script valeurs après un mot

Version imprimable

20/10/2019, 20h08
memderdz

script valeurs après un mot

Bonsoir; scp je voudrais un script qui me fait sortir ces deux lignes pour chaque fichier:

6993.792168,-435.5629799 ,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857,-9.8823341,-12.3940307
48.9024304,46.21985,347.4094277,2.3327781,-3.2195505,155.2927391
c'est à dire les six chiffres apres Polar=
et les dix chiffres apres Hyperpolar=
A partir de fichiers ressemblant a ce qui suit et merci d'avance

fichierAB1:
3,2.086238,-0.041285\C,0,-3.674939,-0.022005,0.042078\C,0,-3.191597,-
1.348375,0.226491\C,0,-4.030645,-2.431853,0.251939\H,0,-2.134802,-1.50
085,0.376174\C,0,-5.884215,-0.983099,-0.073514\C,0,-5.418996,-2.255295
,0.093029\H,0,-3.622292,-3.423786,0.403676\H,0,-6.936392,-0.766484,-0.
19282\H,0,-6.111078,-3.085146,0.106295\C,0,-5.669801,1.425049,-0.29323
8\H,0,-5.485037,2.059604,0.575185\H,0,-5.265464,1.899595,-1.188534\H,0
,-6.742369,1.298673,-0.413774\N,0,-5.05592,0.101413,-0.100401\C,0,1.03
589,1.181848,0.01248\C,0,0.531527,2.519249,0.03756 2\O,0,1.173039,3.560
55,0.05234\N,0,-0.896301,2.505471,0.047791\C,0,-1.615675,3.775709,0.08
171\H,0,-2.219442,3.901838,-0.819997\H,0,-2.250654,3.833254,0.968257\H
,0,-0.863026,4.560566,0.121966\C,0,2.386139,-1.542859,-0.032663\C,0,2.
079466,-2.143644,-1.2533\C,0,2.10136,-2.18171,1.173866\C,0,1.473004,-3
.397854,-1.262955\H,0,2.31553,-1.63332,-2.179749\C,0,1.49554,-3.436128
,1.155002\H,0,2.354698,-1.7008,2.111417\C,0,1.184107,-4.043808,-0.0611
96\H,0,1.236778,-3.872689,-2.207784\H,0,1.278477,-3.941501,2.088609\H,
0,0.725894,-5.025981,-0.072719\S,0,-0.292103,0.009999,-0.007635\\Versi
on=ES64L-G09RevD.01\State=1-A\HF=-1846.3810417\RMSD=5.652e-09\Dipole=-
2.9282718,-1.3308943,-0.0931382\Polar=548.9024304,46.21985,347.4094277
,2.3327781,-3.2195505,155.2927391\HyperPolar=-6993.792168,-435.5629799
,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857,
-9.8823341,-12.3940307\Quadrupole=50.1042017,-13.9408128,-36.1633889,-
8.0283449,1.2594653,-0.8648172\PG=C01 [X(C21H20N3O1S2)]\\@
20/10/2019, 20h19
N_BaH

Bonjour,

Citation:

je voudrais un script

tu veux dire que tu veux faire, ou apprendre à écrire un script qui fait ci et ça, n'est-ce pas ?

nous ne faisons pas le travail des autres, ni leurs devoirs.

tu dois montrer ce que tu as fait, et que tu as cherché une solution.
20/10/2019, 20h27
memderdz
Déjà merci pour votre réponse; non ce n'est pas en devoir que je dois rendre mais plutôt ça m'aide dans mon travail

Le mieux que j'ai fait c'est:

Code:

sed -n '/Polar=/,/Quadrupole/p' nom-du-fichier

mais le problème qui se pose est que des fois polar est en deux lignes du genre
Code:

1 2 121221211pol ar1221212121
donc la commande ne marche pas
20/10/2019, 23h52
balkany

Si Polar et HyperPolar sont toujours précédés par un anti-slash et suivis d'un signe égal, et que le champ contenant le nombre à isoler est bien toujours suivi d'une virgule, alors tu devrais peut-être plutôt utiliser awk :

Code:

awk -F '[=,]' -v 'RS=\\' '/Polar/{print $2}' nom-du-fichier

Si Polar et HyperPolar ne s'écrivent pas toujours de cette manière (en particulier avec les majuscules), alors il faut adapter l'expression régulière /Polar/ dans le programme awk.

