How to add leading zero to bash range?
For example, I need cycle 01,02,03,..,29,30
How can I implement this using bash?
In recent versions of bash you can do:
echo {01..30}
Output:
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Or if it should be comma separated:
echo {01..30} | tr ' ' ','
Which can also be accomplished with parameter expansion:
a=$(echo {01..30})
echo ${a// /,}
Output:
01,02,03,04,05,06,07,08,09,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30
another seq trick will work:
seq -w 30
if you check the man page, you will see the -w option is exactly for your requirement:
-w, --equal-width
equalize width by padding with leading zeroes
You can use seq's format option:
seq -f "%02g" 30
A "pure bash" way would be something like this:
echo {0..2}{0..9}
This will give you the following:
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Removing the first 00 and adding the last 30 is not too hard!
This works:
printf " %02d" $(seq 1 30)
Related
I am pretty new to bash scripting. I have my bash script below and I want to include an if statement when month (ij==09) equals 09 then "i" should be from 01 to 30. I tried several ways but did not work.
How can I include an if statement in the code below to achieve my task.? Any help is appreciated.
Thanks.
#!/bin/bash
for ii in 2007
do
for i in 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 #Day of the Month
do
for ij in 09 10 # Month
do
for j in 0000 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300
do
cdo cat DAS_0125_H.A${ii}${ij}${i}.${j}.002_var.nc outfile_${ii}${ij}${i}.nc
done
done
done
done
The smallest change is adding a continue for day 31 in month 9.
You must test "09" as a string (or as 10#09).
(I also changed cdo ... into echo cdo ...)
for ii in 2007
do
for i in 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 #Day of the Month
do
for ij in 09 10 # Month
do
if [[ "${ij}" == "09" ]] && [[ "${i}" == "31" ]]; then continue; fi
for j in 0000 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300
do
echo "cdo cat DAS_0125_H.A${ii}${ij}${i}.${j}.002_var.nc outfile_${ii}${ij}${i}.nc"
done
done
done
done
It would be easier to read when you use loop through the variales with a seq. You do not want to use `for ((i=1;i<=31;i++)) in view of the leading zeroes.
Also use verbose variable names.
for year in 2007
do
for day in {01..31} # Day of the Month
do
for month in {09,10} # Month
do
if [[ "${month}" == "09" ]] && [[ "${day}" == "31" ]]; then continue; fi
for hour in {00..23}
do
echo cdo cat DAS_0125_H.A${year}${month}${day}.${hour}00.002_var.nc outfile_${year}${month}${day}.nc
done
done
done
done
When the files already exist, you can consider
ls DAS_0125_H.A2007{09,10}{01..31}.{00..23}00.002_var.nc |
sed -r 's/.*([0-9]{8})/cdo cat & outfile_\1.nc/'
When this will show the commands you want, you can execute them by
source <(ls DAS_0125_H.A2007{09,10}{01..31}.{00..23}00.002_var.nc |
sed -r 's/.*([0-9]{8})/cdo cat & outfile_\1.nc/')
I am missing some episodes of the TV series friends, and I would like to know how many files I am missing per season. I would like to print out the last episode of each season and the number of files for each season.
The files have the format:
Friends S01E01 The Pilot.mkv
Friends S10E11 The One Where the Stripper Cries.mkv
The following bash script/oneliner should give you what you need, with details because it might help if you have the last episode of a season but earlier episodes are missing:
#!/bin/bash
ls Friends* | cut -c10-14 | \
awk -F'E' '{arr[$1]=arr[$1]" "$2; num[$1]++;} END { for (i in arr) printf "Season %s (%2d files) : %s\n", i, num[i], arr[i] }' | \
sort
Using awk, arrays with index being the number of the season are incremented to count the number of episodes, and also print the list of episode numbers so you can easily see which ones are missing. I used cut with columns 10 to 14 because in this case, we can safely assume that the numbers are where we want them.
