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Print numbers from table

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

0

How to get four numbers from this table please?

Input:

chil           2   44843.084886752418        44843.080972494339       0.38261625919046571     



o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)

o 0  2 0 1 1  54002.864912732     0.000 = PERIASTRON EPOCH

o 0  3 0 1 1      0.086796671     0.000 = ECCENTRICITY

o 0  4 0 1 1    151.562809500     0.020 = PERIASTRON LONG.

o 0  5 0 1 1    107.927001263     0.053 = K1

o 0  6 0 1 1      0.418206121     0.000 = q = M2/M1,  K2 =  258.071309

o 0  7 0 1 1      0.004220000     0.000 = d omega/dt

o 0  8 0 1 1      0.000000000     0.100 = d P/dt

o 0  9 0 1 1      0.000000000     0.100 = d e/dt

o 0 10 0 1 1      0.000000000     0.100 = d K1/dt

o 0 11 0 1 1      0.000000000     0.100 = d q/dt



o 2  1 0 1 1  99000.000000000  1650.000 = PERIOD(2)

o 2  2 0 1 1  39254.892879388  -984.000 = PERIASTRON EPOCH

o 2  3 0 1 1      0.770000000     0.010 = ECCENTRICITY

o 2  4 0 1 1     68.284996369    -6.000 = PERIASTRON LONG.

Desired output

0.086796671

151.562809500

107.927001263

0.418206121

I unsuccessfully tried

/o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)/ { c=FNR;}

(FNR-c>=2 && c>0) {$2 ~ /<0>/ $3 ~ /<3|4|5|6>/ { print $7}

I want to say: If second column is 0 and the third colums is 3,4,5,6 print column 7

awk

share|improve this question

edited 21 mins ago

Lukáš Altman

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please describe the rule.
  
  – dedowsdi
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I editted the question
  
  – Lukáš Altman
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But second column of your desired output are all 0.
  
  – dedowsdi
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, I am sorry
  
  – Lukáš Altman
  21 mins ago
  

  
  
  
  

add a comment | 

0

How to get four numbers from this table please?

Input:

chil           2   44843.084886752418        44843.080972494339       0.38261625919046571     



o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)

o 0  2 0 1 1  54002.864912732     0.000 = PERIASTRON EPOCH

o 0  3 0 1 1      0.086796671     0.000 = ECCENTRICITY

o 0  4 0 1 1    151.562809500     0.020 = PERIASTRON LONG.

o 0  5 0 1 1    107.927001263     0.053 = K1

o 0  6 0 1 1      0.418206121     0.000 = q = M2/M1,  K2 =  258.071309

o 0  7 0 1 1      0.004220000     0.000 = d omega/dt

o 0  8 0 1 1      0.000000000     0.100 = d P/dt

o 0  9 0 1 1      0.000000000     0.100 = d e/dt

o 0 10 0 1 1      0.000000000     0.100 = d K1/dt

o 0 11 0 1 1      0.000000000     0.100 = d q/dt



o 2  1 0 1 1  99000.000000000  1650.000 = PERIOD(2)

o 2  2 0 1 1  39254.892879388  -984.000 = PERIASTRON EPOCH

o 2  3 0 1 1      0.770000000     0.010 = ECCENTRICITY

o 2  4 0 1 1     68.284996369    -6.000 = PERIASTRON LONG.

Desired output

0.086796671

151.562809500

107.927001263

0.418206121

I unsuccessfully tried

/o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)/ { c=FNR;}

(FNR-c>=2 && c>0) {$2 ~ /<0>/ $3 ~ /<3|4|5|6>/ { print $7}

I want to say: If second column is 0 and the third colums is 3,4,5,6 print column 7

awk

share|improve this question

edited 21 mins ago

Lukáš Altman

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please describe the rule.
  
  – dedowsdi
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I editted the question
  
  – Lukáš Altman
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But second column of your desired output are all 0.
  
  – dedowsdi
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, I am sorry
  
  – Lukáš Altman
  21 mins ago
  

  
  
  
  

add a comment | 

How to get four numbers from this table please?

