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Python: How to slice string using string?

Assuming that the user entered:
"i like eating big apples"
Want to remove "eating" and "apples" together with whatever is in between these two words. Output in this case
"i like"
In another case, if the user entered:
"i like eating apples very much"
Expected output:
"i like very much"
And I want to slice the input starting from "eating" to "apples"
(However, the index cannot be used as you are unsure how long the user is going to type, but it is guaranteed that "eating" and "apples" will be entered)
So, is there any way that we can slide without using the index, instead, we indicate the start and end of the slide with another string?

Slicing a string in python is like this:
mystr = "i like eating big apples"
print(mystr[10:20])
This means between the 10th boundary of characters in the string and the 20th. So it will become: ing big ap.
Now the question is how to find out which index 'eating' starts and 'apple' ends.
Use the .index method to find the beginning of something in a string.
mystr.index('eating') returns 7, so if you print mystr[7:] (which means from the 7th index till the last of the string) you'll have 'eating big apples'.
The second part is a little tricky. If you use mystr.index('apple'), you'll get the beginning of apple, (18), so mystr[7:18] will give you 'eating big '.
In fact you should go some characters further to include the apple word too, which is 5 chars exactly, and this number will be returned by len('apple'). So the final result is:
start = mystr.index('eating')
stop = mystr.index('apple') + len('apple')
print(mystr[start:stop])

You can do the follwoing:
s = "i like eating big apples"
start_ = s.find("eating")
end_ = s.find("apples") + len("apples")
s[start_:end_] # 'eating big apples'
Using find() to find the starting indices of the desired word in the string, and then adjust the start_/end_ to your needs.
To remove the sub string:
s[:start_] + s[end_:] # i like
And for:
s = "i like eating apples very much"
end_ = s.find("apples") + len("apples")
start_ = s.find("eating")
s[:start_] + s[end_:] # 'i like very much'

maybe you can use this:
txt = "Hello, welcome to my world."
x = txt.find("welcome")
print(x)
Which outputs: 7
To find "eating" and "apple"

S = "i like eating big apples"
Index = S.find("eating")
output = S[Index:-1]

Use find() or rfind() method for searching substring's occurrence indices, then paste method's result into slice:
s = "i like eating big apples"
substr = s[s.rfind("eating"):s.rfind("apples")]

You can use str.partition to split string into three parts.
In [112]: s = "i like eating apples very much"
In [113]: h, _, t = s.partition('eating')
In [114]: _, _, t = t.partition('apples')
In [115]: h + t
Out[115]: 'i like very much'
In [116]: s = "i like eating big apples"
In [117]: h, _, t = s.partition('eating')
In [118]: _, _, t = t.partition('apples')
In [119]: h + t
Out[119]: 'i like '

fixing words with spaces using a dictionary look up in python?

I have extracted the list of sentences from a document. I am pre-processing this list of sentences to make it more sensible. I am faced with the following problem
I have sentences such as "more recen t ly the develop ment, wh ich is a po ten t "
I would like to correct such sentences using a look up dictionary? to remove the unwanted spaces.
The final output should be "more recently the development, which is a potent "
I would assume that this is a straight forward task in preprocessing text? I need help with some pointers to look for such approaches. Thanks.

Take a look at word or text segmentation. The problem is to find the most probable split of a string into a group of words. Example:
thequickbrownfoxjumpsoverthelazydog
The most probable segmentation should be of course:
the quick brown fox jumps over the lazy dog
Here's an article including prototypical source code for the problem using Google Ngram corpus:
http://jeremykun.com/2012/01/15/word-segmentation/
The key for this algorithm to work is access to knowledge about the world, in this case word frequencies in some language. I implemented a version of the algorithm described in the article here:
https://gist.github.com/miku/7279824
Example usage:
$ python segmentation.py t hequi ckbrownfoxjum ped
thequickbrownfoxjumped
['the', 'quick', 'brown', 'fox', 'jumped']
Using data, even this can be reordered:
$ python segmentation.py lmaoro fll olwt f pwned
lmaorofllolwtfpwned
['lmao', 'rofl', 'lol', 'wtf', 'pwned']
Note that the algorithm is quite slow - it's prototypical.
Another approach using NLTK:
http://web.archive.org/web/20160123234612/http://www.winwaed.com:80/blog/2012/03/13/segmenting-words-and-sentences/
As for your problem, you could just concatenate all string parts you have to get a single string and the run a segmentation algorithm on it.

