replace full stop by comma
For the below values I have to replace the second and fourth "." by "," in the below file
input
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
desired intermediate output
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
I know awk gsub(/./,",") this replaces everything by comma.But I only need the columns to be seperated by "," . I also wanted to switch the third column in the first place after this.
desired final output
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
linux awk sed
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
|
show 4 more comments
For the below values I have to replace the second and fourth "." by "," in the below file
input
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
desired intermediate output
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
I know awk gsub(/./,",") this replaces everything by comma.But I only need the columns to be seperated by "," . I also wanted to switch the third column in the first place after this.
desired final output
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
linux awk sed
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
1
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
2
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
final expected output is not clear, I think you want a space to separate field after time..awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'
– Sundeep
Feb 10 '17 at 2:35
1
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59
|
show 4 more comments
For the below values I have to replace the second and fourth "." by "," in the below file
input
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
desired intermediate output
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
I know awk gsub(/./,",") this replaces everything by comma.But I only need the columns to be seperated by "," . I also wanted to switch the third column in the first place after this.
desired final output
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
linux awk sed
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
For the below values I have to replace the second and fourth "." by "," in the below file
input
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
desired intermediate output
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
I know awk gsub(/./,",") this replaces everything by comma.But I only need the columns to be seperated by "," . I also wanted to switch the third column in the first place after this.
desired final output
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
linux awk sed
linux awk sed
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
edited Nov 21 '18 at 21:35
Rui F Ribeiro
40.2k1479136
40.2k1479136
asked Feb 10 '17 at 2:19
RKRRKR
23219
asked Feb 10 '17 at 2:19
RKRRKR
23219
asked Feb 10 '17 at 2:19
RKRRKR
23219
23219
1
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
2
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
final expected output is not clear, I think you want a space to separate field after time..awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'
– Sundeep
Feb 10 '17 at 2:35
1
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59
|
show 4 more comments
1
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
2
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
final expected output is not clear, I think you want a space to separate field after time..awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'
– Sundeep
Feb 10 '17 at 2:35
1
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59
1
1
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
2
2
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
final expected output is not clear, I think you want a space to separate field after time..
awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'– Sundeep
Feb 10 '17 at 2:35
final expected output is not clear, I think you want a space to separate field after time..
awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'– Sundeep
Feb 10 '17 at 2:35
1
1
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59
|
show 4 more comments
2 Answers
2
active
oldest
votes
bash-4.1$ cat file
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
bash-4.1$ awk -F. '{print $NF,$1"."$2,$3"."$4}' OFS=, file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
add a comment |
Using sed:
$ sed 's/./,/4; s/./,/2' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
This first replaces the 4th dot with a comma, then the 2nd.
We could have done the substitutions in the opposite order too, but since replacing the 2nd dot removes a dot from the line, the dot originally being 4th becomes the 3rd dot:
$ sed 's/./,/2; s/./,/3' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
The s/.../.../n syntax (where n is a digit) substitutes the n:th matching occurrence on a line.
Moving the last column to the front may be done with
s/^(.*),([^,]*)/2,1/
I.e., match stuff from the start of the line up to a comma, then the comma, then stuff that contains no more commas.
So the full command may be
$ sed 's/./,/4; s/./,/2; s/^(.*),([^,]*)/2,1/' file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
add a comment |
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2 Answers
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2 Answers
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bash-4.1$ cat file
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
bash-4.1$ awk -F. '{print $NF,$1"."$2,$3"."$4}' OFS=, file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
add a comment |
bash-4.1$ cat file
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
bash-4.1$ awk -F. '{print $NF,$1"."$2,$3"."$4}' OFS=, file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
add a comment |
bash-4.1$ cat file
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
bash-4.1$ awk -F. '{print $NF,$1"."$2,$3"."$4}' OFS=, file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
bash-4.1$ cat file
1.351364711.103.7319660.2010-01-01 00:00:00
1.345529841.103.7372875.2010-01-01 00:00:49
1.342955629.103.7455272.2010-01-01 00:01:42
1.339694956.103.7520503.2010-01-01 00:02:28
bash-4.1$ awk -F. '{print $NF,$1"."$2,$3"."$4}' OFS=, file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
answered Feb 10 '17 at 3:43
KamarajKamaraj
2,9641513
2,9641513
add a comment |
add a comment |
Using sed:
$ sed 's/./,/4; s/./,/2' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
This first replaces the 4th dot with a comma, then the 2nd.