+++

Je me rends compte que quand tu parlais de chiffres dans ton premier post, il s'agissait en fait de nombres, comme l'indique clairement le reste de ton post.
Dans ce cas, ça donnerait plutôt :

Code:

awk -F '=' -v 'RS=\\' '/Polar/{print $2}' nom-du-fichier

Je laisse quand même la première « solution » : au cas où tu voudrais isoler un nombre, ça te montrera comment faire… (on se rattrape comme on peut :mrgreen: )
21/10/2019, 06h12
N_BaH

en mettant tout sur une seule ligne, et si HyperPolar et bien toujours après Polar :

Code:

sed -n ':a;$!{N;ba;};s/\n//g; s/.*\\Polar=$[^\\]*$.*\\HyperPolar=$[^\\]*$.*/\1\n\2/p'
21/10/2019, 08h47
memderdz

Citation:

Envoyé par N_BaH

en mettant tout sur une seule ligne, et si HyperPolar et bien toujours après Polar :

Code:

sed -n ':a;$!{N;ba;};s/\n//g; s/.*\\Polar=$[^\\]*$.*\\HyperPolar=$[^\\]*$.*/\1\n\2/p'

Merci pour votre réponse
sur ce fichier ca ne marche pas?

27414883,-2.80068849|H,0,4.89327009,-1.34123111,-2.87880029|H,0,4.3242
1214,0.13713561,-3.61614147|H,0,5.88491835,0.09663742,-2.83152488|S,0,
3.89127457,0.68531615,1.04602614|N,0,2.75224157,0.37073704,-1.34208351
|C,0,2.46629481,0.26170773,0.10229667|C,0,-1.18498553,-0.48084873,1.14
872849|C,0,-2.51925065,-0.24639872,1.11216704|H,0,-3.08287955,-0.21978
07,2.02129511|C,0,-3.2242556,-0.01828171,-0.23784427|C,0,-2.47973629,-
0.05453266,-1.42474961|C,0,-3.12451627,0.15290131,-2.64416588|H,0,-1.4
262708,-0.2396526,-1.39574775|C,0,-5.19376387,0.41681869,-1.42835439|C
,0,-4.49885349,0.39152727,-2.6452469|H,0,-2.57493853,0.12940427,-3.561
94158|H,0,-6.24805815,0.59941779,-1.42374322|H,0,-5.0178681,0.55379693
,-3.56676326|C,0,-5.30306335,0.24604441,0.99393235|H,0,-5.65827947,-0.
73712145,1.22220047|H,0,-4.66559353,0.58704325,1.78276189|H,0,-6.13489
763,0.91169194,0.89460909|N,0,-4.54803605,0.21376345,-0.26693706|C,0,1
.277252,-0.12114555,0.62781309|C,0,0.96240823,-0.23339178,2.12690427|O
,0,1.78752312,0.06085701,3.03032755|N,0,-0.42192908,-0.70691343,2.4054
198|C,0,-1.00824367,0.02300324,3.53871445|H,0,-1.02309561,1.07014608,3
.31923491|H,0,-2.00772113,-0.32064631,3.70559721|H,0,-0.42068753,-0.14
919014,4.41622638|C,0,1.98832937,-0.60842131,-2.12858654|H,0,2.2258831
1,-1.59701338,-1.79518916|H,0,0.94113936,-0.43051314,-1.99958381|H,0,2
.24192071,-0.51045803,-3.16347523|S,0,-0.15652992,-0.54034701,-0.29721
188||Version=EM64W-G09RevE.01|State=1-A|HF=-1648.7772651|RMSD=7.048e-0
09|Dipole=-5.5676511,0.0412021,-2.5393295|Polar=332.7991262,-0.6193637
,99.8786764,6.6211561,0.5968292,235.4350169|HyperPolar=-310.268231,136
.8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,-
521.9044356,35.3931657,-215.8164|Quadrupole=52.8616985,-38.9536683,-13
.9080303,-8.8113919,-9.2244524,2.8179156|PG=C01 [X(C16H18N3O1S2)]||@
21/10/2019, 08h57
N_BaH

non, ça ne fonctionne pas puisque le "séparateur" n'est plus \, mais |.
il faut adapter.

