Develop Reference

Take the path of a file by dropping it on batch or python

in my head this problem seems simple but I cant for the life of me figure it out.
I want to use a function similar to os.replace() to move a file/folder from one location which could vary to one that is set whilst also preserving the name of it.
At this point I couldn't figure it out however to make it slightly more difficult I want to be able to drop a file onto the batch/python script and have the code detect the filepath for the file i dropped on it.
Sorry for the bad explanation in short:
import os
initialfilepath = "The filepath of the file i drop onto the batch/python file"
finalfilepath = "Predetermined/file/path etc"
os.replace(initialfilepath,finalfilepath) <--However i want to preserve the name of the file.
Any help would be greatly appreciated!

With a single line in batch file you can handle all your dropped files:
CMD /k FOR %%s in (%*) do ECHO %%s
In this example it will print all dropped files.
The parameter %* get all file pathnames.
The command FOR as it is shown, split the string by spaces, and handle one by one.
If the file pathname has spaces, so it will come in quotes as one thing.
Note that you don't need CMD /k, but it will keep the console opened at the end.
If you just wanna see before to close, insert PAUSE > nul at the end.
You can group the do commands between ( ... ) and then break the lines.
Consider set #ECHO off in beggining.
Now try it yourself :)
Type FOR /? to see more possibilities to manipulate parameters %~nX, %~dpX, %~xX...

loop in bash/python to call a command including name fixation

I have a line of code that works on a single file:
sextractor ABC347583_ima.fits -CATALOG_NAME ABC347583_ima.cat
this piece of line takes the .fits file and creates a catalog file with the same name but .cat extension
Now, I have over 100+ .fits files(All my files start with the name ABC then numbers and end with _ima) and I would like to make a bash/python script that will read the .fits files 1 by 1 and execute the above code with the corresponding file names as input & output.
Basicly ABC347583_ima.fits , ABC57334_ima.fits etc. will enter and ABC347583_ima.cat , ABC57334_ima.cat etc. will be created.
This is beyond my limited knowlegde and what I only know about this is to use the code with
for i in `cat files`
echo $i
However, this does not exactly match with the command line because of both input & output. Any suggestions about how to pass this will be appreciated.

To iterate in python over all files in a dictonary use os.listdir().
Then you can loop over the filenames.
in the "callthecommandhere" function you can parse the filenames and read the file content and write a new file. I hope I understand you right and that is a help for you.
Like so:
import os
for filename in os.listdir('dirname'):
callthecommandhere(blablahbla, filename, foo)
Br christoph

Python, searching own codes wrote

There are scripts wrote and all of them may of different topics. Some about documents handling, some about text extraction, some about automation.
Sometimes I forgot a usage, for example how to create a .xls file, so I want to search if in the scripts there is a line about how to do it.
What I am doing is to convert all the .py files into .txt, and combine all txt files together. Then I use Word to open this aggregated .txt file and find.
What’s the better way to search specific lines in own written codes?
**converting .py to .txt:
folder = "C:\\Python27\\"
for a in os.listdir(folder):
root, ext = os.path.splitext(a)
if ext == ".py":
os.rename(folder + a, folder + root + ".txt")
**putting all .txt together:
base_folder = "C:\\TXTs\\"
all_files = []
for each in os.listdir(base_folder):
if each.endswith('.txt'):
kk = os.path.join(base_folder, each)
all_files.append(kk)
with open(base_folder + " final.txt", 'w') as outfile:
for fname in all_files:
with open(fname) as infile:
for line in infile:
outfile.write(line)

Keep all of your code in a single directory tree, e.g. code.
If your operating system doesn't have a decent search tool like grep, install "the silver searcher". (You will also need a decent terminal emulator.)
For example (I'm using FreeBSD here), I tend to keep all my source code under src/. If I want to know in which scripts I use xls files, I type:
ag xls src/
Which retuns:
src/progs/uren/project-uren.py
24: print("Usage: {} urenbriefjesNNNN.xlsm project".format(binary),
This tells me to look at line 24 of the file src/progs/uren/project-uren.py.
If I e.g. search for Counter:
ag Counter src/
I get multiple hits:
src/scripts/public/csv2tbl.py
13:from collections import Counter
45: letters, sep = Counter(), Counter()
src/scripts/public/dvd2webm.py
15:from collections import Counter
94: rv = Counter(re.findall('crop=(\d+:\d+:\d+:\d+)', proc.stderr))
src/scripts/public/scripts-tests.py
14:from collections import Counter
26: v = Counter(rndcaps(100000)).values()

You can install rStudio, an open source IDE for the r language. In the Edit menu there is a Find in File... feature you can use just like find and replace in a word document. It will go through files in the directory you point it to...I have not yet had problems searching scripts as they are, untransformed to txt. It will search for terms or regex expressions....it is pretty useful!
As is R!

cat *.py | grep xls if you're on Linux.
Otherwise it may be helpful to keep some sort of README file with your python scripts. I, personally, prefer Markdown:
## Scripts
They do stuff
### Script A
Does stuff A, call `script_a.py -h` for more info
### Script B
Does stuff B, call `script_b.py -h` for more info
It compiles to this:
Scripts
They do stuff
Script A
Does stuff A, call script_a.py -h for more info
Script B
Does stuff B, call script_b.py -h for more info
It takes basically no time to write and Markdown can be easily used on sites such as SO, Github, Reddit and others. This very answer, in fact, is written in Markdown. But if you can't be bothered with Markdown, a simple README.txt is still much better than nothing.

