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
--
http://mail.python.org/mailman/listinfo/python-list