Bonjour,

Juste pour information, pour les 2 samples juste concaténé dans un fichier:

Code:

1
2
3
4
5
6
7
8
9
10
11
$  awk -F '[\\\\|]' 'BEGIN{RS="@"}{print $(NF-5)"\n"$(NF-4)}' samples
Polar=548.9024304,46.21985,347.4094277
,2.3327781,-3.2195505,155.2927391
HyperPolar=-6993.792168,-435.5629799
,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857,
-9.8823341,-12.3940307
Polar=332.7991262,-0.6193637
,99.8786764,6.6211561,0.5968292,235.4350169
HyperPolar=-310.268231,136
.8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,-
521.9044356,35.3931657,-215.8164

Mais, est-ce qu'ils sont toujours à ces positions... ?

21/10/2019, 11h14
balkany

Ce sont les sauts de ligne résiduels qui ne te plaisent pas dans ma proposition ? :pleure:
Tu peux faire ça sinon :

Code:

awk -F '=' -v 'RS=\\' '/Polar/{gsub("\n","",$2); print $2}' nom-du-fichier

Et si le délimiteur de lignes change, il suffit de reporter ce changement au niveau du RS= :

Code:

awk -F '=' -v 'RS=|' '/Polar/{gsub("\n","",$2); print $2}' nom-du-fichier

Citation:
Envoyé par disedorgue
Bonjour,

Juste pour information, pour les 2 samples juste concaténé dans un fichier:
Code:

1 2 3 4 5 6 7 8 9 10 11 $ awk -F '[\\\\|]' 'BEGIN{RS="@"}{print $(NF-5)"\n"$(NF-4)}' samples Polar=548.9024304,46.21985,347.4094277 ,2.3327781,-3.2195505,155.2927391 HyperPolar=-6993.792168,-435.5629799 ,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857, -9.8823341,-12.3940307 Polar=332.7991262,-0.6193637 ,99.8786764,6.6211561,0.5968292,235.4350169 HyperPolar=-310.268231,136 .8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,- 521.9044356,35.3931657,-215.8164
Mais, est-ce qu'ils sont toujours à ces positions... ?
oui ils sont a la meme position a la fin du fichier

Citation:

Envoyé par balkany

Ce sont les sauts de ligne résiduels qui ne te plaisent pas dans ma proposition ? :pleure:
Tu peux faire ça sinon :

Code:

awk -F '=' -v 'RS=\\' '/Polar/{gsub("\n","",$2); print $2}' nom-du-fichier

Et si le délimiteur de lignes change, il suffit de reporter ce changement au niveau du RS= :

Code:

awk -F '=' -v 'RS=|' '/Polar/{gsub("\n","",$2); print $2}' nom-du-fichier

Merci ca marche et si je veux lancer la commade sur tout les fichiers *.out d'un répertoire pour un résultat d'un seul coup de la sorte

nom-du -fichier
332.7991262,-0.6193637 ,99.8786764,6.6211561,0.5968292,235.4350169
-310.268231,136 .8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,- 521.9044356,35.3931657,-215.8164
nom-du -fichier
332.7991262,-0.6193637 ,99.8786764,6.6211561,0.5968292,235.4350169
-310.268231,136 .8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,- 521.9044356,35.3931657,-215.8164
nom-du -fichier
348.3308051,0.2456 78,128.56784,6.4628527,0.6555265,249.8750908
-367.8828343,1 13.0258747,-32.8454065,11.6893551,-301.6998263,60.0115679,-41.2481129, -542.9139361,21.9230514,-209.6434909

21/10/2019, 13h08
balkany

À condition que tous tes fichiers *.out aient le même séparateur de lignes, par exemple | :

Code:

awk -F '=' -v 'RS=|' 'FNR==1{print FILENAME} /Polar/{gsub("\n","",$2); print $2}' /chemin-vers-le-répertoire/*.out

Il existe aussi une syntaxe propre à gawk (GNU awk, vers lequel pointe probablement awk chez toi : pour le vérifier, ls -l /usr/bin/awk), qui évite le test FNR==1

Code:

awk -F '=' -v 'RS=|' 'BEGINFILE{print FILENAME} /Polar/{gsub("\n","",$2); print $2}' /chemin-vers-le-répertoire/*.out

Bonjour :coucou:

Code:

1
2
3
$ sed -z 's/\n//g;s@\\@\n@g' 20191021_fichierAB1.txt |sed '/Polar/!d'
Polar=548.9024304,46.21985,347.4094277,2.3327781,-3.2195505,155.2927391
HyperPolar=-6993.792168,-435.5629799,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857,-9.8823341,-12.3940307

Code:

1
2
3
$ sed -z 's/\n//g;s@|@\n@g' 20191021_fichier2.txt |sed '/Polar/!d'
Polar=332.7991262,-0.6193637,99.8786764,6.6211561,0.5968292,235.4350169
HyperPolar=-310.268231,136.8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,-521.9044356,35.3931657,-215.8164

À noter:

option -z pour que tout le fichier soit dans l'espace de travail.
On peut utiliser "grep" en second étage à la place de sed.