Your goal is to improve text, not necessarily to make it perfect; so the approach you outline makes sense in my opinion. I would keep it simple and use a "greedy" approach: Start with the first fragment and stick pieces to it as long as the result is in the dictionary; if the result is not, spit out what you have so far and start over with the next fragment. Yes, occasionally you'll make a mistake with cases like the me thod, so if you'll be using this a lot, you could look for something more sophisticated. However, it's probably good enough.
Mainly what you require is a large dictionary. If you'll be using it a lot, I would encode it as a "prefix tree" (a.k.a. trie), so that you can quickly find out if a fragment is the start of a real word. The nltk provides a Trie implementation.
Since this kind of spurious word breaks are inconsistent, I would also extend my dictionary with words already processed in the current document; you may have seen the complete word earlier, but now it's broken up.

--Solution 1:
Lets think of these chunks in your sentence as beads on an abacus, with each bead consisting of a partial string, the beads can be moved left or right to generate the permutations. The position of each fragment is fixed between two adjacent fragments.
In current case, the beads would be :
(more)(recen)(t)(ly)(the)(develop)(ment,)(wh)(ich)(is)(a)(po)(ten)(t)
This solves 2 subproblems:
a) Bead is a single unit,so We do not care about permutations within the bead i.e. permutations of "more" are not possible.
b) The order of the beads is constant, only the spacing between them changes. i.e. "more" will always be before "recen" and so on.
Now, generate all the permutations of these beads , which will give output like :
morerecentlythedevelopment,which is a potent
morerecentlythedevelopment,which is a poten t
morerecentlythedevelop ment, wh ich is a po tent
morerecentlythedevelop ment, wh ich is a po ten t
morerecentlythe development,whichisapotent
Then score these permutations based on how many words from your relevant dictionary they contain, most correct results can be easily filtered out.
more recently the development, which is a potent will score higher than morerecentlythedevelop ment, wh ich is a po ten t
Code which does the permutation part of the beads:
import re
def gen_abacus_perms(frags):
if len(frags) == 0:
return []
if len(frags) == 1:
return [frags[0]]
prefix_1 = "{0}{1}".format(frags[0],frags[1])
prefix_2 = "{0} {1}".format(frags[0],frags[1])
if len(frags) == 2:
nres = [prefix_1,prefix_2]
return nres
rem_perms = gen_abacus_perms(frags[2:])
res = ["{0}{1}".format(prefix_1, x ) for x in rem_perms] + ["{0} {1}".format(prefix_1, x ) for x in rem_perms] + \
["{0}{1}".format(prefix_2, x ) for x in rem_perms] + ["{0} {1}".format(prefix_2 , x ) for x in rem_perms]
return res
broken = "more recen t ly the develop ment, wh ich is a po ten t"
frags = re.split("\s+",broken)
perms = gen_abacus_perms(frags)
print("\n".join(perms))
demo:http://ideone.com/pt4PSt
--Solution#2:
I would suggest an alternate approach which makes use of text analysis intelligence already developed by folks working on similar problems and having worked on big corpus of data which depends on dictionary and grammar .e.g. search engines.
I am not well aware of such public/paid apis, so my example is based on google results.
Lets try to use google :
You can keep putting your invalid terms to Google, for multiple passes, and keep evaluating the results for some score based on your lookup dictionary.
here are two relevant outputs by using 2 passes of your text :
This outout is used for a second pass :
Which gives you the conversion as ""more recently the development, which is a potent".
To verify the conversion, you will have to use some similarity algorithm and scoring to filter out invalid / not so good results.
One raw technique could be using a comparison of normalized strings using difflib.
>>> import difflib
>>> import re
>>> input = "more recen t ly the develop ment, wh ich is a po ten t "
>>> output = "more recently the development, which is a potent "
>>> input_norm = re.sub(r'\W+', '', input).lower()
>>> output_norm = re.sub(r'\W+', '', output).lower()
>>> input_norm
'morerecentlythedevelopmentwhichisapotent'
>>> output_norm
'morerecentlythedevelopmentwhichisapotent'
>>> difflib.SequenceMatcher(None,input_norm,output_norm).ratio()
1.0

