Programming Forums > Script Programming > Sed and Awk

bigalnz · Jan 21st, 2015, 3:58 PM

Hi,

A friend showed my how to create a file that contains:

$0$0$0$0
$0$0$0$1
....
$9$9$9$9

with the line
(Toggle Plain Text)

printf '%s\n' $\{0..9}$\{0..9}$\{0..9}$\{0..9}

I now now needing to create a similar file that is all the combinations of:

$d$d$m$m

Where dd = days of month and months of year, ie dd = 01-31 and mm = 01-12
ie

$0$1$0$1 # first jan
$0$1$0$2 # second jan
..
$3$1$1$2 # 31st dec

I also want to create mmyy file (where yy = 00 to 99)

$0$1$0$1 # short date format for Jan 2000 or Jan 1900
...
$1$2$9$9 # short date format for Dec 1999

I assume I need a counter, but my bash isnt good enough to work in a counter.

Could someone please assist?

Many thanks

-Al

sharted · Jan 23rd, 2015, 8:26 AM

As this is the sed and awk forum, here are the loops you need in awk:

day and month:

(Toggle Plain Text)

bigalnz $] cat day_month.awk
#!/bin/gawk -f

BEGIN{

    #
    days_per_month[1]=31;
    days_per_month[2]=28;
    days_per_month[3]=31;
    days_per_month[4]=30;
    days_per_month[5]=31;
    days_per_month[6]=30;
    days_per_month[7]=31;
    days_per_month[8]=31;
    days_per_month[9]=30;
    days_per_month[10]=31;
    days_per_month[11]=30;
    days_per_month[12]=31;

    for(month=1; month<=12; month=month+1){
        n_days=days_per_month[month];
        for(day=1; day<=n_days; day=day+1){
            d_tens=day/10;
            d_units=day%10;
            m_tens=month/10;
            m_units=month%10;
            printf("$%d$%d$%d$%d\n", d_tens, d_units, m_tens, m_units);

        }
    }
}[bigalnz $] ./day_month.awk
$0$1$0$1                              
$0$2$0$1                              
$0$3$0$1                              
$0$4$0$1                              
$0$5$0$1                              
$0$6$0$1                              
$0$7$0$1                              
$0$8$0$1           

clipped 

$2$6$1$2
$2$7$1$2
$2$8$1$2
$2$9$1$2
$3$0$1$2
$3$1$1$2
[bigalnz $]

month and year:

(Toggle Plain Text)

[bigalnz $] cat ./month_year.awk
#!/bin/gawk -f

BEGIN{

    for(year=00; year<=99; year=year+1){
        for(month=1; month<=12; month=month+1){
            m_tens=month/10;
            m_units=month%10;
            y_tens=year/10
            y_units=year%10

            printf("$%d$%d$%d$%d\n", m_tens, m_units, y_tens,y_units);
            #print y_tens " " y_units ;

        }
    }
}[ bigalnz $] ./month_year.awk 
$0$1$0$0
$0$2$0$0
$0$3$0$0
$0$4$0$0
$0$5$0$0
$0$6$0$0

clipped 

$0$3$9$9
$0$4$9$9
$0$5$9$9
$0$6$9$9
$0$7$9$9
$0$8$9$9
$0$9$9$9
$1$0$9$9
$1$1$9$9
$1$2$9$9
[bigalnz $]

If this is no good for you, you can loop in bash with the following construction:

(Toggle Plain Text)

for (( i=0 ; i < 100 ; i++ )) ; do 
  echo $i 
done ;

Have another go and if you're still stuck, post back.

hth.

danielbmartin · Jan 27th, 2015, 12:54 PM

The solution given by sharted is clear and readable.

I offer one here which is relatively concise but admittedly less readable.
The significance of my solution is that it takes a different approach.
Rather than use a table of month lengths, it makes a "calendar" in which
every month has 31 days, and then "knocks out" the 7 unwanted days.

(Toggle Plain Text)

# Build a calendar such that each month has 31 days,
#   then "knock out" the superflous dates.
awk 'BEGIN{for (j=1;j<=31;j++) d=d "$"int(j/10)"$"j%10" "; split(d,day," ");
           for (j=1;j<=12;j++) {w=d; gsub(/ /,day[j]"\n",w); c=c w};
           gsub(/.2.9.0.2\n|.3.0.0.2\n|.3.1.0.2\n|.3.1.0.4\n|\
                 .3.1.0.6\n|.3.1.0.9\n|.3.1.1.1\n/,"",c);
           print c}' >$OutFile

Daniel B. Martin

danielbmartin · Jan 29th, 2015, 12:36 PM

As a matter of personal coding style I try to avoid explicit loops.
Here is an even more concise (and less readable) solution.