Citation:

Envoyé par Flodelarab

Bonjour :coucou:

Code:

1
2
3
$ sed -z 's/\n//g;s@\\@\n@g' 20191021_fichierAB1.txt |sed '/Polar/!d'
Polar=548.9024304,46.21985,347.4094277,2.3327781,-3.2195505,155.2927391
HyperPolar=-6993.792168,-435.5629799,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857,-9.8823341,-12.3940307

Code:

1
2
3
$ sed -z 's/\n//g;s@|@\n@g' 20191021_fichier2.txt |sed '/Polar/!d'
Polar=332.7991262,-0.6193637,99.8786764,6.6211561,0.5968292,235.4350169
HyperPolar=-310.268231,136.8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,-521.9044356,35.3931657,-215.8164

À noter:

option -z pour que tout le fichier soit dans l'espace de travail.
On peut utiliser "grep" en second étage à la place de sed.

Ici, j'adore la simplicité :applo::applo::applo:

Merci pour toutes vos réponses j'utilise :

Code:

1
2
3
4
5
6
7
8
9
10
for i in $(pwd)/*.out;
do
echo $i
echo awk -F '=' -v 'RS=\\' '/Polar/{gsub("\n","",$2); print $2}' $i
echo $i  
grep -A 1 Tot= $i
grep Approx $i
echo $i
awk -F '=' -v 'RS=|' '/Polar/{gsub("\n","",$2); print $2}' $i
done

Same player shoot again.

Code:

1
2
3
$ sed -z 'y/\n\\/ \n/;' 20191021_fichierAB1.txt | grep 'Polar'                                                         
Polar=548.9024304,46.21985,347.4094277 ,2.3327781,-3.2195505,155.2927391
HyperPolar=-6993.792168,-435.5629799 ,-543.1264696,-243.8574187,52.2389104,9.825482,22.3514152,-39.3158857, -9.8823341,-12.3940307

Code:

1
2
3
$ sed -z 'y/\n|/ \n/;' 20191021_fichier2.txt | grep 'Polar'                                                            
Polar=332.7991262,-0.6193637 ,99.8786764,6.6211561,0.5968292,235.4350169
HyperPolar=-310.268231,136 .8596018,-26.8950518,-39.5777845,-294.0528687,77.9741809,-11.1576499,- 521.9044356,35.3931657,-215.8164

21/10/2019, 20h49
memderdz

Citation:

Envoyé par disedorgue

Ici, j'adore la simplicité :applo::applo::applo:

ca ne marche pas si polar est écris en deux lignes

498553,-0.48084873,1.14872849|C,0,-2.51925065,-0.24639872,1.11216704|
H,0,-3.08287955,-0.2197807,2.02129511|C,0,-3.2242556,-0.01828171,-0.23
784427|C,0,-2.47973629,-0.05453266,-1.42474961|C,0,-3.12451627,0.15290
131,-2.64416588|H,0,-1.4262708,-0.2396526,-1.39574775|C,0,-5.19376387,
0.41681869,-1.42835439|C,0,-4.49885349,0.39152727,-2.6452469|H,0,-2.57
493853,0.12940427,-3.56194158|H,0,-6.24805815,0.59941779,-1.42374322|H
,0,-5.0178681,0.55379693,-3.56676326|C,0,-5.30306335,0.24604441,0.9939
3235|H,0,-5.65827947,-0.73712145,1.22220047|H,0,-4.66559353,0.58704325
,1.78276189|H,0,-6.13489763,0.91169194,0.89460909|N,0,-4.54803605,0.21
376345,-0.26693706|C,0,1.277252,-0.12114555,0.62781309|C,0,0.96240823,
-0.23339178,2.12690427|O,0,1.78752312,0.06085701,3.03032755|N,0,-0.421
92908,-0.70691343,2.4054198|C,0,-1.00824367,0.02300324,3.53871445|H,0,
-1.02309561,1.07014608,3.31923491|H,0,-2.00772113,-0.32064631,3.705597
21|H,0,-0.42068753,-0.14919014,4.41622638|C,0,1.98832937,-0.60842131,-
2.12858654|H,0,2.22588311,-1.59701338,-1.79518916|H,0,0.94113936,-0.43
051314,-1.99958381|H,0,2.24192071,-0.51045803,-3.16347523|S,0,-0.15652
992,-0.54034701,-0.29721188||Version=EM64W-G09RevE.01|State=1-A|HF=-16
55.5964412|RMSD=6.670e-009|Dipole=-5.0355349,0.0580285,-2.2618645|Pola
r=399.0819696,0.3780517,137.7508108,6.8861467,-0.2766246,273.2366043|H
yperPolar=-2843.1582646,160.9794985,-31.8127931,20.485959,-572.6839217
,101.8582483,-40.642245,-877.6781298,34.7148673,-268.1840978|Quadrupol
e=50.8926976,-37.59692,-13.2957776,-7.0281935,-7.7076326,1.8553663|PG=
C01 [X(C16H18N3O1S2)]||@
21/10/2019, 20h53
Flodelarab

