I have 104 files that need to be concatenated in groups of 10. The last group would be 4 files. Order of concatenation is important.
I was planning to sort the files (standard name sort works) and then pipe into a concatenation command in groups of 10, but I'd like to do it in zsh.
I can get the first 10 files in zsh with:
cat *([1,10]) > newfile1
How can I get the next 10 into newfile2, then the next....
I often have a need for this task, so I'm trying to stay generic. Is there an easy way to create this loop in zsh?
I am new to zsh and would very much appreciate explaining any suggestion.
Thank you!
Related
I have a huge table I want to extract information from. Firstly, I want to extract a certain line based on a pattern -> I've done that successfully with grep. However this line has loads of columns and I'm interested only in a couple of them that have a certain pattern in them (partial match - beginning of the string). Is it possible to extract only the columns and the number of the column (the nth column) for some partial matches? Hope I was clear enough.
Languages: Preferably in bash but I can also work in R, alternatively I'm open to suggestions if you think another language can be more helpful.
Thanks!
Awk is perfect for stuff like this. To help you write a script I think we need more details. But I'm guessing you'll want to use the print feature of awk. To print out the nth column of a file "your_file" do:
awk '{print $n}' your_file
In solving your problem you may also want to loop over all N columns which you can do via:
for i in {1..N} ;
do
awk -v col=${i} '{print $col}' your_file ;
done
So I was trying to do some research on it, but I could not find the answer. So I know that ls -l returns all things in the folder alphabetically, whilst ls -alt returns a list of files by their modification date, though without respect to alphabetical ordering.
I tried doing ls -l -alt, and also ls -alt -l, still no luck. What is the correct way to group them together?
Edit: With example.
Say I have the following list of directories:
aalexand bam carson duong garrett hollande jjackson ksmith mkumba olandt rcs solorzan truong yoo
aalfs battiste chae echo ghamilto holly jkelly kturner mls old.2016 reichman sophia twong zbib
I want to order them by alphabet, so say aalexand comes first. However, if aalfs has been modified last. So in other words has been changed more recently (not really sure how to structure this with proper grammar) it should appear first.
So if this were like a SQL query then we order by date last modified, group by directory name.
I am not sure what you want to do.
But, first of all: ls -l -alt is a double use of the -l parameter (take a look at man ls for more information about the parameters).
ls -l (l stands for list) just lists only one file per line (if you don't need the extra information like permissions, use -1 instead of -l). The -a includes hidden files. -t is for sorting by modified time. You cannot sort by name AND by time, except if two files would have the same name, which is not posible. Could you please explain your wish further?
Maybe you include a short example list of files including their modified time and your desired output, maybe then I can understand.
I'm trying to concatenate horizontally a number of files (1000) *.txt in a folder.
How can I loop over the files using the "paste" function?
NB: all the *.txt files are in the same directory.
Why loop? You can use wildcards.
paste *.txt > combined.txt
In general, it would be a question of just calling paste *.txt (and redirecting the output: paste *.txt > output.txt, as #zx did). Try it, but you'll be generating some enormously long lines. If paste can`t handle the line length you'll be generating, you'll have to reproduce its effect using a scripting language that has no line length limit, like perl or python.
Another possible sticking point is if your shell can't handle this many arguments in the expansion of the glob *.txt. Again, you can solve that with a script. It's easy to do so if that's your situation, let us know here.
PS. Given what paste does, looping is not going to do it for you: You (presumably) need the file contents side by side in the output, not one after the other.
I have a text file (more correctly, a “German style“ CSV file, i.e. semicolon-separated, decimal comma) which has a date and the value of a measurement on each line.
There are stretches of faulty values which I want to remove before further work. I'd like to store these cuts in some script so that my corrections are documented and I can replay those corrections if necessary.
