On Tue, 23 Aug 2005 19:20:15 +0200, Mohammed Altaj <[EMAIL PROTECTED]> wrote:
>
>Dear All
>
>This is my problem again , I tried to sort it out , but i couldn't , I
>am reading data from file using readlines , my input like :
>
>0 1 2 4
>1 2 4
>2 3
>3 4
>
>What i am doing is , starting with the first element in the first line (
>which is 0 in this case )and do search in the other lines , if i found
>another 0 , i will save (print out) all the elements except 0 , (in this
>is case i have no 0 elsewhere) so i will print only 0. Now do search by
>the 2nd element in the first line (which is 1 in this case) , in the 2nd
>line we have 1 , so i should save(print out) all elements except 1 which
>are 2 4 , and so on for the rest of the first line , and for the rest of
>the file , my out put should be
>
>0 1 2 4 2 1 4 3 4 1 2 3
>1 2 3 4 3
>2 3 4
>3 4
>
>I managed to do all these things , but i did it in the way that i am
>reading my data as strings ( no space between numbers) something like
>
>0124
>124
>23
>34
>
>what i would like to know or to do is , how can i deal with my data
>after reading it as strings(i need the space between numbers) because i
>had problem when dealing with number larger than 9 , example :
>
>0 1 5 9
>1 12 10
>4 6 7
>10 9
>
>so , when i remove the space between numbers , i loose all my data , i
>mean it will look like
>0159
>11210
>467
>509
>
>
>This is my code :
>
>
>def belong_to(x,a):
> c=-1
> for i in range(len(a)-1):
> if x==int(a[i]):
This was selecting single characters at a[i], so it
was a misleading clue re your actual requirements ;-)
> 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()
>
>
>
>Thank you all
>
You could try dealing with the data as lists of numbers, e.g, (slight mod from
my previous)
>>> # 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
... 0 1 5 9
... 1 12 10
... 4 6 7
... 10 9
... """)
>>> import sys
>>> out_file = sys.stdout
>>>
>>> lines = [map(int,line.split()) for line in in_file] # make lines as int
>>> lists
>>> for i, line in enumerate(lines):
... out = []
... for digit in line:
... out.append(digit)
... for followingline in lines[i+1:]:
... if digit in followingline:
... out.extend([x for x in followingline if x != digit])
... out_file.write(' '.join(map(str, out))+"\n")
...
0 1 5 9 2 1 4 3 3 2 4 4 1 2 3 6 7
1 0 5 9 12 10 2 3 4 3 6 7
2 3 4
3 4 6 7
0 1 12 10 5 9 10
1 12 10 9
4 6 7
10 9
(The output now has to have spaces as well, to delimit the numbers).
Regards,
Bengt Richter
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