Ce n'est pas parce que c'est écrit en deux lignes mais parce que les lettres majuscules et les lettres minuscules ne sont pas les mêmes.

Code:

sed -z 'y/\n|/ \n/;' fichier.txt | grep -i 'Polar'
21/10/2019, 20h54
memderdz

j'u aussi ces deux commandes la

Code:

tr -d '\n' < data.txt | sed 's/\\/\n\\/g' | fgrep '\Polar=' | sed 's/\\Polar=//'

Code:

tr -d '\n' < data.txt | sed 's/\\/\n\\/g' | fgrep '\HyperPolar=' | sed 's/\\HyperPolar=//'

comment changer le séparateur entre | et \ pour régler le problème si polar est écris en deux lignes
21/10/2019, 21h43
balkany

Autant pour moi, je ne me rends compte que maintenant que ce que je t'ai donné ne marchait pas dans le cas d'un « Polar » en plusieurs lignes. Décidément…
Donc voici le correctif :

Code:

awk -F '=' -v 'RS=|' '{gsub("\n","")} /Polar/{print $2}' nom-du-fichier
22/10/2019, 10h54
memderdz

dans ce cas ca ne donne qu'une seule ligne?

9663742,-2.83152488|S,0,3.89127457,0.68531615,1.04602614|N,0,2.7522415
7,0.37073704,-1.34208351|C,0,2.46629481,0.26170773,0.10229667|C,0,-1.1
8498553,-0.48084873,1.14872849|C,0,-2.51925065,-0.24639872,1.11216704|
H,0,-3.08287955,-0.2197807,2.02129511|C,0,-3.2242556,-0.01828171,-0.23
784427|C,0,-2.47973629,-0.05453266,-1.42474961|C,0,-3.12451627,0.15290
131,-2.64416588|H,0,-1.4262708,-0.2396526,-1.39574775|C,0,-5.19376387,
0.41681869,-1.42835439|C,0,-4.49885349,0.39152727,-2.6452469|H,0,-2.57
493853,0.12940427,-3.56194158|H,0,-6.24805815,0.59941779,-1.42374322|H
,0,-5.0178681,0.55379693,-3.56676326|C,0,-5.30306335,0.24604441,0.9939
3235|H,0,-5.65827947,-0.73712145,1.22220047|H,0,-4.66559353,0.58704325
,1.78276189|H,0,-6.13489763,0.91169194,0.89460909|N,0,-4.54803605,0.21
376345,-0.26693706|C,0,1.277252,-0.12114555,0.62781309|C,0,0.96240823,
-0.23339178,2.12690427|O,0,1.78752312,0.06085701,3.03032755|N,0,-0.421
92908,-0.70691343,2.4054198|C,0,-1.00824367,0.02300324,3.53871445|H,0,
-1.02309561,1.07014608,3.31923491|H,0,-2.00772113,-0.32064631,3.705597
21|H,0,-0.42068753,-0.14919014,4.41622638|C,0,1.98832937,-0.60842131,-
2.12858654|H,0,2.22588311,-1.59701338,-1.79518916|H,0,0.94113936,-0.43
051314,-1.99958381|H,0,2.24192071,-0.51045803,-3.16347523|S,0,-0.15652
992,-0.54034701,-0.29721188||Version=EM64W-G09RevE.01|State=1-A|HF=-16
55.5964412|RMSD=6.670e-009|Dipole=-5.0355349,0.0580285,-2.2618645|Pola
r=399.0819696,0.3780517,137.7508108,6.8861467,-0.2766246,273.2366043|H
yperPolar=-2843.1582646,160.9794985,-31.8127931,20.485959,-572.6839217
,101.8582483,-40.642245,-877.6781298,34.7148673,-268.1840978|Quadrupol
e=50.8926976,-37.59692,-13.2957776,-7.0281935,-7.7076326,1.8553663|PG=
C01 [X(C16H18N3O1S2)]||@