# Build a calendar such that each month has 31 days,
# then "knock out" the 7 superfluous dates.
# This method uses no explicit loops.

(Toggle Plain Text)

seq 0 371  \
|awk -F "" '{sum[k=1+int($0/31)]++; $0=sprintf("%02d%02d",sum[k],k);
  if ($0!~/2902|3002|3102|3104|3106|3109|3111/) print $1"$"$2"$"$3"$"$4}'  \
>$OutFile

Daniel B. Martin

bigalnz · Feb 1st, 2015, 1:22 AM

Thanks guys. Both work very well.

Jan 21st, 2015, 3:58 PM	#1
bigalnz Newbie Join Date: Jan 2011 Posts: 4 Rep Power: 0	Date Combinations Hi, A friend showed my how to create a file that contains: $0$0$0$0 $0$0$0$1 .... $9$9$9$9 with the line (Toggle Plain Text) printf '%s\n' $\{0..9}$\{0..9}$\{0..9}$\{0..9} I now now needing to create a similar file that is all the combinations of: $d$d$m$m Where dd = days of month and months of year, ie dd = 01-31 and mm = 01-12 ie $0$1$0$1 # first jan $0$1$0$2 # second jan .. $3$1$1$2 # 31st dec I also want to create mmyy file (where yy = 00 to 99) $0$1$0$1 # short date format for Jan 2000 or Jan 1900 ... $1$2$9$9 # short date format for Dec 1999 I assume I need a counter, but my bash isnt good enough to work in a counter. Could someone please assist? Many thanks -Al

Jan 23rd, 2015, 8:26 AM	#2
sharted Hobbyist Programmer Join Date: Feb 2012 Posts: 123 Rep Power: 6	Re: Date Combinations As this is the sed and awk forum, here are the loops you need in awk: day and month: (Toggle Plain Text) bigalnz $] cat day_month.awk #!/bin/gawk -f BEGIN{ # days_per_month[1]=31; days_per_month[2]=28; days_per_month[3]=31; days_per_month[4]=30; days_per_month[5]=31; days_per_month[6]=30; days_per_month[7]=31; days_per_month[8]=31; days_per_month[9]=30; days_per_month[10]=31; days_per_month[11]=30; days_per_month[12]=31; for(month=1; month<=12; month=month+1){ n_days=days_per_month[month]; for(day=1; day<=n_days; day=day+1){ d_tens=day/10; d_units=day%10; m_tens=month/10; m_units=month%10; printf("$%d$%d$%d$%d\n", d_tens, d_units, m_tens, m_units); } } }[bigalnz $] ./day_month.awk $0$1$0$1 $0$2$0$1 $0$3$0$1 $0$4$0$1 $0$5$0$1 $0$6$0$1 $0$7$0$1 $0$8$0$1 clipped $2$6$1$2 $2$7$1$2 $2$8$1$2 $2$9$1$2 $3$0$1$2 $3$1$1$2 [bigalnz $] bigalnz $] cat day_month.awk #!/bin/gawk -f BEGIN{ # days_per_month[1]=31; days_per_month[2]=28; days_per_month[3]=31; days_per_month[4]=30; days_per_month[5]=31; days_per_month[6]=30; days_per_month[7]=31; days_per_month[8]=31; days_per_month[9]=30; days_per_month[10]=31; days_per_month[11]=30; days_per_month[12]=31; for(month=1; month<=12; month=month+1){ n_days=days_per_month[month]; for(day=1; day<=n_days; day=day+1){ d_tens=day/10; d_units=day%10; m_tens=month/10; m_units=month%10; printf("$%d$%d$%d$%d\n", d_tens, d_units, m_tens, m_units); } } }[bigalnz $] ./day_month.awk $0$1$0$1 $0$2$0$1 $0$3$0$1 $0$4$0$1 $0$5$0$1 $0$6$0$1 $0$7$0$1 $0$8$0$1 clipped $2$6$1$2 $2$7$1$2 $2$8$1$2 $2$9$1$2 $3$0$1$2 $3$1$1$2 [bigalnz $] month and year: (Toggle Plain Text) [bigalnz $] cat ./month_year.awk #!/bin/gawk -f BEGIN{ for(year=00; year<=99; year=year+1){ for(month=1; month<=12; month=month+1){ m_tens=month/10; m_units=month%10; y_tens=year/10 y_units=year%10 printf("$%d$%d$%d$%d\n", m_tens, m_units, y_tens,y_units); #print y_tens " " y_units ; } } }[ bigalnz $] ./month_year.awk $0$1$0$0 $0$2$0$0 $0$3$0$0 $0$4$0$0 $0$5$0$0 $0$6$0$0 clipped $0$3$9$9 $0$4$9$9 $0$5$9$9 $0$6$9$9 $0$7$9$9 $0$8$9$9 $0$9$9$9 $1$0$9$9 $1$1$9$9 $1$2$9$9 [bigalnz $] [bigalnz $] cat ./month_year.awk #!/bin/gawk -f BEGIN{ for(year=00; year<=99; year=year+1){ for(month=1; month<=12; month=month+1){ m_tens=month/10; m_units=month%10; y_tens=year/10 y_units=year%10 printf("$%d$%d$%d$%d\n", m_tens, m_units, y_tens,y_units); #print y_tens " " y_units ; } } }[ bigalnz $] ./month_year.awk $0$1$0$0 $0$2$0$0 $0$3$0$0 $0$4$0$0 $0$5$0$0 $0$6$0$0 clipped $0$3$9$9 $0$4$9$9 $0$5$9$9 $0$6$9$9 $0$7$9$9 $0$8$9$9 $0$9$9$9 $1$0$9$9 $1$1$9$9 $1$2$9$9 [bigalnz $] If this is no good for you, you can loop in bash with the following construction: (Toggle Plain Text) for (( i=0 ; i < 100 ; i++ )) ; do echo $i done ; for (( i=0 ; i < 100 ; i++ )) ; do echo $i done ; Have another go and if you're still stuck, post back. hth.

