manipulating string question

[python101]

[code=python]
#assume that I have a string

stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'

# I would like to make it
stringA = '1$ASD DSA D2$ASFSADSA FSASADSAF 3$AS'
#whatever there is 'AS' it should become 'i$AS', where i is its occurrence in the string
#how can I do that?
[/code]

[bartonc]

[code=python]
#assume that I have a string

stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'

# I would like to make it
stringA = '1$ASD DSA D2$ASFSADSA FSASADSAF 3$AS'
#whatever there is 'AS' it should become 'i$AS', where i is its occurrence in the string
#how can I do that?
[/code]

[CODE=python]
>>> stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
>>> i = 1
>>> resList = []
>>> for anStr in stringA.split() :
... thisItem = ""
... mark = 0
... n = anStr.count("AS ")
... for j in range(n):
... k = anStr.find("AS" )
... thisItem += anStr[mark:k] + "%d$AS" %i
... i += 1
... mark += k + 2
... thisItem += anStr[mark:]
... if n == 0:
... thisItem = anStr
... resList.append( thisItem)
...
>>> resList
['1$ASD', 'DSA', 'D2$ASFSADSA', 'FSAFSADSAF', '3$AS']
>>> " ".join(resL ist)
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'
>>> [/CODE]

[bartonc]

[code=python]
#assume that I have a string

stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'

# I would like to make it
stringA = '1$ASD DSA D2$ASFSADSA FSASADSAF 3$AS'
#whatever there is 'AS' it should become 'i$AS', where i is its occurrence in the string
#how can I do that?
[/code]

Always more ways than one to skin a cat:[CODE=python]
>>> import re
>>> stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
>>> marks = re.finditer("AS ", stringA)
>>> res = ""
>>> mark = 0
>>> i = 1
>>> for m in marks:
... res += stringA[mark:m.start()] + "%d$AS" %i
... mark = m.end()
... i += 1
...
>>> res
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'
>>> [/CODE]

[rhitam30111985]

here is another way:
[CODE=python]
stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
a=list(stringA)
b=[]

count=1
for i in range(len(a)):
if a[i]=='A' and a[i+1]=='S':
b.append(str(co unt))
b.append('$')
b.append(a[i])
count +=1
else:
b.append(a[i])
stringA=''.join (b)
print stringA
[/CODE]

[bvdet]

Nice solutions guys. For the exercise, here's another:[code=Python]def indexList(s, item, i=0):
i_list = []
while True:
try:
i = s.index(item, i)
i_list.append(i )
i += 1
except:
break
return i_list

def str_replace_mul ti(s, tar, sub):
"""
The target substring is to be replaced by an index
number representing its occurrence + '$' + the target.
'AS', replaced by i$AS'
'sub' must be a string that can be evaluated
"""
s1 = s
indices = indexList(strin gA, tar)
for i, item in enumerate(indic es):
item += len(s1)-len(stringA)
s1 = ''.join([s1[:max(0,item)], eval(sub), s1[item+len(tar):]])
return s1[/code]

>>> stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
>>> str_replace_mul ti(stringA, 'AS', "'%d$AS' % (i+1)")
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'
>>>

I keep finding uses for function indexList() :)

[rogerlew]

And another... with re.
It had to be done!

import re
def repl(match):
    globals()['count']+=1
    return '$%i%s'%(globals()['count'],
                    match.groups()[0])
         
globals()['count']=0 
print re.sub('(AS)',repl,stringA)

>>> $1ASD DSA D$2ASFSADSA FSAFSADSAF $3AS

[python101]

Thank everyone's, especially rhitam30111985; s code

here is another way:
[CODE=python]
stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
a=list(stringA)
b=[]

count=1
for i in range(len(a)):
if a[i]=='A' and a[i+1]=='S':
b.append(str(co unt))
b.append('$')
b.append(a[i])
count +=1
else:
b.append(a[i])
stringA=''.join (b)
print stringA
[/CODE]

[bartonc]

Thank everyone's, especially rhitam30111985; s code

Which will throw an error if the very last character is an "A".

[rhitam30111985]

hey barton u are right .. didnt notice that .. ok how about this?

[CODE=python]
stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
a=list(stringA)
b=[]

count=1
if a[-1]=='A':
a.append(' ')# :-P
for i in range(len(a)):
if a[i]=='A' and a[i+1]=='S':
b.append(str(co unt))
b.append('$')
b.append(a[i])
count +=1
else:
b.append(a[i])
stringA=''.join (b)#while printing, the extra space wont be noticed
print stringA
[/CODE]

[bartonc]

hey barton u are right .. didnt notice that .. ok how about this?

Yep. That would do it. But wouldn't the result contain the extra character?

I still like my Regular Expression method best.

[rhitam30111985]

yep.. it will contain the extra chcracter.. which is just a space... so not really a problem we can put a check at the end for the last character if it is a space.. then we can use the list.pop() or list.remove() method to get rid of it... .. as for regular expressions.. i personally stay away from them.... they r just beyond my comprehension.. :-(

[bartonc]

yep.. it will contain the extra chcracter.. which is just a space... so not really a problem .. as for regular expressions.. i personally stay away from them.... they r just beyond my comprehension..

This (slightly simplified) one seems pretty basic:[CODE=python]
import re

stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
matches = re.finditer("AS ", stringA)
res = ""
mark = 0 # marks where we left off in the string
for i, match in enumerate(match es):
res += stringA[mark:match.star t()] + "%d$AS" %i # or ] + str(i) + "$AS"
mark = match.end()

print res
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'[/CODE]

[bartonc]

And another... with re.
It had to be done!

import re
def repl(match):
    globals()['count']+=1
    return '$%i%s'%(globals()['count'],
                    match.groups()[0])
         
globals()['count']=0 
print re.sub('(AS)',repl,stringA)

>>> $1ASD DSA D$2ASFSADSA FSAFSADSAF $3AS

Let Python search for the variable outside the scope of the function instead of returning the dictionary reference (just thinking out loud, here)... I think I really like your way. Just for completeness, I'll post to see how it looks:

import re
def repl(match):
    global count
    count += 1

Yep. I like your way better.