Change just the date format


 
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# 1  
Old 04-15-2016
Change just the date format

need some more help please

i have a large file and
i want to change just the date format
with awk or sed

from this

yyyy-mm-dd

Code:
 2016-04-15  8.30 
 2016-04-15  7.30 
 2016-04-13  7.30 
 2016-04-11  8.30 
 2016-04-11  7.30 
 2016-04-08  8.30 
 2016-04-08  7.30 
 2016-04-06  7.30 
 2016-04-04  8.30

to this

dd-mm-yyyy

Code:
15-04-2016  8.30
15-04-2016  7.30
13-04-2016  7.30
11-04-2016  8.30
11-04-2016  7.30
08-04-2016  8.30
08-04-2016  7.30
06-04-2016  7.30
04-04-2016  8.30

leaving the times as they are

thanks

Last edited by bob123; 04-15-2016 at 06:56 PM..
# 2  
Old 04-15-2016
try:
Code:
sed 's/\([0-9][0-9][0-9][0-9]\)-\([0-9][0-9]\)-\([0-9][0-9]\)/\3-\2-\1/' infile

These 2 Users Gave Thanks to rdrtx1 For This Post:
# 3  
Old 04-15-2016
@rdrtx1

works great Smilie

thanks
# 4  
Old 04-16-2016
Hi.

Using a utility from the package dateutils, dconv:
Code:
#!/usr/bin/env bash

# @(#) s1       Demonstrate date manipulation, format changes, dconv.

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C dateutils.dconv

FILE=${1-data1}

pl " Input data file $FILE:"
cat $FILE

pl " Expected output:"
cat expected-output.txt

# 2016-04-15  8.30 -> 15-04-2016  8.30
pl " Results:"
dateutils.dconv -S -i "%F" -f "%d-%m-%Y" < $FILE

exit 0

producing:
Code:
$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 3.16.0-4-amd64, x86_64
Distribution        : Debian 8 (jessie) 
bash GNU bash 4.3.30
dateutils.dconv dconv 0.3.1

-----
 Input data file data1:
 2016-04-15  8.30 
 2016-04-15  7.30 
 2016-04-13  7.30 
 2016-04-11  8.30 
 2016-04-11  7.30 
 2016-04-08  8.30 
 2016-04-08  7.30 
 2016-04-06  7.30 
 2016-04-04  8.30

-----
 Expected output:
15-04-2016  8.30
15-04-2016  7.30
13-04-2016  7.30
11-04-2016  8.30
11-04-2016  7.30
08-04-2016  8.30
08-04-2016  7.30
06-04-2016  7.30
04-04-2016  8.30

-----
 Results:
 15-04-2016  8.30 
 15-04-2016  7.30 
 13-04-2016  7.30 
 11-04-2016  8.30 
 11-04-2016  7.30 
 08-04-2016  8.30 
 08-04-2016  7.30 
 06-04-2016  7.30 
 04-04-2016  8.30

The dateutils package can be found in repositories Debian, MacOS ("brew"), SuSE, etc., and also at dateutils (verified 2016.04.16)

Best wishes ... cheers, drl
This User Gave Thanks to drl For This Post:
# 5  
Old 04-16-2016
Quote:
Originally Posted by bob123
need some more help please

i have a large file and
i want to change just the date format
with awk or sed
from this
yyyy-mm-dd

Code:
 2016-04-15  8.30 
 2016-04-15  7.30 
 2016-04-13  7.30 
 2016-04-11  8.30 
 2016-04-11  7.30 
 2016-04-08  8.30 
 2016-04-08  7.30 
 2016-04-06  7.30 
 2016-04-04  8.30

to this

dd-mm-yyyy

Code:
15-04-2016  8.30
15-04-2016  7.30
13-04-2016  7.30
11-04-2016  8.30
11-04-2016  7.30
08-04-2016  8.30
08-04-2016  7.30
06-04-2016  7.30
04-04-2016  8.30

leaving the times as they are
thanks
Hello bob123,

Could you please try following and let me know if this helps you on same.
Code:
awk '{split($1, A,"-");$1=A[3] "-" A[2]"-" A[1]}; 1'  Input_file

Output will be as follows.
Code:
15-04-2016 8.30
15-04-2016 7.30
13-04-2016 7.30
11-04-2016 8.30
11-04-2016 7.30
08-04-2016 8.30
08-04-2016 7.30
06-04-2016 7.30
04-04-2016 8.30

Thanks,
R. Singh
# 6  
Old 04-16-2016
Quote:
Originally Posted by RavinderSingh13
Hello bob123,

Could you please try following and let me know if this helps you on same.
Code:
awk '{split($1, A,"-");$1=A[3] "-" A[2]"-" A[1]}; 1'  Input_file

Output will be as follows.
Code:
15-04-2016 8.30
15-04-2016 7.30
13-04-2016 7.30
11-04-2016 8.30
11-04-2016 7.30
08-04-2016 8.30
08-04-2016 7.30
06-04-2016 7.30
04-04-2016 8.30

Thanks,
R. Singh
yes that works great as well
thank you Smilie
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