Input:

chil           2   44843.084886752418        44843.080972494339       0.38261625919046571     



o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)

o 0  2 0 1 1  54002.864912732     0.000 = PERIASTRON EPOCH

o 0  3 0 1 1      0.086796671     0.000 = ECCENTRICITY

o 0  4 0 1 1    151.562809500     0.020 = PERIASTRON LONG.

o 0  5 0 1 1    107.927001263     0.053 = K1

o 0  6 0 1 1      0.418206121     0.000 = q = M2/M1,  K2 =  258.071309

o 0  7 0 1 1      0.004220000     0.000 = d omega/dt

o 0  8 0 1 1      0.000000000     0.100 = d P/dt

o 0  9 0 1 1      0.000000000     0.100 = d e/dt

o 0 10 0 1 1      0.000000000     0.100 = d K1/dt

o 0 11 0 1 1      0.000000000     0.100 = d q/dt



o 2  1 0 1 1  99000.000000000  1650.000 = PERIOD(2)

o 2  2 0 1 1  39254.892879388  -984.000 = PERIASTRON EPOCH

o 2  3 0 1 1      0.770000000     0.010 = ECCENTRICITY

o 2  4 0 1 1     68.284996369    -6.000 = PERIASTRON LONG.

Desired output

0.086796671

151.562809500

107.927001263

0.418206121

I unsuccessfully tried

/o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)/ { c=FNR;}

(FNR-c>=2 && c>0) {$2 ~ /<0>/ $3 ~ /<3|4|5|6>/ { print $7}

I want to say: If second column is 0 and the third colums is 3,4,5,6 print column 7

awk

share|improve this question

edited 21 mins ago

Lukáš Altman

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

chil           2   44843.084886752418        44843.080972494339       0.38261625919046571     



o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)

o 0  2 0 1 1  54002.864912732     0.000 = PERIASTRON EPOCH

o 0  3 0 1 1      0.086796671     0.000 = ECCENTRICITY

o 0  4 0 1 1    151.562809500     0.020 = PERIASTRON LONG.

o 0  5 0 1 1    107.927001263     0.053 = K1

o 0  6 0 1 1      0.418206121     0.000 = q = M2/M1,  K2 =  258.071309

o 0  7 0 1 1      0.004220000     0.000 = d omega/dt

o 0  8 0 1 1      0.000000000     0.100 = d P/dt

o 0  9 0 1 1      0.000000000     0.100 = d e/dt

o 0 10 0 1 1      0.000000000     0.100 = d K1/dt

o 0 11 0 1 1      0.000000000     0.100 = d q/dt



o 2  1 0 1 1  99000.000000000  1650.000 = PERIOD(2)

o 2  2 0 1 1  39254.892879388  -984.000 = PERIASTRON EPOCH

o 2  3 0 1 1      0.770000000     0.010 = ECCENTRICITY

o 2  4 0 1 1     68.284996369    -6.000 = PERIASTRON LONG.

0.086796671

151.562809500

107.927001263

0.418206121

/o 0  1 0 1 1      5.732820000     0.000 = PERIOD(0)/ { c=FNR;}

(FNR-c>=2 && c>0) {$2 ~ /<0>/ $3 ~ /<3|4|5|6>/ { print $7}

share|improve this question

edited 21 mins ago

Lukáš Altman

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

share|improve this question

edited 21 mins ago

Lukáš Altman

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

asked 54 mins ago

Lukáš AltmanLukáš Altman

855

2 Answers
2

active

oldest

votes

0

Your command can be just simplified to just below. You were missing a group operator to combine conditions to match values from $2 and $3. You can put them together with a &&

awk '$NF == "PERIOD(0)" { c=FNR; } 

     ( FNR-c>=2 && c>0 ) && ( $2 ~ /<0>/ && $3 ~ /<3|4|5|6>/ ) { print $7 }' file

share|improve this answer

edited 11 mins ago

answered 17 mins ago

InianInian

5,8681633

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I got these numbers 54002.864912732 0.086796671 151.562809500 107.927001263 39254.892879388 0.770000000 68.284996369 3.000000000 and lots of spaces.
  
