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Replace only specified occurrences of string - Python Regex

I am playing around with re module in python, what i am stuck at is I want to replace only specified occurrences of the string.
for example
import re
string = "aabbaabbaabbabbaabbaa"
#I want to replace only 3rd time 'bb' appeared in the string with white space
string = re.sub("bb"," ",string,3) #if iI do this all first 3 occurrences got replaced
print(string)
output
aa aa aa aabbaabbaa
any idea how to to replace only 3rd occurrence
so the output would look like This
aabbaabbaa aabbaabbaa

This may not be the perfect way but it is a solution:
string = re.sub('bb',' ',string, 3)
string = re.sub(' ','bb',string,2)
This is just an alternative solution I can think of.

Modify the regex so it only matches the third occurrence?
re.sub(r'(.*?bb.*?bb.*?)bb', r'\1 ', string, 1)
This could be extended to a large number of repetitions like r'(.*?(bb.*?){9999})bb'

Trouble using regex to locate special characters

I am using beautifulsoup and selenium to collect some data from a page. After narrowing the data down the string that I want, it gives me 'First Blood○○○○○●○○○○'. My goal is to determine the position of the filled in dot (so 5 in this case if we are counting from 0).
I started by trying to remove all of the non-special characters using:
test = re.sub(r'[a-z]+', '', collectStatistics[5], re.I)
Which gave me 'F B○○○○○●○○○○' so I am guessing F B are also special characters. I have no clue how to go about writing a regex that will detect the filled in circle so any advice would be appreciated.
Thanks in advance :)

I think regexes (regices?) are overkill here.
First, cut off everything after the filled dot:
line = line.split('●')[0] # Split on filled dots, then take only the first part
Now, count the empty dots:
result = line.count('○') # Count occurrences

It founds F and B because your regex finds lowercase letters.If you want to find all of them change regex to [a-zA-Z]+
import re
collectStatistics = "First Blood○○○○○●○○○○"
test = re.sub(r'[a-zA-Z]+', '', collectStatistics,re.I)
print (test)
OUTPUT :
○○○○○●○○○○

Is it possible to search and replace a string with "any" characters?

There are probably several ways to solve this problem, so I'm open to any ideas.
I have a file, within that file is the string "D133330593" Note: I do have the exact position within the file this string exists, but I don't know if that helps.
Following this string, there are 6 digits, I need to replace these 6 digits with 6 other digits.
This is what I have so far:
def editfile():
f = open(filein,'r')
filedata = f.read()
f.close()
#This is the line that needs help
newdata = filedata.replace( -TOREPLACE- ,-REPLACER-)
#Basically what I need is something that lets me say "D133330593******"
#->"D133330593123456" Note: The following 6 digits don't need to be
#anything specific, just different from the original 6
f = open(filein,'w')
f.write(newdata)
f.close()

Use the re module to define your pattern and then use the sub() function to substitute occurrence of that pattern with your own string.
import re
...
pat = re.compile(r"D133330593\d{6}")
re.sub(pat, "D133330593abcdef", filedata)
The above defines a pattern as -- your string ("D133330593") followed by six decimal digits. Then the next line replaces ALL occurrences of this pattern with your replacement string ("abcdef" in this case), if that is what you want.
If you want a unique replacement string for each occurrence of pattern, then you could use the count keyword argument in the sub() function, which allows you to specify the number of times the replacement must be done.
Check out this library for more info - https://docs.python.org/3.6/library/re.html

Let's simplify your problem to you having a string:
s = "zshisjD133330593090909fdjgsl"
and you wanting to replace the 6 characters after "D133330593" with "123456" to produce:
"zshisjD133330594123456fdjgsl"
To achieve this, we can first need to find the index of "D133330593". This is done by just using str.index:
i = s.index("D133330593")
Then replace the next 6 characters, but for this, we should first calculate the length of our string that we want to replace:
l = len("D133330593")
then do the replace:
s[:i+l] + "123456" + s[i+l+6:]
which gives us the desired result of:
'zshisjD133330593123456fdjgsl'
I am sure that you can now integrate this into your code to work with a file, but this is how you can do the heart of your problem .
Note that using variables as above is the right thing to do as it is the most efficient compared to calculating them on the go. Nevertheless, if your file isn't too long (i.e. efficiency isn't too much of a big deal) you can do the whole process outlined above in one line:
s[:s.index("D133330593")+len("D133330593")] + "123456" + s[s.index("D133330593")+len("D133330593")+6:]
which gives the same result.

How to find and count emoticons in a string using python?

This topic has been addressed for text based emoticons at link1, link2, link3. However, I would like to do something slightly different than matching simple emoticons. I'm sorting through tweets that contain the emoticons' icons. The following unicode information contains just such emoticons: pdf.
Using a string with english words that also contains any of these emoticons from the pdf, I would like to be able to compare the number of emoticons to the number of words.
The direction that I was heading down doesn't seem to be the best option and I was looking for some help. As you can see in the script below, I was just planning to do the work from the command line:
$cat <file containing the strings with emoticons> | ./emo.py
emo.py psuedo script:
import re
import sys
for row in sys.stdin:
print row.decode('utf-8').encode("ascii","replace")
#insert regex to find the emoticons
if match:
#do some counting using .split(" ")
#print the counting
The problem that I'm running into is the decoding/encoding. I haven't found a good option for how to encode/decode the string so I can correctly find the icons. An example of the string that I want to search to find the number of words and emoticons is as follows:
"Smiley emoticon rocks! I like you."
The challenge: can you make a script that counts the number of words and emoticons in this string? Notice that the emoticons are both sitting next to the words with no space in between.

