What are the differences between sort file1 -o file2 and sort file1 > file2 ? So far from what I have done they do the same thing but perhaps I'm missing something.
Following two commands are similar as long as file1 and file2 are different.
sort file1 -o file2 # Output redirection within sort command
sort file1 > file2 # Output redirection via shell
Let's see what happens when input and output files are same file i.e. you try to sort in-place
sort file -o file # Works perfectly fine and does in-place sorting
sort file > file # Surprise! Generates empty file. Data is lost :(
In summary, above two redirection methods are similar but not the same
Test
$ cat file
2
5
1
4
3
$ sort file -o file
$ cat file
1
2
3
4
5
$ sort file > file
$ cat file
$ ls -s file
0 file
The result is the same but in the case of -o file2 the resulting file is created by sort directly while in the other case, it is created by bash and filled with the standard output of sort. The xfopen defined in line 450 of sort.c in coreutils treats both cases (stdout and -o filename) equally.
Redirecting the standard output of sort is more generic as it could be redirected to another program with a | in place of a >, which the -o option makes more difficult to do (but not impossible)
The -o option is handy for in place sorting as the redirection to the same file will lead to a truncated file because it is created (and truncated) by the shell prior to the invocation of sort.
There is not much difference > is a standard unix output redirection function. That is to say 'write your output that you would otherwise display on the terminal to the given file' The -o option is more specific to the sort function. It is a way to again say 'write the output to this given file'
The > can be used where a tool does not specifically have a write to file argument or option.
Related
I want to have a command in a variable that runs a program and specifies the output filename for it depending on the number of files exits (to work on a new file each time).
Here is what I have:
export MY_COMMAND="myprogram -o ./dir/outfile-0.txt"
However I would like to make this outfile number increases each time MY_COMMAND is being executed. You may suppose myprogram creates the file soon enough before the next call. So the number can be retrieved from the number of files exists in the directory ./dir/. I do not have access to change myprogram itself or the use of MY_COMMAND.
Thanks in advance.
Given that you can't change myprogram — its -o option will always write to the file given on the command line, and assuming that something also out of your control is running MY_COMMAND so you can't change the way that MY_COMMAND gets called, you still have control of MY_COMMAND
For the rest of this answer I'm going to change the name MY_COMMAND to callprog mostly because it's easier to type.
You can define callprog as a variable as in your example export callprog="myprogram -o ./dir/outfile-0.txt", but you could instead write a shell script and name that callprog, and a shell script can do pretty much anything you want.
So, you have a directory full of outfile-<num>.txt files and you want to output to the next non-colliding outfile-<num+1>.txt.
Your shell script can get the numbers by listing the files, cutting out only the numbers, sorting them, then take the highest number.
If we have these files in dir:
outfile-0.txt
outfile-1.txt
outfile-5.txt
outfile-10.txt
ls -1 ./dir/outfile*.txt produces the list
./dir/outfile-0.txt
./dir/outfile-1.txt
./dir/outfile-10.txt
./dir/outfile-5.txt
(using outfile and .txt means this will work even if there are other files not name outfile)
Scrape out the number by piping it through the stream editor sed … capture the number and keep only that part:
ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:'
(I'm using colon : instead of the standard slash / so I don't have to escape the directory separator in dir/outfile)
Now you just need to pick the highest number. Sort the numbers and take the top
| sort -rn | head -1
Sorting with -n is numeric, not lexigraphic sorting, -r reverses so the highest number will be first, not last.
Putting it all together, this will list the files, edit the names keeping only the numeric part, sort, and get just the first entry. You want to assign that to a variable to work with it, so it is:
high=$(ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' | sort -rn | head -1)
In the shell (I'm using bash) you can do math on that, $[high + 1] so if high is 10, the expression produces 11
You would use that as the numeric part of your filename.