Jan 27th, 2015, 12:54 PM	#3
danielbmartin Newbie Join Date: Apr 2010 Location: North Carolina Posts: 12 Rep Power: 0	Re: Date Combinations The solution given by sharted is clear and readable. I offer one here which is relatively concise but admittedly less readable. The significance of my solution is that it takes a different approach. Rather than use a table of month lengths, it makes a "calendar" in which every month has 31 days, and then "knocks out" the 7 unwanted days. (Toggle Plain Text) # Build a calendar such that each month has 31 days, # then "knock out" the superflous dates. awk 'BEGIN{for (j=1;j<=31;j++) d=d "$"int(j/10)"$"j%10" "; split(d,day," "); for (j=1;j<=12;j++) {w=d; gsub(/ /,day[j]"\n",w); c=c w}; gsub(/.2.9.0.2\n\|.3.0.0.2\n\|.3.1.0.2\n\|.3.1.0.4\n\|\ .3.1.0.6\n\|.3.1.0.9\n\|.3.1.1.1\n/,"",c); print c}' >$OutFile # Build a calendar such that each month has 31 days, # then "knock out" the superflous dates. awk 'BEGIN{for (j=1;j<=31;j++) d=d "$"int(j/10)"$"j%10" "; split(d,day," "); for (j=1;j<=12;j++) {w=d; gsub(/ /,day[j]"\n",w); c=c w}; gsub(/.2.9.0.2\n\|.3.0.0.2\n\|.3.1.0.2\n\|.3.1.0.4\n\|\ .3.1.0.6\n\|.3.1.0.9\n\|.3.1.1.1\n/,"",c); print c}' >$OutFile Daniel B. Martin

Jan 29th, 2015, 12:36 PM	#4
danielbmartin Newbie Join Date: Apr 2010 Location: North Carolina Posts: 12 Rep Power: 0	Re: Date Combinations As a matter of personal coding style I try to avoid explicit loops. Here is an even more concise (and less readable) solution. # Build a calendar such that each month has 31 days, # then "knock out" the 7 superfluous dates. # This method uses no explicit loops. (Toggle Plain Text) seq 0 371 \ \|awk -F "" '{sum[k=1+int($0/31)]++; $0=sprintf("%02d%02d",sum[k],k); if ($0!~/2902\|3002\|3102\|3104\|3106\|3109\|3111/) print $1"$"$2"$"$3"$"$4}' \ >$OutFile seq 0 371 \ \|awk -F "" '{sum[k=1+int($0/31)]++; $0=sprintf("%02d%02d",sum[k],k); if ($0!~/2902\|3002\|3102\|3104\|3106\|3109\|3111/) print $1"$"$2"$"$3"$"$4}' \ >$OutFile Daniel B. Martin

Feb 1st, 2015, 1:22 AM	#5
bigalnz Newbie Join Date: Jan 2011 Posts: 4 Rep Power: 0	Re: Date Combinations Thanks guys. Both work very well.

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Date Combinations