किसी वर्ण में किसी स्थान पर किसी वर्ण को प्रतिस्थापित करना

Question

मेरे पास फाइल इस प्रकार है और कॉमा को दूसरी स्थिति में बदलना होगा जैसे #

1K,1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
1K,2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
1K,3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
1K,8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
1K,9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10,,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11,,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12,,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18,,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03

मैंने यह कोशिश की, लेकिन अल्पविराम से पहले वर्णों को हटाना:

sed 's/^\(.\{2\}\),/\#/' input_file

प्राप्त परिणाम:

#1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
#2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
#3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
#8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
#9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10,,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11,,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12,,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18,,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03

वांछित परिणाम:

1K#1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
1K#2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
1K#3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
1K#8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
1K#9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10,,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11,,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12,,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18,,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03

Answer 1

वास्तव में, जिस अल्पविराम ,को प्रतिस्थापित किया जाना #है, वह स्थिति 3 में है :

sed 's/^\(..\),/\1#/' input_file

• ^ - स्ट्रिंग एंकर की शुरुआत

• (..) - 1 पर कब्जा समूह 1 2 वर्णों से युक्त

• \1 - 1 कैप्चर किए गए समूह का संदर्भ

---

उत्पादन:

1K#1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
1K#2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
1K#3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
1K#8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
1K#9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10,,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11,,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12,,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18,,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03

Answer 2

sed 's/^\(..\),\([0-9]\)/\1#\2/' input_file

लगता है कि आप regexp से परिचित नहीं हैं। यह प्रश्न उतना ही आसान है जितना कि एक प्रश्न के बिना पूछे जाने वाला प्रश्न।

Answer 3

use below command to get the desired result

input file u.txt

1K,1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
1K,2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
1K,3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
1K,8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
1K,9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10,,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11,,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12,,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18,,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03


command: sed "s/,/#/1" u.txt

sed "s/,/#/1" u.txt 
1K#1,,,,,,,,,,,0.10000000E+00,0.2837726134152E+08,0.6756896377655E+02
1K#2,,,,,,,,,,,0.10000000E+00,0.2837950666514E+08,0.6757007624345E+02
1K#3,,,,,,,,,,,0.10000000E+00,0.2837975330813E+08,0.6756827852038E+02
1K#8,,,,,,,,,,,0.10000000E+00,0.2610685746631E+08,0.1832495587770E+03
1K#9,,,,,,,,,,,0.10000000E+00,0.2610716318002E+08,0.1836118051089E+03
1K10#,,,,,,,,,,0.10000000E+00,0.2609596327361E+08,0.1822385749588E+03
1K11#,,,,,,,,,,0.10000000E+00,0.2610713453046E+08,0.1841214115744E+03
1K12#,,,,,,,,,,0.10000000E+00,0.2610673990054E+08,0.1846882770826E+03
1K18#,,,,,,,,,,0.10000000E+00,0.2610076439025E+08,0.1854595804085E+03

Answer 4

यदि आप एक पैटर्न बना सकते हैं जिसमें हर "के," उपस्थिति "के #" होनी चाहिए, तो आप निम्नलिखित sedप्रतिस्थापन कर सकते हैं :

sed 's/K,/K#/g' input_file

Comments

इस विशिष्ट मामले के लिए, मुझे लगता है कि आपका उत्तर काफी सरल है, और दूसरों की तुलना में बेहतर है, जिसमें मेरा भी शामिल है।

— ब्रूस