The whole shell script then just needs to use that number in the filename. Here it is, with lines broken for better readability:
#!/bin/sh
high=$(ls -1 ./dir/outfile*.txt \
| sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' \
| sort -rn | head -1)
echo "myprogram -o ./dir/outfile-$[high + 1].txt"
Of course you wouldn't echo myprogram, you'd just run it.
you could do this in a bash function under your .bashrc by using wc to get the number of files in the dir and then adding 1 to the result
yourfunction () {
dir=/path/to/dir
filenum=$(expr $(ls $dir | wc -w) + 1)
myprogram -o $dir/outfile-${filenum}.txt
}
this should get the number of files in $dir and append 1 to that number to get the number you need for the filename. if you place it in your .bashrc or under .bash_aliases and source .bashrc then it should work like any other shell command
You can try exporting a function for MY_COMMAND to run.
next_outfile () {
my_program -o ./dir/outfile-${_next_number}.txt
((_next_number ++ ))
}
export -f next_outfile
export MY_COMMAND="next_outfile" _next_number=0
This relies on a "private" global variable _next_number being initialized to 0 and not otherwise modified.
I searched online but I didn't find anything that could answer my question.
I'm using a java tool in Ubuntu Linux, calling it with bash command; this tool has two paths for two different input files:
java -Xmx8G -jar picard.jar FastqToSam \
FASTQ=6484_snippet_1.fastq \ #first read file of pair
FASTQ2=6484_snippet_2.fastq \ #second read file of pair
[...]
What I'd like to do is for example, instead of specify the path of a single FASTQ, specify the path of two different files.
So instead of having cat file1 file2 > File and using File as input of FASTQ, I'd like that this operation would be executed on the fly and create the File on the fly, without saving it on the file system (that would be what happens with the command cat file1 file2 > File).
I hope that I've been clear in explaining my question, in case just ask me and I'll try to explain better.
Most well-written shell commands which accept a file name argument also usually accept a list of file name arguments. Like cat file or cat file1 file2 etc.
If the program you are trying to use doesn't support this, and cannot easily be fixed, perhaps your OS or shell makes /dev/stdin available as a pseudo-file.
cat file1 file2 | java -mumble -crash -burn FASTQ=/dev/stdin
Some shells also have process substitutions, which (typically) look to the calling program like a single file containing whatever the process substitution produces on standard output.
java -mumble -crash -burn FASTQ=<(cat file1 file2) FASTQ2=<(cat file3 file4)
If neither of these work, a simple shell script which uses temporary files and deletes them when it's done is a tried and true solution.
#!/bin/sh
: ${4?Need four file name arguments, will process them pairwise}
t=$(mktemp -d -t fastqtwoness.XXXXXXX) || exit
trap 'rm -rf $t' EXIT HUP INT TERM # remove in case of failure or when done
cat "$1" "$2" >$t/1.fastq
cat "$3" "$4" >$t/2.fastq
exec java -mumble -crash -burn FASTQ=$t/1.fastq FASTQ2=$t/2.fastq
As per my understanding, the pipe operator in Linux takes standard output of the one command and channelizes it to the standard input of the next command. But I have faced one anomaly.
I am trying to get the content of a file in the standard ouput as below.
cat file1
It displays the content. Let's say the content is another file named file2.
Now I want to display the content of file2.
So to take the advantage of pipe operator, I am trying to execute as below
cat file1 | cat
The first cat command should pipe the output (here "file2"). The cat in the subsequent command must accept it from the standard input (here the value is "file2") and print the content of file2.
But it displays "file2" only instead of its contents.
What you should do:
cat `cat file1`
From man cat:
Concatenate FILE(s), or standard input, to standard output.
In other words, if there is a filename provided as parameter, it will display its content to the standard output, otherwise it will just redirect to standard input to the standard output.
In your case, the filename is read from the standard input and it is interpreted as a string to concatenate to the standard output.
The backquotes are used to inject the standard output of a command, i.e:
cat `cat file1`
is equivalent to
cat file2
which will dump file2 content to standard output.
You could use xargs:
cat file1 | xargs cat
XARGS General Commands Manual
As an easy example, consider the following command:
$ sort file.txt
This will output the file's data in sorted order. How do I put that data right back into the same file? I want to update the file with the sorted results.
This is not the solution:
$ sort file.txt > file.txt
... as it will cause the file to come out blank. Is there a way to update this file without creating a temporary file?