I would recommend stripping away the spaces and looking for dictionary words to break it down into. There are a few things you can do to make it more accurate. To make it get the first word in text with no spaces, try taking the entire string, and going through dictionary words from a file (you can download several such files from http://wordlist.sourceforge.net/), the longest ones first, than taking off letters from the end of the string you want to segment. If you want it to work on a big string, you can make it automatically take off letters from the back so that the string you are looking for the first word in is only as long as the longest dictionary word. This should result in you finding the longest words, and making it less likely to do something like classify "asynchronous" as "a synchronous". Here is an example that uses raw input to take in the text to correct and a dictionary file called dictionary.txt:
dict = open("dictionary.txt",'r') #loads a file with a list of words to break string up into
words = raw_input("enter text to correct spaces on: ")
words = words.strip() #strips away spaces
spaced = [] #this is the list of newly broken up words
parsing = True #this represents when the while loop can end
while parsing:
if len(words) == 0: #checks if all of the text has been broken into words, if it has been it will end the while loop
parsing = False
iterating = True
for iteration in range(45): #goes through each of the possible word lengths, starting from the biggest
if iterating == False:
break
word = words[:45-iteration] #each iteration, the word has one letter removed from the back, starting with the longest possible number of letters, 45
for line in dict:
line = line[:-1] #this deletes the last character of the dictionary word, which will be a newline. delete this line of code if it is not a newline, or change it to [1:] if the newline character is at the beginning
if line == word: #this finds if this is the word we are looking for
spaced.append(word)
words = words[-(len(word)):] #takes away the word from the text list
iterating = False
break
print ' '.join(spaced) #prints the output
If you want it to be even more accurate, you could try using a natural language parsing program, there are several available for python free online.

Here's something really basic:
chunks = []
for chunk in my_str.split():
chunks.append(chunk)
joined = ''.join(chunks)
if is_word(joined):
print joined,
del chunks[:]
# deal with left overs
if chunks:
print ''.join(chunks)
I assume you have a set of valid words somewhere that can be used to implement is_word. You also have to make sure it deals with punctuation. Here's one way to do that:
def is_word(wd):
if not wd:
return False
# Strip of trailing punctuation. There might be stuff in front
# that you want to strip too, such as open parentheses; this is
# just to give the idea, not a complete solution.
if wd[-1] in ',.!?;:':
wd = wd[:-1]
return wd in valid_words

You can iterate through a dictionary of words to find the best fit. Adding the words together when a match is not found.
def iterate(word,dictionary):
for word in dictionary:
if words in possibleWord:
finished_sentence.append(words)
added = True
else:
added = False
return [added,finished_sentence]
sentence = "more recen t ly the develop ment, wh ich is a po ten t "
finished_sentence = ""
sentence = sentence.split()
for word in sentence:
added,new_word = interate(word,dictionary)
while True:
if added == False:
word += possible[sentence.find(possibleWord)]
iterate(word,dictionary)
else:
break
finished_sentence.append(word)
This should work. For the variable dictionary, download a txt file of every single english word, then open it in your program.

my index.py file be like
from wordsegment import load, segment
load()
print(segment('morerecentlythedevelopmentwhichisapotent'))
my index.php file be like
<html>
<head>
<title>py script</title>
</head>
<body>
<h1>Hey There!Python Working Successfully In A PHP Page.</h1>
<?php
$python = `python index.py`;
echo $python;
?>
</body>
</html>
Hope this will work

