9.69812 7.7105
7.71334 8.28308 8.05211 7.57295
9.07925 9.17628 8.96038 8.1141
10.1289 8.9778 9.17119 7.52616
11.5821 9.54838 8.76748 8.19344
9.18122 9.71467 10.1511 13.4099
$
$ tr ' ' '\n' < file | grep '^\*'                             
*smt_0018
*smlf_0042
*sugk_0053

# 10  
Old 06-01-2010
Hi,
I did not understand how it differs.

$ cat sample
*sml_0018 *smmt_0031 sdmt_0042 *srmt_0053
1 1 1 1
11.2902 12.4807 10.7395 7.9672
8.98292 9.03125 9.69812 7.7105
7.71334 8.28308 8.05211 7.57295
9.07925 9.17628 8.96038 8.1141
10.1289 8.9778 9.17119 7.52616
11.5821 9.54838 8.76748 8.19344
9.18122 9.71467 10.1511 13.4099



$ awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^*/) A[i]; if (i in A) printf $i"\t"} print ""}' sample
*sml_0018 *smmt_0031 sdmt_0042 *srmt_0053
1 1 1 1
11.2902 12.4807 10.7395 7.9672
8.98292 9.03125 9.69812 7.7105
7.71334 8.28308 8.05211 7.57295
9.07925 9.17628 8.96038 8.1141
10.1289 8.9778 9.17119 7.52616
11.5821 9.54838 8.76748 8.19344
9.18122 9.71467 10.1511 13.4099
Last edited by jacks; 06-01-2010 at 08:28 AM..
# 11  
Old 06-01-2010
Using your latest sample I get:
Code:
$ awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^*/) A[i]; if (i in A) printf $i"\t"} print ""}' sample
*sml_0018       *smmt_0031      *srmt_0053
1       1       1
11.2902 12.4807 7.9672
8.98292 9.03125 7.7105
7.71334 8.28308 7.57295
9.07925 9.17628 8.1141
10.1289 8.9778  7.52616
11.5821 9.54838 8.19344
9.18122 9.71467 13.4099

What do you get if you escape the *:
Code:
awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^\*/) A[i]; if (i in A) printf $i"\t"} print ""}' infile

Are you on Solaris? Then you need to use nawk or /usr/xpg4/bin/awk
This User Gave Thanks to Scrutinizer For This Post:
jacks
# 12  
Old 06-01-2010
Oh my, didn't realize it's a tab delimited file, try:
Code:
tr '\t' '\n' < file | grep '^\*' > newfile

Please use code tags.
These 2 Users Gave Thanks to Franklin52 For This Post:
bankimmehta jacks
# 13  
Old 06-01-2010
Hi Scrutinizer,

This works nice :

awk '{for (i=1;i<=NF;i++) {if (NR==1 && $i~/^\*/) A[i]; if (i in A) printf $i"\t"} print ""}' infile

Can you please tell me why it worked this time ?


Thanks to you and Franklin.
# 14  
Old 06-01-2010
Hi Jacks,

Glad we found the culprit. * is a wildcard character (it repeats the preceding character). If one needs to use it literally (to match the character "*" it would typically need to be escaped by a backslash character. However since in this particular the * is first character the pattern it could not possibly be a wildcard and hence it should not need an escape. But not every implementation of awk sees it that way apparantly.
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