The output is as follows:
Season 01 ( 9 files) : 01 02 03 04 05 06 07 08 09
Season 02 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 03 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 04 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 05 ( 9 files) : 01 03 04 05 06 07 08 09 10
Season 06 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 07 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 08 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 09 (10 files) : 01 02 03 04 05 06 07 08 09 10
Season 10 ( 7 files) : 01 02 03 04 05 06 10
The following bash scipt will work:
#!/bin/bash
for i in {01..10}
do
ls Friends\ S$i* | tail -n 1
ls Friends\ S$i* | wc -l
printf "\n"
done
It will produce results as follows:
Friends S01E24 The One Where Rachel Finds Out.mkv
24
Friends S02E24 The One with Barry and Mindy's Wedding.mkv
24
I am trying to remove leading zeroes from a BASH array... I have an array like:
echo "${DATES[#]}"
returns
01 02 02 03 04 07 08 09 10 11 13 14 15 16 17 18 20 21 22 23
I'd like to remove the leading zeroes from the dates and store back into array or another array, so i can iterate in another step... Any suggestions?
I tried this,
for i in "${!DATES[#]}"
do
DATESLZ["$i"]=(echo "{DATES["$i"]}"| sed 's/0*//' )
done
but failed (sorry, i'm an old Java programmer who was tasked to do some BASH scripts)
Use parameter expansion:
DATES=( ${DATES[#]#0} )
With bash arithmetic, you can avoid the octal woes by specifying your numbers are base-10:
day=08
((day++)) # bash: ((: 08: value too great for base (error token is "08")
((day = 10#$day + 1))
echo $day # 9
printf "%02d\n" $day # 09
You can use bash parameter expansion (see http://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html) like this:
echo ${DATESLZ[#]#0}
If: ${onedate%%[!0]*} will select all 0's in front of the string $onedate.
we could remove those zeros by doing this (it is portable):
echo "${onedate#"${onedate%%[!0]*}"}"
For your case (only bash):
#!/bin/bash
dates=( 01 02 02 08 10 18 20 21 0008 00101 )
for onedate in "${dates[#]}"; do
echo -ne "${onedate}\t"
echo "${onedate#"${onedate%%[!0]*}"}"
done
Will print:
$ script.sh
01 1
02 2
02 2
08 8
10 10
18 18
20 20
21 21
0008 8
00101 101
I'd like to know if there's a way to get the top 5 keywords by grouping them by days of the current month.
I'd like to receive a dataset like the following as result.
Supposing today is 4 of December i want to retrieve data for days from 1 to 4 of December, limiting the number of keywords for day to 5:
Day Keyword Visits
----------------------
01 keyword1 703
01 keyword2 688
01 keyword3 115
02 keyword1 109
02 keyword2 66
02 keyword3 53
02 keyword4 40
02 keyword5 23
03 keyword1 23
03 keyword2 19
03 keyword3 17
04 keyword1 14
04 keyword2 14
What i've currently done is setting the following parameters:
(you can test it here if you have a ganalytics account: http://code.google.com/intl/it-IT/apis/analytics/docs/gdata/gdataExplorer.html)
Dimensions: ga:day,ga:keyword
Metrics: ga:visits
Filters: ga:medium==organic;ga:keyword!=(not provided)
Sort: ga:day,-ga:visits,ga:keyword
Now i just need a method to limit the number of keywords for day (if possible).
cat file
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16
I tried with
column -c 4 file
to get an output with 4 columns, but it didn't work - I just get the same as the input.
Do I misunderstand the column man-page?
A second question: what format should the argument to the -s flag have?
Give this a try:
fold -w 12 file
The number 12 is the number of data columns * the number of characters in a column (two digits + one space). The -w option is for designating a screen width in terms of character columns.
The column command won't work for this because it's intended to format newspaper-style columns.
This comes close to working the way you want:
sed 's/ /\n/g' file | column -xc 35
The "35" is somewhat arbitrary, but any value from 32 to 39 will work in this case. It's related to the width of the fields (2 characters which is less than the width of a tab stop), the number of fields desired per line and the width of tab stops (8 characters). So, basically, 8 * 4 is 32.
Here's a demonstration of the -s option (which is used with -t):
$ echo -e "a;b|c\naaaaa;bbbbb|ccccc"|column -t -s ';|'
a b c
aaaaa bbbbb ccccc
Without using column, the output looks like:
$ echo -e "a;b|c\naaaaa;bbbbb|ccccc"
a;b|c
aaaaa;bbbbb|ccccc
Let's guess you want:
01 02 03 04
05 06 07 08
09 10 11 12
13 14 15 16
In this case:
$ xargs -n4 < file