  – Lukáš Altman
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LukášAltman: I copy pasted your input in question and ran the command which produced the output as expected. You need to show us the actual file you are running this command on
  
  – Inian
  11 mins ago
  

  
  
  
  

add a comment | 

0

awk '$2==0 && $3~/^[3456]$/{print $7}' data

share|improve this answer

edited 2 mins ago

answered 17 mins ago

dedowsdidedowsdi

43415

add a comment | 

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0

Your command can be just simplified to just below. You were missing a group operator to combine conditions to match values from $2 and $3. You can put them together with a &&

awk '$NF == "PERIOD(0)" { c=FNR; } 

     ( FNR-c>=2 && c>0 ) && ( $2 ~ /<0>/ && $3 ~ /<3|4|5|6>/ ) { print $7 }' file

share|improve this answer

edited 11 mins ago

answered 17 mins ago

InianInian

5,8681633

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I got these numbers 54002.864912732 0.086796671 151.562809500 107.927001263 39254.892879388 0.770000000 68.284996369 3.000000000 and lots of spaces.
  
  – Lukáš Altman
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LukášAltman: I copy pasted your input in question and ran the command which produced the output as expected. You need to show us the actual file you are running this command on
  
  – Inian
  11 mins ago
  

  
  
  
  

add a comment | 

0

Your command can be just simplified to just below. You were missing a group operator to combine conditions to match values from $2 and $3. You can put them together with a &&

awk '$NF == "PERIOD(0)" { c=FNR; } 

     ( FNR-c>=2 && c>0 ) && ( $2 ~ /<0>/ && $3 ~ /<3|4|5|6>/ ) { print $7 }' file

share|improve this answer

edited 11 mins ago

answered 17 mins ago

InianInian

5,8681633

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I got these numbers 54002.864912732 0.086796671 151.562809500 107.927001263 39254.892879388 0.770000000 68.284996369 3.000000000 and lots of spaces.
  
  – Lukáš Altman
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LukášAltman: I copy pasted your input in question and ran the command which produced the output as expected. You need to show us the actual file you are running this command on
  
  – Inian
  11 mins ago
  

  
  
  
  

add a comment | 

Your command can be just simplified to just below. You were missing a group operator to combine conditions to match values from $2 and $3. You can put them together with a &&

awk '$NF == "PERIOD(0)" { c=FNR; } 

     ( FNR-c>=2 && c>0 ) && ( $2 ~ /<0>/ && $3 ~ /<3|4|5|6>/ ) { print $7 }' file

share|improve this answer

edited 11 mins ago

answered 17 mins ago

InianInian

5,8681633

awk '$NF == "PERIOD(0)" { c=FNR; } 

     ( FNR-c>=2 && c>0 ) && ( $2 ~ /<0>/ && $3 ~ /<3|4|5|6>/ ) { print $7 }' file

share|improve this answer

edited 11 mins ago

answered 17 mins ago

InianInian

5,8681633

answered 17 mins ago

InianInian

5,8681633

answered 17 mins ago

InianInian

5,8681633

0

awk '$2==0 && $3~/^[3456]$/{print $7}' data

share|improve this answer

edited 2 mins ago

answered 17 mins ago

dedowsdidedowsdi

43415

add a comment | 

0

awk '$2==0 && $3~/^[3456]$/{print $7}' data

share|improve this answer

edited 2 mins ago

answered 17 mins ago

dedowsdidedowsdi

43415

add a comment | 

awk '$2==0 && $3~/^[3456]$/{print $7}' data

share|improve this answer

edited 2 mins ago

answered 17 mins ago

dedowsdidedowsdi

43415

awk '$2==0 && $3~/^[3456]$/{print $7}' data

share|improve this answer

edited 2 mins ago

answered 17 mins ago

dedowsdidedowsdi

43415

answered 17 mins ago

dedowsdidedowsdi

43415

answered 17 mins ago

dedowsdidedowsdi

43415