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Grepping exact pattern and deleting the rows

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# 1  
Old 05-02-2016
Grepping exact pattern and deleting the rows
Hi All,

I have the Input as below:

Code:
21011513    6030    5570    1710       0       0    5140       0    3430    3430       0
21049513  152340  138260  101210       0       0  134440       0   33880   33880       0
31003514   16100   13280    7580    3250    1530   15650    8090    8100    8100       0
31513514   28720   23310   11840    6140    2330   27180   15960   15340   15340       0
16161515    6380    5370    3750     820     530    5950    1790    2200    2200
45141518    7580    6430      40       0     610    6730    3510    6690    6690       0
39019519  152060  130190   92940       0       0  132790       0   39850   39850       0

I want to grep the lines containing 513 and 515 in the positions 6 to 9. and i want to delete those rows. So my output should look like below.

Code:
31003514   16100   13280    7580    3250    1530   15650    8090    8100    8100       0
31513514   28720   23310   11840    6140    2330   27180   15960   15340   15340       0
45141518    7580    6430      40       0     610    6730    3510    6690    6690       0
39019519  152060  130190   92940       0       0  132790       0   39850   39850       0

I have tried the below code:


nawk -F, '$1 !~ /513|515/' sample > op2

With this code i am getting the output as below:

Code:
31003514   16100   13280    7580    3250    1530   15650    8090    8100    8100       0
45141518    7580    6430      40       0     610    6730    3510    6690    6690       0
39019519  152060  130190   92940       0       0  132790       0   39850   39850       0

Since the Row 4 in my Input contains '513' in the positions 3 to 5, that row also getting ignored.

Is there any better way to achieve the desired output ?

Thanks in Advance.

Regards,
am24
# 2  
Old 05-02-2016
Hello am24,

Could you please try following and let me know if this helps.
Code:
awk '($1 !~ /513$/ && $1 !~ /515$/)'  Input_file

Output will be as follows.
Code:
31003514   16100   13280    7580    3250    1530   15650    8090    8100    8100       0
31513514   28720   23310   11840    6140    2330   27180   15960   15340   15340       0
45141518    7580    6430      40       0     610    6730    3510    6690    6690       0
39019519  152060  130190   92940       0       0  132790       0   39850   39850       0

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
am24
# 3  
Old 05-02-2016
Why use -F, when there is no comma as field separator?

Code:
$ awk '$1 !~ /51[35]$/' infile
31003514   16100   13280    7580    3250    1530   15650    8090    8100    8100       0
31513514   28720   23310   11840    6140    2330   27180   15960   15340   15340       0
45141518    7580    6430      40       0     610    6730    3510    6690    6690       0
39019519  152060  130190   92940       0       0  132790       0   39850   39850       0

This User Gave Thanks to zaxxon For This Post:
am24
# 4  
Old 05-02-2016
We are talking of positions 6 - 8, aren't we? Try
Code:
grep -v '^.....51[35] ' file

# 5  
Old 05-02-2016
Hi Ravinder/Zaxxon

Thank you so much for the quick response. Both of your codes worked fine. I have got the expected output. Thanks Zaxxon for reminding me about the Field Separator.

Just to understand, the '$' in the code indicates the end of the column right ? Please correct me if i am wrong.

What if i want to ignore any row that contains '513' in the mid position of the column ?

Say like below:
31513514 28720 23310 11840 6140 2330 27180 15960 15340 15340 0

in the above 513 is in positions 3 to 5. How can i ignore this particular row?

Regards,
am24
# 6  
Old 05-02-2016
Hello am24,

Could you please try following and let me know if this helps you.
Code:
awk '(substr($1,3,3) != 51[35])'  Input_file

Thanks,
R. Singh
# 7  
Old 05-02-2016
Thanks Rudi.

I have tried the code given by you. It worked fine. Also i got the logic. The ^ symbol is for the exact match and each '.' is referring to the position starting from 1.

Am I correct ?

If i am correct then, the solution for teh second question i asked will be
grep -v '^..513... ' sample > op6

Thanks again :-)

Regards,
am24

---------- Post updated at 07:15 AM ---------- Previous update was at 07:10 AM ----------

Hi Ravinder,

Thanks. I have tried the code. But getting an error message as below.

Code:
nawk '(substr($1,3,3) != 51[35])'  sample > op8
awk: syntax error near line 1
awk: bailing out near line 1

Regards,
am24
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