Discovering Poetic Form with NLTK and CMU Dict

Edit: This code has been worked on and released as a basic module: https://github.com/hyperreality/Poetry-Tools
I'm a linguist who has recently picked up python and I'm working on a project which hopes to automatically analyze poems, including detecting the form of the poem. I.e. if it found a 10 syllable line with 0101010101 stress pattern, it would declare that it's iambic pentameter. A poem with 5-7-5 syllable pattern would be a haiku.
I'm using the following code, part of a larger script, but I have a number of problems which are listed below the program:
corpus in the script is simply the raw text input of the poem.
import sys, getopt, nltk, re, string
from nltk.tokenize import RegexpTokenizer
from nltk.util import bigrams, trigrams
from nltk.corpus import cmudict
from curses.ascii import isdigit
...
def cmuform():
tokens = [word for sent in nltk.sent_tokenize(corpus) for word in nltk.word_tokenize(sent)]
d = cmudict.dict()
text = nltk.Text(tokens)
words = [w.lower() for w in text]
regexp = "[A-Za-z]+"
exp = re.compile(regexp)
def nsyl(word):
lowercase = word.lower()
if lowercase not in d:
return 0
else:
first = [' '.join([str(c) for c in lst]) for lst in max(d[lowercase])]
second = ''.join(first)
third = ''.join([i for i in second if i.isdigit()]).replace('2', '1')
return third
#return max([len([y for y in x if isdigit(y[-1])]) for x in d[lowercase]])
sum1 = 0
for a in words:
if exp.match(a):
print a,nsyl(a),
sum1 = sum1 + len(str(nsyl(a)))
print "\nTotal syllables:",sum1
I guess that the output that I want would be like this:
1101111101
0101111001
1101010111
The first problem is that I lost the line breaks during the tokenization, and I really need the line breaks to be able to identify form. This should not be too hard to deal with though. The bigger problems are that:
I can't deal with non-dictionary words. At the moment I return 0 for them, but this will confound any attempt to identify the poem, as the syllabic count of the line will probably decrease.
In addition, the CMU dictionary often says that there is stress on a word - '1' - when there is not - '0 - . Which is why the output looks like this: 1101111101, when it should be the stress of iambic pentameter: 0101010101
So how would I add some fudging factor so the poem still gets identified as iambic pentameter when it only approximates the pattern? It's no good to code a function that identifies lines of 01's when the CMU dictionary is not going to output such a clean result. I suppose I'm asking how to code a 'partial match' algorithm.

Welcome to stack overflow. I'm not that familiar with Python, but I see you have not received many answers yet so I'll try to help you with your queries.
First some advice: You'll find that if you focus your questions your chances of getting answers are greatly improved. Your post is too long and contains several different questions, so it is beyond the "attention span" of most people answering questions here.
Back on topic:
Before you revised your question you asked how to make it less messy. That's a big question, but you might want to use the top-down procedural approach and break your code into functional units:
split corpus into lines
For each line: find the syllable length and stress pattern.
Classify stress patterns.
You'll find that the first step is a single function call in python:
corpus.split("\n");
and can remain in the main function but the second step would be better placed in its own function and the third step would require to be split up itself, and would probably be better tackled with an object oriented approach. If you're in academy you might be able to convince the CS faculty to lend you a post-grad for a couple of months and help you instead of some workshop requirement.
Now to your other questions:
Not loosing line breaks: as #ykaganovich mentioned, you probably want to split the corpus into lines and feed those to the tokenizer.
Words not in dictionary/errors: The CMU dictionary home page says:
Find an error? Please contact the developers. We will look at the problem and improve the dictionary. (See at bottom for contact information.)
There is probably a way to add custom words to the dictionary / change existing ones, look in their site, or contact the dictionary maintainers directly.
You can also ask here in a separate question if you can't figure it out. There's bound to be someone in stackoverflow that knows the answer or can point you to the correct resource.
Whatever you decide, you'll want to contact the maintainers and offer them any extra words and corrections anyway to improve the dictionary.
Classifying input corpus when it doesn't exactly match the pattern: You might want to look at the link ykaganovich provided for fuzzy string comparisons. Some algorithms to look for:
Levenshtein distance: gives you a measure of how different two strings are as the number of changes needed to turn one string into another. Pros: easy to implement, Cons: not normalized, a score of 2 means a good match for a pattern of length 20 but a bad match for a pattern of length 3.
Jaro-Winkler string similarity measure: similar to Levenshtein, but based on how many character sequences appear in the same order in both strings. It is a bit harder to implement but gives you normalized values (0.0 - completely different, 1.0 - the same) and is suitable for classifying the stress patterns. A CS postgrad or last year undergrad should not have too much trouble with it ( hint hint ).
I think those were all your questions. Hope this helps a bit.

To preserve newlines, parse line by line before sending each line to the cmu parser.
For dealing with single-syllable words, you probably want to try both 0 and 1 for it when nltk returns 1 (looks like nltk already returns 0 for some words that would never get stressed, like "the"). So, you'll end up with multiple permutations:
1101111101
0101010101
1101010101
and so forth. Then you have to pick ones that look like a known forms.
For non-dictionary words, I'd also fudge it the same way: figure out the number of syllables (the dumbest way would be by counting the vowels), and permutate all possible stresses. Maybe add some more rules like "ea is a single syllable, trailing e is silent"...
I've never worked with other kinds of fuzzying, but you can check https://stackoverflow.com/questions/682367/good-python-modules-for-fuzzy-string-comparison for some ideas.

