split function

[bruno modulix]

Mohammed Altaj wrote:
> Dear All
> 
> What i want to do is , my input is like
> 0 2
> 0 3
> 0 4
> 1 2
> 1 4
> 2 3
> 3 4
> 
> I am comparing and put the number in group , like ,the first three lines
> , all has zero as first input for each line, so the out put should look
> like
> 0 2 3 4
> and so on
> 1 2 4
> 2 3
> 3 4
> 
> I managed to do what i need , but i did in this case , there is no space
> between numbers , like
> 02
> 03
> 04
> 12
> 14
> 23
> 34
> 
> so , how can i do this with spaces between numbers

Remove them ?

> This is my code

Your code does not seems to work as expected. I get:
0 3 4 2 1 3
0 4 3 2
0 4 1
1 4 2 3
1 4
2 3
3

> 
> def belong_to(x,a):          
>     c=-1
>     for i in range(len(a)-1):
>         if x==int(a[i]):
>             c=i
>     return c

def belong_to(x, line):
  for i, c in enumerate(line.strip()):
    if x == int(c):
      return i
  return -1

def belong_to(x, line):
   return line.find(str(x))

belong_to = lambda x, line: line.find(str(x))

Now, is it useful to define a function for such a trivial test ?


> def list_belong(x,a):             # This function to check if this line
>     c=-1                                  # line has been searched
> before or not
>     for i in range(len(a)):
>         if a[i]==x:
>             c=1
>             break
>     return c

def list_belong(line, lines):
   return line in lines

... is it really useful to define a function for such a trivial test ?

> x=0
> occur=[]
> 
> in_file=open('data.dat','r')
> out_file=open('result.dat','w')
> fileList = in_file.readlines()
> for k in fileList:
>     v=k
>     occur.append(k)
>     n=len(v)-1
>     for i in range(n):
>         temp=int(v[i])
>         print temp,
>         out_file.write(str(temp))
>         for line in fileList:
>             if v!=line:
>                 if list_belong(line,occur)!=1:
>                     if belong_to(temp,line) != -1:
>                         j=belong_to(temp,line)
>                         for i in range(len(line)-1):
>                             if i!=j:
>                                 print line[i],
>                                 out_file.write(line[i])

ouch :(

>                               
>     print
>     out_file.write("\n")
> 
> out_file.close()
> in_file.close()
> 

May I suggest a much more simple version that conform to your specs and
solves the space problem ?

from itertools import groupby

in_file=open('data.dat','r')

# strip EOLs and get rid of whitespaces
lines = [line.strip().replace(' ', '') for line in_file.readlines()]
in_file.close()

out_file=open('result.dat','w')

# group lines by first char
for key, groups in groupby(lines, lambda line: line[0]):
    buf = "%s %s" % (key, " ".join([group[1] for group in groups]))
    print buf
    out_file.write("%s\n" % buf)

out_file.close()


Python is meant to make your life easier...

-- 
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'onurb at xiludom.gro'.split('@')])"