We could have done the substitutions in the opposite order too, but since replacing the 2nd dot removes a dot from the line, the dot originally being 4th becomes the 3rd dot:
$ sed 's/./,/2; s/./,/3' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
The s/.../.../n syntax (where n is a digit) substitutes the n:th matching occurrence on a line.
Moving the last column to the front may be done with
s/^(.*),([^,]*)/2,1/
I.e., match stuff from the start of the line up to a comma, then the comma, then stuff that contains no more commas.
So the full command may be
$ sed 's/./,/4; s/./,/2; s/^(.*),([^,]*)/2,1/' file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
add a comment |
Using sed:
$ sed 's/./,/4; s/./,/2' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
This first replaces the 4th dot with a comma, then the 2nd.
We could have done the substitutions in the opposite order too, but since replacing the 2nd dot removes a dot from the line, the dot originally being 4th becomes the 3rd dot:
$ sed 's/./,/2; s/./,/3' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
The s/.../.../n syntax (where n is a digit) substitutes the n:th matching occurrence on a line.
Moving the last column to the front may be done with
s/^(.*),([^,]*)/2,1/
I.e., match stuff from the start of the line up to a comma, then the comma, then stuff that contains no more commas.
So the full command may be
$ sed 's/./,/4; s/./,/2; s/^(.*),([^,]*)/2,1/' file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
add a comment |
Using sed:
$ sed 's/./,/4; s/./,/2' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
This first replaces the 4th dot with a comma, then the 2nd.
We could have done the substitutions in the opposite order too, but since replacing the 2nd dot removes a dot from the line, the dot originally being 4th becomes the 3rd dot:
$ sed 's/./,/2; s/./,/3' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
The s/.../.../n syntax (where n is a digit) substitutes the n:th matching occurrence on a line.
Moving the last column to the front may be done with
s/^(.*),([^,]*)/2,1/
I.e., match stuff from the start of the line up to a comma, then the comma, then stuff that contains no more commas.
So the full command may be
$ sed 's/./,/4; s/./,/2; s/^(.*),([^,]*)/2,1/' file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
Using sed:
$ sed 's/./,/4; s/./,/2' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
This first replaces the 4th dot with a comma, then the 2nd.
We could have done the substitutions in the opposite order too, but since replacing the 2nd dot removes a dot from the line, the dot originally being 4th becomes the 3rd dot:
$ sed 's/./,/2; s/./,/3' file
1.351364711,103.7319660,2010-01-01 00:00:00
1.345529841,103.7372875,2010-01-01 00:00:49
1.342955629,103.7455272,2010-01-01 00:01:42
1.339694956,103.7520503,2010-01-01 00:02:28
The s/.../.../n syntax (where n is a digit) substitutes the n:th matching occurrence on a line.
Moving the last column to the front may be done with
s/^(.*),([^,]*)/2,1/
I.e., match stuff from the start of the line up to a comma, then the comma, then stuff that contains no more commas.
So the full command may be
$ sed 's/./,/4; s/./,/2; s/^(.*),([^,]*)/2,1/' file
2010-01-01 00:00:00,1.351364711,103.7319660
2010-01-01 00:00:49,1.345529841,103.7372875
2010-01-01 00:01:42,1.342955629,103.7455272
2010-01-01 00:02:28,1.339694956,103.7520503
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
edited Feb 3 at 20:13
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
answered Feb 3 at 20:07
KusalanandaKusalananda
130k17246406
130k17246406
add a comment |
add a comment |
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1
So, the 2nd and 4th periods become commas?
– Jeff Schaller
Feb 10 '17 at 2:24
Yup exactly the second and fourth becomes commas
– RKR
Feb 10 '17 at 2:25
2
If that's the case, the question should clearly say so.
– Jeff Schaller
Feb 10 '17 at 2:26
final expected output is not clear, I think you want a space to separate field after time..
awk -F. '{print $5 " " $1 "." $2 "," $3 "." $4}'– Sundeep
Feb 10 '17 at 2:35
1
@RKR why don't you post a original file contents and the command you used to extract the required fields. so we can offer a slight changes in your command itself.
– Kamaraj
Feb 10 '17 at 2:59