This is my first post on stackoverflow.
And I'm a python newbie, so please excuse any deficits in code style.
But I too am attempting to extract accurate metre from poems.
And the code included in this question helped me, so I post what I came up with that builds on that foundation. It is one way to extract the stress as a single string, correct with a 'fudging factor' for the cmudict bias, and not lose words that are not in the cmudict.
import nltk
from nltk.corpus import cmudict
prondict = cmudict.dict()
#
# parseStressOfLine(line)
# function that takes a line
# parses it for stress
# corrects the cmudict bias toward 1
# and returns two strings
#
# 'stress' in form '0101*,*110110'
# -- 'stress' also returns words not in cmudict '0101*,*1*zeon*10110'
# 'stress_no_punct' in form '0101110110'
def parseStressOfLine(line):
stress=""
stress_no_punct=""
print line
tokens = [words.lower() for words in nltk.word_tokenize(line)]
for word in tokens:
word_punct = strip_punctuation_stressed(word.lower())
word = word_punct['word']
punct = word_punct['punct']
#print word
if word not in prondict:
# if word is not in dictionary
# add it to the string that includes punctuation
stress= stress+"*"+word+"*"
else:
zero_bool=True
for s in prondict[word]:
# oppose the cmudict bias toward 1
# search for a zero in array returned from prondict
# if it exists use it
# print strip_letters(s),word
if strip_letters(s)=="0":
stress = stress + "0"
stress_no_punct = stress_no_punct + "0"
zero_bool=False
break
if zero_bool:
stress = stress + strip_letters(prondict[word][0])
stress_no_punct=stress_no_punct + strip_letters(prondict[word][0])
if len(punct)>0:
stress= stress+"*"+punct+"*"
return {'stress':stress,'stress_no_punct':stress_no_punct}
# STRIP PUNCTUATION but keep it
def strip_punctuation_stressed(word):
# define punctuations
punctuations = '!()-[]{};:"\,<>./?##$%^&*_~'
my_str = word
# remove punctuations from the string
no_punct = ""
punct=""
for char in my_str:
if char not in punctuations:
no_punct = no_punct + char
else:
punct = punct+char
return {'word':no_punct,'punct':punct}
# CONVERT the cmudict prondict into just numbers
def strip_letters(ls):
#print "strip_letters"
nm = ''
for ws in ls:
#print "ws",ws
for ch in list(ws):
#print "ch",ch
if ch.isdigit():
nm=nm+ch
#print "ad to nm",nm, type(nm)
return nm
# TESTING results
# i do not correct for the '2'
line = "This day (the year I dare not tell)"
print parseStressOfLine(line)
line = "Apollo play'd the midwife's part;"
print parseStressOfLine(line)
line = "Into the world Corinna fell,"
print parseStressOfLine(line)
"""
OUTPUT
This day (the year I dare not tell)
{'stress': '01***(*011111***)*', 'stress_no_punct': '01011111'}
Apollo play'd the midwife's part;
{'stress': "0101*'d*01211***;*", 'stress_no_punct': '010101211'}
Into the world Corinna fell,
{'stress': '01012101*,*', 'stress_no_punct': '01012101'}

Error: match word in file

There are two sentences in "test_tweet1.txt"
#francesco_con40 2nd worst QB. DEFINITELY Tony Romo. The man who likes to share the ball with everyone. Including the other team.
#mariakaykay aga tayo tomorrow ah. :) Good night, Ces. Love you! >:D<
In "Personal.txt"
The Game (rapper)
The Notorious B.I.G.
The Undertaker
Thor
Tiësto
Timbaland
T.I.
Tom Cruise
Tony Romo
Trajan
Triple H
My codes:
import re
popular_person = open('C:/Users/Personal.txt')
rpopular_person = popular_person.read()
file1 = open("C:/Users/test_tweet1.txt").readlines()
array = []
count1 = 0
for line in file1:
array.append(line)
count1 = count1 + 1
print "\n",count1, line
ltext1 = line.split(" ")
for i,text in enumerate(ltext1):
if text in rpopular_person:
print text
text2 = ' '.join(ltext1)
Results from the codes showed:
1 #francesco_con40 2nd worst QB. DEFINITELY Tony Romo. The man who likes to share the ball with everyone. Including the other team.
Tony
The
man
to
the
the
2 #mariakaykay aga tayo tomorrow ah. :) Good night, Ces. Love you! >:D<
aga
I tried to match word from "test_tweet1.txt" with "Personal.txt".
Expected result:
Tony
Romo
Any suggestion?