Si c'est à moi que tu t'adresses, non, avec ce que je t'ai donné en #21, on obtient bien deux lignes :

Code:

1
2
3
4
$ awk -F '=' -v 'RS=|' '{gsub("\n","")} /Polar/{print $2}' f
399.0819696,0.3780517,137.7508108,6.8861467,-0.2766246,273.2366043
-2843.1582646,160.9794985,-31.8127931,20.485959,-572.6839217,101.8582483,-40.642245,-877.6781298,34.7148673,-268.1840978
$

22/10/2019, 12h08
memderdz

ok merci sinon comment je peux transformer mon paragraphe à (| début de nouvelle ligne)

|S,0,3.89127457,0.68531615,1.04602614
|N,0,2.75224157,0.37073704,-1.34208351
|C,0,2.46629481,0.26170773,0.10229667
|C,0,-1.18498553,-0.48084873,1.14872849
|C,0,-2.51925065,-0.24639872,1.11216704
|H,0,-3.08287955,-0.2197807,2.02129511
|C,0,-3.2242556,-0.01828171,-0.23784427
|C,0,-2.47973629,-0.05453266,-1.42474961
|C,0,-3.12451627,0.15290131,-2.64416588
|H,0,-1.4262708,-0.2396526,-1.39574775
|C,0,-5.19376387,0.41681869,-1.42835439
|C,0,-4.49885349,0.39152727,-2.6452469
|H,0,-2.57493853,0.12940427,-3.56194158
|H,0,-6.24805815,0.59941779,-1.42374322
|H,0,-5.0178681,0.55379693,-3.56676326
|C,0,-5.30306335,0.24604441,0.99393235
|H,0,-5.65827947,-0.73712145,1.22220047
|H,0,-4.66559353,0.58704325,1.78276189
|H,0,-6.13489763,0.91169194,0.89460909
|N,0,-4.54803605,0.21376345,-0.26693706
|C,0,1.277252,-0.12114555,0.62781309
|C,0,0.96240823,-0.23339178,2.12690427
|O,0,1.78752312,0.06085701,3.03032755
|N,0,-0.42192908,-0.70691343,2.4054198
|C,0,-1.00824367,0.02300324,3.53871445
|H,0,-1.02309561,1.07014608,3.31923491
|H,0,-2.00772113,-0.32064631,3.70559721
|H,0,-0.42068753,-0.14919014,4.41622638
|C,0,1.98832937,-0.60842131,-2.12858654
|H,0,2.22588311,-1.59701338,-1.79518916
|H,0,0.94113936,-0.43051314,-1.99958381
|H,0,2.24192071,-0.51045803,-3.16347523
|S,0,-0.15652992,-0.54034701,-0.29721188
|
|Version=EM64W-G09RevE.01
|State=1-A
|HF=-1655.5964412
|RMSD=6.670e-009
|Dipole=-5.0355349,0.0580285,-2.2618645
|Polar=399.0819696,0.3780517,137.7508108,6.8861467,-0.2766246,273.2366043
|HyperPolar=-2843.1582646,160.9794985,-31.8127931,20.485959,-572.6839217,101.8582483,-40.642245,-877.6781298,34.7148673,-268.1840978
|Quadrupole=50.8926976,-37.59692,-13.2957776,-7.0281935,-7.7076326,1.8553663
|PG=C01 [X(C16H18N3O1S2)]||@
22/10/2019, 13h46
balkany

En reprenant l'expression précédente avec une petite modif :

Code:

awk -F '=' -v 'RS=|' '{gsub("\n",""); print RS $0}' f

Fuseau horaire GMT +2. Il est actuellement 00h39.