manipulating string question

[python101]

#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?
 

[bartonc]

#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?
 

>>> 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(resList)
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'
>>> 

[bartonc]

#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?
 

Always more ways than one to skin a cat:

>>> 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'
>>> 

[rhitam30111985]

here is another way:

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(count))
        b.append('$')
        b.append(a[i])    
        count +=1
    else:
        b.append(a[i])
stringA=''.join(b)
print stringA
 

[bvdet]

Nice solutions guys. For the exercise, here's another:

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_multi(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(stringA, tar)
    for i, item in enumerate(indices):
        item += len(s1)-len(stringA)
        s1 = ''.join([s1[:max(0,item)], eval(sub), s1[item+len(tar):]])
    return s1

>>> stringA = 'ASD DSA DASFSADSA FSAFSADSAF AS'
>>> str_replace_multi(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:

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(count))
        b.append('$')
        b.append(a[i])    
        count +=1
    else:
        b.append(a[i])
stringA=''.join(b)
print stringA
 

[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?

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(count))
        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
 

[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:

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(matches):
    res += stringA[mark:match.start()] + "%d$AS" %i  # or ] + str(i) + "$AS"
    mark = match.end()
 
print res
'1$ASD DSA D2$ASFSADSA FSAFSADSAF 3$AS'

[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.