Your problem seems to be that rpopular_person is just a single string. Therefore, when you ask things like 'to' in rpopular_person, you get a value of True, because the characters 't', 'o' appear in sequence. I am assuming that the same goes for 'the' elsewhere in Personal.txt.
What you want to do is split up Personal.txt into individual words, the way you're splitting your tweets. You can also make the resulting list of words into a set, since that'll make your lookup much faster. Something like this:
people = set(popular_person.read().split())
It's also worth noticing that I'm calling split(), with no arguments. This splits on all whitespace--newlines, tabs, and so on. This way you get everything separately like you intend. Or, if you don't actually want this (since this will give you results of "The" all the time based on your edited contents of Personal.txt), make it:
people = set(popular_person.read().split('\n'))
This way you split on newlines, so you only look for full name matches.
You're not getting "Romo" because that's not a word in your tweet. The word in your tweet is "Romo." with a period. This is very likely to remain a problem for you, so what I would do is rearrange your logic (assuming speed isn't an issue). Rather than looking at each word in your tweet, look at each name in your Personal.txt file, and see if it's in your full tweet. This way you don't have to deal with punctuation and so on. Here's how I'd rewrite your functionality:
rpopular_person = set(personal.split())
with open("Personal.txt") as p:
people = p.read().split('\n') # Get full names rather than partial names
with open("test_tweet1.txt") as tweets:
for tweet in tweets:
for person in people:
if person in tweet:
print person

you need to split rpopular_person to get it to match words instead of substrings
rpopular_person = open('C:/Users/Personal.txt').read().split()
this gives:
Tony
The
the reason Romo isn't showing up is that on your line split you have "Romo." Maybe you should look for rpopular_person in the lines, instead of the other way around. Maybe something like this
popular_person = open('C:/Users/Personal.txt').read().split("\n")
file1 = open("C:/Users/test_tweet1.txt")
array = []
for count1, line in enumerate(file1):
print "\n", count1, line
for person in popular_person:
if person in line:
print person

How to save a regular expression user input value (Python)

I am making a simple chat bot in Python. It has a text file with regular expressions which help to generate the output. The user input and the bot output are separated by a | character.
my name is (?P<'name'>\w*) | Hi {'name'}!
This works fine for single sets of input and output responses, however I would like the bot to be able to store the regex values the user inputs and then use them again (i.e. give the bot a 'memory'). For example, I would like to have the bot store the value input for 'name', so that I can have this in the rules:
my name is (?P<'word'>\w*) | You said your name is {'name'} already!
my name is (?P<'name'>\w*) | Hi {'name'}!
Having no value for 'name' yet, the bot will first output 'Hi steve', and once the bot does have this value, the 'word' rule will apply. I'm not sure if this is easily feasible given the way I have structured my program. I have made it so that the text file is made into a dictionary with the key and value separated by the | character, when the user inputs some text, the program compares whether the user input matches the input stored in the dictionary, and prints out the corresponding bot response (there is also an 'else' case if no match is found).
I must need something to happen at the comparing part of the process so that the user's regular expression text is saved and then substituted back into the dictionary somehow. All of my regular expressions have different names associated with them (there are no two instances of 'word', for example...there is 'word', 'word2', etc), I did this as I thought it would make this part of the process easier. I may have structured the thing completely wrong to do this task though.
Edit: code
import re
io = {}
with open("rules.txt") as brain:
for line in brain:
key, value = line.split('|')
io[key] = value
string = str(raw_input('> ')).lower()+' word'
x = 1
while x == 1:
for regex, output in io.items():
match = re.match(regex, string)
if match:
print(output.format(**match.groupdict()))
string = str(raw_input('> ')).lower()+' word'
else:
print ' Sorry?'
string = str(raw_input('> ')).lower()+' word'

