split function

[Bengt Richter]

On Mon, 22 Aug 2005 22:14:55 +0200, Mohammed Altaj <mohammed at aims.ac.za> wrote:

>
>Dear all
>
>Sorry , I confused between two things , what i said in the last e-mail i
>already managed to do using C code , But what i need to do using python
>is : my input data  :
>
>0 2 3 4
>1 2 4
>2 3
>3 4
>
>what i suppose to do is ,  using the first line and start searching
>number by number ,first i have 0 search in the rest of lines if there is
>0 print out the all numbers except 0 , after that , start searching
>using the 2ed element in the first line which is 2 , in the 2ed line we
>have 1 , 4 . in the 3rd line we have 3 , in the 4th line we do not have
>2. And so on for 3 and 4 , and also for the 2nd , 3rd lines , so the
>output should be
>
>0 2 1 4 3 3 2 4 4 1 2 3
>1 2 3 4 3
>2 3
>3 4
>
>And i managed to do this , but i did in the case of no space between
>numbers,when i am reading from file , like
>0234
>124
>23
>34
>
>I want my code be able to deal with the space between numbers , and this
>is my code again
>def belong_to(x,a):
>    c=-1
>    for i in range(len(a)-1):
>        if x==int(a[i]):
>            c=i
>    return c
>
>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
>
>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])
>                               
>                                
>                              
>    print
>    out_file.write("\n")
>
>out_file.close()
>in_file.close()
>
>
>Thanks
>

Does this do what you want? (except that it writes to stdout and doesn't print)?
(note that I get 234 instead of 23 on the 3rd line of output)

 >>> # test with string file input and stdout output
 ... import StringIO
 >>> in_file = StringIO.StringIO("""\
 ... 0 2 3 4
 ... 1 2 4
 ... 2 3
 ... 3 4
 ... """)
 >>> import sys
 >>> out_file = sys.stdout
 >>>
 >>> lines = [''.join(line.split()) for line in in_file] # clean out spaces and '\n'
 >>> for i, line in enumerate(lines):
 ...     for digit in line:
 ...         out_file.write(digit)
 ...         for followingline in lines[i+1:]:
 ...             if digit not in followingline: continue
 ...             line_sans_digit = followingline.replace(digit,'')
 ...             out_file.write(line_sans_digit)
 ...     out_file.write("\n")
 ...
 021433244123
 12343
 234
 34

Also note that I eliminated all duplicate digits from a selected line, rather than just the
first match. Easy to change.

Regards,
Bengt Richter