List months ending on the same day of the week - bash

Question

This is an assignment. I have the following code:
#! /bin/bash
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
for m in {1..12}; do
if [ $m -eq 12 ] # december exception
then echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y" +%A)
break
fi
echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y - 1 days" +%A) # print the last day of the week for the month
done
It lists the last day of the week for every month:
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
Now I need to reverse it, so that it lists months ending on every day of the week like so:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
I'm thinking of a nested loop,
for d in {1..7};
And storing months in an array?

Answer 1

#! /usr/bin/env bash
# if year is not specified use the current year
declare -r year="${1:-$(date +%Y)}"
# associative array (aka hash table)
declare -A months_per_day=()
for m in {01..12}; do
day_month=$(LANG=C date -d "${year}-${m}-01 +1 month -1 day" +"%A %b")
months_per_day[${day_month% *}]+=" ${day_month#* }"
done
for day in Sunday Monday Tuesday Wednesday Thursday Friday Saturday; do
echo "${day} -${months_per_day[${day}]:-}"
done
Output:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec

Answer 2

Using GNU awk for time functions, with lots of intermediate and descriptively named variables to make it easy to understand:
$ cat tst.sh
#!/usr/bin/env bash
awk -v year="$1" '
BEGIN {
OFS = " - "
year = (year == "" ? strftime("%Y") : year)
secsInDay = 24*60*60
for ( mthNr=1; mthNr<=12; mthNr++ ) {
lastDayEpochSecs = mktime(year " " (mthNr+1) " 1 12 0 0") - secsInDay
mthAbbrDayName = strftime("%b %A", lastDayEpochSecs)
split(mthAbbrDayName,m)
mthAbbr = m[1]
dayName = m[2]
mthNr2mthAbbr[mthNr] = mthAbbr
mthAbbr2dayName[mthAbbr] = dayName
dayName2mthAbbrs[dayName] = \
(dayName in dayName2mthAbbrs ? dayName2mthAbbrs[dayName] " " : "" ) mthAbbr
}
for ( mthNr=1; mthNr<=12; mthNr++ ) {
mthAbbr = mthNr2mthAbbr[mthNr]
dayName = mthAbbr2dayName[mthAbbr]
print mthAbbr, dayName
}
print "\n--------\n"
for ( dayName in dayName2mthAbbrs ) {
mthAbbrs = dayName2mthAbbrs[dayName]
print dayName, mthAbbrs
}
}
'
$ ./tst.sh
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
--------
Tuesday - May
Friday - Sep
Sunday - Jul
Thursday - Mar Jun
Saturday - Apr Dec
Monday - Jan Feb Oct
Wednesday - Aug Nov
The above will be much faster than calling date multiple times in a shell loop and is trivial to modify to do anything else you need.

Answer 3

This answer refactors your implementation as a reusable getlastday function. Then, we go through the loop 7 times matching the getlastday to the matching months, and print it out:
#!/bin/bash
getlastday() {
if [ $m -eq 12 ]; then
echo $(date -d "$(($m%12+1))/1/$y" +%A)
return
fi
echo $(date -d "$(($m%12+1))/1/$y - 1 days" +%A) # print the last day of the week for the month
}
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
for d in Monday Tuesday Wednesday Thursday Friday Saturday Sunday; do
months=()
for m in {1..12}; do
lastday=$(getlastday)
if [ $lastday != $d ]; then continue; fi
months+=($(date -d $m/1/$y +%b))
done
echo $d - ${months[#]}
done
[EDIT: version 2]
As per #markp-fuso's comment below the above has an inefficiency in that there are 84 iterations to build the list.
The following is an improvement where we construct an res result array with 7 entries in it. One for each day in the week. Then we use +%u to get the day number instead of the day string. This will give us the index to the res result append where to append the month to:
#!/bin/bash
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
res=("Sunday -" "Monday -" "Tuesday -" "Wednesday -" "Thursday -" "Friday -" "Saturday -")
for m in {1..12}; do
mstr=$(date -d $m/1/$y +%b)
if [ $m -eq 12 ]; then
d=$(date -d "$(($m%12+1))/1/$y" +%u)
else
d=$(date -d "$(($m%12+1))/1/$y - 1 days" +%u)
fi
res[$d]="${res[$d]} $mstr"
done
for d in {0..6}; do
echo "${res[$d]}"
done

Answer 4

Using jq your task can be solved:
INPUT='
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
'
jq -Rrs '
(split("\n") | map(split(" - "))) as $input | # split lines and split each line by "-"
reduce ("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday") as $day # iterate over all weekdays
([]; . + [ ($input | map(select(.[1] == $day)) | # select entries for $day
.[0][1] + " - " + (map(.[0]) | join(" "))) # generate output for $day
]
) | .[]
' <<< "$INPUT"
Output
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec

Answer 5

Try this Shellcheck-clean code:
#! /bin/bash -p
year=${1-$(date +%Y)}
months=( Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec )
day_months=( 'Sunday -' 'Monday -' 'Tuesday -' 'Wednesday -'
'Thursday -' 'Friday -' 'Saturday -' )
for m in {1..12}; do
last_day=$(date -d "$year-$m-1 +1 month -1 day" +%w)
day_months[last_day]+=" ${months[m-1]}"
done
printf '%s\n' "${day_months[#]}"

Answer 6

declare -A a
for i in '1/1 + 1 year' {2..12}/1; do
d=($(LC_ALL=C date '+%A %B' -d "$i - 1 second"))
a["$d"]+=${a["$d"]:+ }${d[1]}
done
for i in {Sun,Mon,Tues,Wednes,Thurs,Fri,Satur}day; do
echo "$i - ${a["$i"]}"
done
This prints the list for the current year:
Sunday - July
Monday - January February October
Tuesday - May
Wednesday - August November
Thursday - March June
Friday - September
Saturday - December April

Answer 7

technically the only info you'll need is what Jan ends on, and either (a) whether year is leap or not, or (b) whether Feb ends on same day of week as Jan.
The rest are easily deduced without running date 12 times

Answer 8

Using ksh93:
#!/bin/ksh93
# if year is not specified use the current year
year="${1:-$(date +%Y)}"
typeset -A days
for month in Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
do
lastday=$(printf "%(%A)T" "final day $month $year")
days[$lastday]+=" $month"
done
for day in Sunday Monday Tuesday Wednesday Thursday Friday Saturday
do
printf "%-9s -%s\n" "$day" "${days[$day]:-}"
done
For 2022 produces the following output:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
For 2023 produces the following output:
Sunday - Apr Dec
Monday - Jul
Tuesday - Jan Feb Oct
Wednesday - May
Thursday - Aug Nov
Friday - Mar Jun
Saturday - Sep