I had some difficulty to understand the principle of your algorithm because I'm not used to employ the named groups.
The following code is the way I would solve your problem, I hope it will give you some ideas.
I think that having only one dictionary isn't a good principle, it increases the complexity of reasoning and of the algorithm. So I based the code on two dictionaries: direg and memory
Theses two dictionaries have keys that are indexes of groups, not all the indexes, only some particular ones, the indexes of the groups being the last in each individual patterns.
Because, for the fun, I decided that the regexes must be able to have several groups.
What I call individual patterns in my code are the following strings:
"[mM]y name [Ii][sS] (\w*)"
"[Ii]n repertory (\w*) I [wW][aA][nN][tT] file (\w*)"
"[Ii] [wW][aA][nN][tT] to ([ \w]*)"
You see that the second individual pattern has 2 capturing groups: consequently there are 3 individual patterns, but a total of 4 groups in all the individual groups.
So the creation of the dictionaries needs some additional care to take account of the fact that the index of the last matching group ( which I use with help of the attribute of name lastindex of a regex MatchObject ) may not correspond to the numbering of individual regexes present in the regex pattern: it's harder to explain than to understand. That's the reason why I count in the function distr() the occurences of strings {0} {1} {2} {3} {4} etc whose number MUST be the same as the number of groups defined in the corresponding individual pattern.
I found the suggestion of Laurence D'Oliveiro to use '||' instead of '|' as separator interesting.
My code simulates a session in which several inputs are done:
import re
regi = ("[mM]y name [Ii][sS] (\w*)"
"||Hi {0}!"
"||You said that your name was {0} !!!",
"[Ii]n repertory (\w*) I [wW][aA][nN][tT] file (\w*)"
"||OK here's your file {0}\\{1} :"
"||I already gave you the file {0}\\{1} !",
"[Ii] [wW][aA][nN][tT] to ([ \w]*)"
"||OK, I will do {0}"
"||You already did {0}. Do yo really want again ?")
direg = {}
memory = {}
def distr(regi,cnt = 0,di = direg,mem = memory,
regnb = re.compile('{\d+}')):
for i,el in enumerate(regi,start=1):
sp = el.split('||')
cnt += len(regnb.findall(sp[1]))
di[cnt] = sp[1]
mem[cnt] = sp[2]
yield sp[0]
regx = re.compile('|'.join(distr(regi)))
print 'direg :\n',direg
print
print 'memory :\n',memory
for inp in ('I say that my name is Armano the 1st',
'In repertory ONE I want file SPACE',
'I want to record music',
'In repertory ONE I want file SPACE',
'I say that my name is Armstrong',
'But my name IS Armstrong now !!!',
'In repertory TWO I want file EARTH',
'Now my name is Helena'):
print '\ninput ==',inp
mat = regx.search(inp)
if direg[mat.lastindex]:
print 'output ==',direg[mat.lastindex]\
.format(*(d for d in mat.groups() if d))
direg[mat.lastindex] = None
memory[mat.lastindex] = memory[mat.lastindex]\
.format(*(d for d in mat.groups() if d))
else:
print 'output ==',memory[mat.lastindex]\
.format(*(d for d in mat.groups() if d))
if not memory[mat.lastindex].startswith('Sorry'):
memory[mat.lastindex] = 'Sorry, ' \
+ memory[mat.lastindex][0].lower()\
+ memory[mat.lastindex][1:]
result
direg :
{1: 'Hi {0}!', 3: "OK here's your file {0}\\{1} :", 4: 'OK, I will do {0}'}
memory :
{1: 'You said that your name was {0} !!!', 3: 'I already gave you the file {0}\\{1} !', 4: 'You already did {0}. Do yo really want again ?'}
input == I say that my name is Armano the 1st
output == Hi Armano!
input == In repertory ONE I want file SPACE
output == OK here's your file ONE\SPACE :
input == I want to record music
output == OK, I will do record music
input == In repertory ONE I want file SPACE
output == I already gave you the file ONE\SPACE !
input == I say that my name is Armstrong
output == You said that your name was Armano !!!
input == But my name IS Armstrong now !!!
output == Sorry, you said that your name was Armano !!!
input == In repertory TWO I want file EARTH
output == Sorry, i already gave you the file ONE\SPACE !
input == Now my name is Helena
output == Sorry, you said that your name was Armano !!!

OK, let me see if I understand this:
You want to a dictionary of key-value pairs. This will be the “memory” of the chatbot.
You want to apply regular-expression rules to user input. But which rules might apply is conditional on which keys are already present in the memory dictionary: if “name” is not yet defined, then the rule that defines “name” applies; but if it is, then the rule that mentions “word” applies.
Seems to me you need more information attached to your rules. For example, the “word” rule you gave above shouldn’t actually add “word” to the dictionary, otherwise it would only apply once (imagine if the user keeps trying to say “my name is x” more than twice).
Does that give you a bit more idea about how to proceed?
Oh, by the way, I think “|” is a poor choice for a separator character, because it can occur in regular expressions. Not sure what to suggest: how about “||”?