Python: filtering table

xgreen · 07-09-2005, 08:27 PM

i've a file contain a 'ls -l' output. i need to filter out all text file (*.txt)

e.g of ls -l output in file named "test":

code :
--------
y=file("/home/nazmin/temp/test","r")

for i in y:
lines=i.strip()
print lines
y.close()

output
--------
-rw-r--r-- 1 nazmin users 25K 2005-05-27 21:59 linux_serial.txt
drwx------ 2 nazmin users 112 2005-06-08 20:57 mail/
-rw-r--r-- 1 nazmin users 258 2005-06-05 01:50 metallica
-rw-r--r-- 1 nazmin users 5.0K 2005-05-21 10:18 msg00123.html

output required :

linux_serial.txt

carl.waldbieser · 07-10-2005, 12:58 AM

This could be tricky, but if you can assume that the user, group , and file names don't have spaces in them, you can split each line on the spaces and take the 8th field. If the last 4 characters are ".txt" you could print it.

Example:

Code:

filename = line.split(' ')[7]
if filename[-4:] == ".txt":
    print filename

If the fields themselves can have spaces in them, then the problem becomes trickier.

One thing that strinkes me as odd, though-- why not just use "ls *.txt"?

CroMagnon · 07-10-2005, 01:46 AM

Why not just split on the colon in the time field? I'm pretty sure a colon wont be showing up anywhere before that point:

Code:

files = [ x.strip().split(':',1)[1][3:] for x in lines if x.strip()[-4:] == '.txt' ]

xgreen · 07-10-2005, 05:38 AM

thanks guys... i've tired your suggestion..but no luck but i managed to write the code (i think the code quite long for the job ...need to find better ways of doing it :-) )

Code:

y=file("/home/nazmin/testing/test","r")
for x in y:
        lines=x.strip()
        list=[]
        for i in lines:
                  list.append(i)

       c=list.index(":")
       new_list=list[c+4:]
	
      txt=new_list[-4:]
      new_txt="".join(txt)
      if new_txt == '.txt':
                new_word="".join(new_list)
                print new_word	
y.close()

CroMagnon · 07-10-2005, 04:07 PM

That's exactly what my line does, except in a much more convoluted way. The only difference is, my line doesn't actually produce the output, it makes a list of the filenames. So you could do this:

Code:

y=file("/home/nazmin/testing/test","r")
files = [ x.strip().split(':',1)[1][3:] for x in y if x.strip()[-4:] == '.txt' ]
for f in files:
    print f

xgreen · 07-11-2005, 06:19 AM

CroMagnon, thanks for the code & it is working. could you explain the meaning of this code.

files = [ x.strip().split(':',1)[1][3:] for x in y if x.strip()[-4:] == '.txt' ]

CroMagnon · 07-11-2005, 03:50 PM

Sure. It might be easier to explain in smaller chunks. Excuse me if I tell you stuff you already know...

Code:

[x for x in range(10)]

This builds a list by walking through the list provided by range(10), effectively making a list like this: [0,1,2,3,4,5,6,7,8,9]

Not very useful by itself

But then we can add a condition, like this:

Code:

[x for x in range(10) if x < 5]

This walks through the list and only includes the value if it meets the condition, so we get [0,1,2,3,4].
Finally, we can modify the value before it is added to the list, like this:

Code:

[x * 2 for x in range(10) if x < 5]

So it takes each value that meets the condition, and multiplies it by 2, so we get [0,2,4,6,8].

So, on to my code:

Code:

files = [ x.strip().split(':',1)[1][3:] for x in y if x.strip()[-4:] == '.txt' ]

x.strip() - removes the spaces
.split(':',1) - breaks the string into two pieces, around the colon (the 1 means to only split once, not for every colon, just in case)
Because the result of split is a list, and we want the data after the colon, I add [1] to get the right list element, then we use [3:] to chop the first 3 characters off the start of the string.
The condition at the end just stops the program from operating on any line that doesn't end with ".txt".

Hope that helps, let me know if you have any questions.

xgreen · 07-13-2005, 06:07 AM

CroMagnon,

thanks for the superb explanation...really helping