The lines look like this:
28.01.2005 14:48:38;5,166
28.01.2005 14:50:38;2,916
28.01.2005 14:52:38;0,000
28.01.2005 14:54:38;0,000
(long stretch of values that should be removed; could also be something else beside 0)
01.02.2005 00:11:43;0,000
01.02.2005 00:13:43;1,333
01.02.2005 00:15:43;3,250
Now I'd like to store a list of begin and end patterns like 28.01.2005 14:52:38 + 01.02.2005 00:11:43, and the script would cut the lines matching these begin/end pairs and everything that's between them.
I'm thinking about hacking an awk script, but perhaps I'm missing an already existing tool.
Have a look at sed:
sed '/start_pat/,/end_pat/d'
will delete lines between start_pat and end_pat (inclusive).
To delete multiple such pairs, you can combine them with multiple -e options:
sed -e '/s1/,/e1/d' -e '/s2/,/e2/d' -e '/s3/,/e3/d' ...
Firstly, why do you need to keep a record of what you have done? Why not keep a backup of the original file, or take a diff between the old & new files, or put it under source control?
For the actual changes I suggest using Vim.
The Vim :global command (abbreviated to :g) can be used to run :ex commands on lines that match a regex. This is in many ways more powerful than awk since the commands can then refer to ranges relative to the matching line, plus you have the full text processing power of Vim at your disposal.
For example, this will do something close to what you want (untested, so caveat emptor):
:g!/^\d\d\.\d\d\.\d\d\d\d/ -1 write tmp.txt >> | delete
This matches lines that do NOT start with a date (the ! negates the match), appends the previous line to the file tmp.txt, then deletes the current line.
You will probably end up with duplicate lines in tmp.txt, but they can be removed by running the file through uniq.
you are also use awk
awk '/start/,/end/' file
I would seriously suggest learning the basics of perl (i.e. not the OO stuff). It will repay you in bucket-loads.
It is fast and simple to write a bit of perl to do this (and many other such tasks) once you have grasped the fundamentals, which if you are used to using awk, sed, grep etc are pretty simple.
You won't have to remember how to use lots of different tools and where you would previously have used multiple tools piped together to solve a problem, you can just use a single perl script (usually much faster to execute).
And, perl is installed on virtually every unix/linux distro now.
(that sed is neat though :-)
use grep -L (print none matching lines)
Sorry - thought you just wanted lines without 0,000 at the end
I have two files ...
file1:
002009092312291100098420090922111
010555101070002956200453T+00001190.81+00001295.920010.87P
010555101070002956200449J+00003128.85+00003693.90+00003128
010555101070002956200176H+00000281.14+00000300.32+00000281
file2:
002009092410521000098420090709111
010560458520002547500432M+00001822.88+00001592.96+00001822
010560458520002547500432D+00000106.68+00000114.77+00000106
In both files in every record starting with 01, the string from 3rd char to 25th char, i.e up to alphabet is the key.
Based on this key, I have to compare two files, and if there is any record matching in file 2, then I have to replace that record in file1, or else append it if it won't match.
Well, this is a fairly unspecific (and basic) programming question. We'll be better able to help us if you explain exactly what you did and where you got stuck.
Also, it looks a bit like homework, and people are wary of giving too much help on homework problems, as it might look like cheating.
To get you started:
I'd recommend Perl to solve this, but awk or another scripting language will also do. I'd recommend against sh/bash, as they are weak on text manipulation; also combining grep et al will become rather cumbersome.
First write a Perl program that filters records starting with 01. Then extract the key and put it into a hash (a Perl structure). Then output a new, combined file as required.
Using awk get the fields from 3-25 but doing something like
awk -F "" '/^01/{print $1}' file_name | cut -c 3-25 and match the first two fields with 01 from both files and get all the lines in two different buffers and compare both the buffers using for line in in a shell script.
Whenever the line in second buffer matches the first one grep the line in second buffer in first file and replace the line in first file with the line in second. I think you need to work a bit around the logic.