Sure, I could do something like this:
sort file.txt > temp.txt; mv temp.txt file.txt
But I would rather keep the results in memory until processing is done, and then write them back to the same file. sort actually has a flag that will allow this to be possible:
sort file.txt -o file.txt
...but I'm looking for a solution that doesn't rely on the binary having a special flag to account for this, as not all are guaranteed to. Is there some kind of linux command that will hold the data until the processing is finished?
For sort, you can use the -o option.
For a more general solution, you can use sponge, from the moreutils package:
sort file.txt | sponge file.txt
As mentioned below, error handling here is tricky. You may end up with an empty file if something goes wrong in the steps before sponge.
This is a duplicate of this question, which discusses the solutions above: How do I execute any command editing its file (argument) "in place" using bash?
You can do it with sed (with its r command), and Process Substitution:
sed -ni r<(sort file) file
In this way, you're telling sed not to print the (original) lines (-n option) and to append the file generated by <(sort file).
The well known -i option is the one which does the trick.
Example
$ cat file
b
d
c
a
e
$ sed -ni r<(sort file) file
$ cat file
a
b
c
d
e
Try vim-way:
$ ex -s +'%!sort' -cxa file.txt
When I am trying to sort a file and save the sorted output in itself, like this
sort file1 > file1;
the contents of the file1 is getting erased altogether, whereas when i am trying to do the same with 'tee' command like this
sort file1 | tee file1;
it works fine [ed: "works fine" only for small files with lucky timing, will cause lost data on large ones or with unhelpful process scheduling], i.e it is overwriting the sorted output of file1 in itself and also showing it on standard output.
Can someone explain why the first case is not working?
As other people explained, the problem is that the I/O redirection is done before the sort command is executed, so the file is truncated before sort gets a chance to read it. If you think for a bit, the reason why is obvious - the shell handles the I/O redirection, and must do that before running the command.
The sort command has 'always' (since at least Version 7 UNIX) supported a -o option to make it safe to output to one of the input files:
sort -o file1 file1 file2 file3
The trick with tee depends on timing and luck (and probably a small data file). If you had a megabyte or larger file, I expect it would be clobbered, at least in part, by the tee command. That is, if the file is large enough, the tee command would open the file for output and truncate it before sort finished reading it.
It doesn't work because '>' redirection implies truncation, and to avoid keeping the whole output of sort in the memory before re-directing to the file, bash truncates and redirects output before running sort. Thus, contents of the file1 file will be truncated before sort will have a chance to read it.
It's unwise to depend on either of these command to work the way you expect.
The way to modify a file in place is to write the modified version to a new file, then rename the new file to the original name:
sort file1 > file1.tmp && mv file1.tmp file1
This avoids the problem of reading the file after it's been partially modified, which is likely to mess up the results. It also makes it possible to deal gracefully with errors; if the file is N bytes long, and you only have N/2 bytes of space available on the file system, you can detect the failure creating the temporary file and not do the rename.
Or you can rename the original file, then read it and write to a new file with the same name:
mv file1 file1.bak && sort file1.bak > file1
Some commands have options to modify files in place (for example, perl and sed both have -i options (note that the syntax of sed's -i option can vary). But these options work by creating temporary files; it's just done internally.
Redirection has higher precedence. So in the first case, > file1 executes first and empties the file.
The first command doesn't work (sort file1 > file1), because when using the redirection operator (> or >>) shell creates/truncates file before the sort command is even invoked, since it has higher precedence.
The second command works (sort file1 | tee file1), because sort reads lines from the file first, then writes sorted data to standard output.
So when using any other similar command, you should avoid using redirection operator when reading and writing into the same file, but you should use relevant in-place editors for that (e.g. ex, ed, sed), for example:
ex '+%!sort' -cwq file1
or use other utils such as sponge.
Luckily for sort there is the -o parameter which write results to the file (as suggested by #Jonathan), so the solution is straight forward: sort -o file1 file1.
Bash open a new empty file when reads the pipe, and then calls to sort.
In the second case, tee opens the file after sort has already read the contents.
You can use this method
sort file1 -o file1
This will sort and store back to the original file. Also, you can use this command to remove duplicated line:
sort -u file1 -o file1