The technical term for what you're trying to do is a "full text file search". Googling this together with your operating system name will give you many methods. Here is one for Windows: https://www.howtogeek.com/99406/how-to-search-for-text-inside-of-any-file-using-windows-search/.
If you're on MacOS I recommend looking into BASH command line syntax to do a bit more complex automation tasks (although what you need is also perfectly covered in Spotlight search). On Windows 10 you could check out the new Linux Subsystem that gives you the same syntax [1]. Composing small commands together using pipes and xargs in command line is a very powerful automation tool. For what you're asking I still think a full text search is the most straightforward solution, but since you're already into programming I thought I bring this up.
To demonstrate, the task you describe would be something like
find . -name "*.py" | xargs -I {} grep -H "xls" {}
This would search your working directory (and all subdirectories) for python files (using . as its first argument to find, which refers to the directory you're currently in, shown by pwd), and then search each of those python files for the string "xls". xargs takes all lines from standard input (which the pipe | gets from the last command) and converts them into command line parameters. grep -H searches files for the specified string and prints the occurrences together with the file name.
[1] I'm assuming you're not on Linux already since you like to use MS Office.

Python: How to get the URL to a file when the file is received from a pipe?

I created, in Python, an executable whose input is the URL to a file and whose output is the file, e.g.,
file:///C:/example/folder/test.txt --> url2file --> the file
Actually, the URL is stored in a file (url.txt) and I run it from a DOS command line using a pipe:
type url.txt | url2file
That works great.
I want to create, in Python, an executable whose input is a file and whose output is the URL to the file, e.g.,
a file --> file2url --> URL
Again, I am using DOS and connecting executables via pipes:
type url.txt | url2file | file2url
Question: file2url is receiving a file. How do I get the file's URL (or path)?

In general, you probably can't.
If the url is not stored in the file, I seems very difficult to get the url. Imagine someone reads a text to you. Without further information you have no way to know what book it comes from.
However there are certain usecases where you can do it.
Pipe the url together with the file.
If you need the url and you can do that, try to keep the url together with the file. Make url2file pipe your url first and then the file.
Restructure your pipeline
Maybe you don't need to find the url for the file, if you restructure your pipeline.
Index your files
If only a certain files could potentially be piped into file2url, you could precalculate a hash for all files and store it in your program together with the url. In python you would do this using a dict where the key is the file (as a string) and the value is the url. You could use pickle to write the dict object to a file and load it at the start of your program.
Then you could simply lookup the url from this dict.
You might want to research how databases or search functions in explorers handle indexing or alternative solutions.
Searching for the file
You could use one significant line of the file and use something like grep or head on linux to search all files of your computer for this line. Note that grep and head are programs, not python functions. For DOS, you might need to google the equivalent programs.
FYI: grep searches for one line of text inside a file.
head puts out the first few lines of a file. I suggest comparing only the first few lines of files to avoid searching through huge file.
Searching all files on the computer might take very long.
You could only search files with the same size as your piped input.
Use url.txt
If file2url knows the location of the file url.txt, then you could look up all files in url.txt until you find a file identical to the file that was piped into your program. You could combine this with the hashing/ indexing solution.

'file2url' receives the data via standard input (like keyboard).
The data is transferred by the kernel and it doesn't necessarily have to have any file-system representation. So if there's no file there's no URL or path to that for you to get.

Let's try to do it by obvious way:
$ cat test.py | python test.py
import sys
print ''.join(sys.stdin.readlines())
print sys.stdin.name
<stdin>
So, filename is "< stdin>" because, for the python there is no filename - only input.
Another way is a system-dependent. Find a command line, which was used, for example, but no garantee that is will be works.

is there a built-in python analog to unix 'wc' for sniffing a file?

Everyone's done this--from the shell, you need some details about a text file (more than just ls -l gives you), in particular, that file's line count, so:
# > wc -l iris.txt
149 iris.txt
i know that i can access shell utilities from python, but i am looking for a python built-in, if there is one.
The crux of my question is getting this information without opening the file (hence my reference to the unix utility *wc -*l)
(is 'sniffing' the correct term for this--i.e., peeking at a file w/o opening it?')

You can always scan through it quickly, right?
lc = sum(1 for l in open('iris.txt'))

No, I would not call this "sniffing". Sniffing typically refers to looking at data as it passes through, like Ethernet packet capture.
You cannot get the number of lines from a file without opening it. This is because the number of lines in the file is actually the number of newline characters ("\n" on linux) in the file, which you have to read after open()ing it.