First, there is no need to encode here at all. You're got a Unicode string, and the re engine can handle Unicode, so just use it.
A character class can include a range of characters, by specifying the first and last with a hyphen in between. And you can specify Unicode characters that you don't know how to type with \U escape sequences. So:
import re
s=u"Smiley emoticon rocks!\U0001f600 I like you.\U0001f601"
count = len(re.findall(ru'[\U0001f600-\U0001f650]', s))
Or, if the string is big enough that building up the whole findall list seems wasteful:
emoticons = re.finditer(ru'[\U0001f600-\U0001f650]', s)
count = sum(1 for _ in emoticons)
Counting words, you can do separately:
wordcount = len(s.split())
If you want to do it all at once, you can use an alternation group:
word_and_emoticon_count = len(re.findall(ru'\w+|[\U0001f600-\U0001f650]', s))
As #strangefeatures points out, Python versions before 3.3 allowed "narrow Unicode" builds. And, for example, most CPython Windows builds are narrow. In narrow builds, characters can only be in the range U+0000 to U+FFFF. There's no way to search for these characters, but that's OK, because they're don't exist to search for; you can just assume they don't exist if you get an "invalid range" error compiling the regexp.
Except, of course, that there's a good chance that wherever you're getting your actual strings from, they're UTF-16-BE or UTF-16-LE, so the characters do exist, they're just encoded into surrogate pairs. And you want to match those surrogate pairs, right? So you need to translate your search into a surrogate-pair search. That is, convert your high and low code points into surrogate pair code units, then (in Python terms) search for:
(lead == low_lead and lead != high_lead and low_trail <= trail <= DFFF or
lead == high_lead and lead != low_lead and DC00 <= trail <= high_trail or
low_lead < lead < high_lead and DC00 <= trail <= DFFF)
You can leave off the second condition in the last case if you're not worried about accepting bogus UTF-16.
If it's not obvious how that translates into regexp, here's an example for the range [\U0001e050-\U0001fbbf] in UTF-16-BE:
(\ud838[\udc50-\udfff])|([\ud839-\ud83d].)|(\ud83e[\udc00-\udfbf])
Of course if your range is small enough that low_lead == high_lead this gets simpler. For example, the original question's range can be searched with:
\ud83d[\ude00-\ude50]
One last trick, if you don't actually know whether you're going to get UTF-16-LE or UTF-16-BE (and the BOM is far away from the data you're searching): Because no surrogate lead or trail code unit is valid as a standalone character or as the other end of a pair, you can just search in both directions:
(\ud838[\udc50-\udfff])|([\ud839-\ud83d][\udc00-\udfff])|(\ud83e[\udc00-\udfbf])|
([\udc50-\udfff]\ud838)|([\udc00-\udfff][\ud839-\ud83d])|([\udc00-\udfbf]\ud83e)

My solution includes the emoji and regex modules. The regex module supports recognizing grapheme clusters (sequences of Unicode codepoints rendered as a single character), so we can count emojis like 👨‍👩‍👦‍👦 once, although it consists of 4 emojis.
import emoji
import regex
def split_count(text):
emoji_counter = 0
data = regex.findall(r'\X', text)
for word in data:
if any(char in emoji.UNICODE_EMOJI for char in word):
emoji_counter += 1
# Remove from the given text the emojis
text = text.replace(word, '')
words_counter = len(text.split())
return emoji_counter, words_counter
Testing:
line = "hello 👩🏾‍🎓 emoji hello 👨‍👩‍👦‍👦 how are 😊 you today🙅🏽🙅🏽"
counter = split_count(line)
print("Number of emojis - {}, number of words - {}".format(counter[0], counter[1]))
Output:
Number of emojis - 5, number of words - 7

If you are trying to read unicode characters outside the ascii range, don't convert into the ascii range. Just leave it as unicode and work from there (untested):
import sys
count = 0
emoticons = set(range(int('1f600',16), int('1f650', 16)))
for row in sys.stdin:
for char in row:
if ord(char) in emoticons:
count += 1
print "%d emoticons found" % count
Not the best solution, but it should work.

This is my solution using re:
import re
text = "your text with emojis"
em_count = len(re.findall(r'[^\w\s,.]', text))
print(em_count)

python: regular expressions, how to match a string of undefind length which has a structure and finishes with a specific group

I need to create a regexp to match strings like this 999-123-222-...-22
The string can be finished by &Ns=(any number) or without this... So valid strings for me are
999-123-222-...-22
999-123-222-...-22&Ns=12
999-123-222-...-22&Ns=12
And following are not valid:
999-123-222-...-22&N=1
I have tried testing it several hours already... But did not manage to solve, really need some help

Not sure if you want to literally match 999-123-22-...-22 or if that can be any sequence of numbers/dashes. Here are two different regexes:
/^[\d-]+(&Ns=\d+)?$/
/^999-123-222-\.\.\.-22(&Ns=\d+)?$/
The key idea is the (&Ns=\d+)?$ part, which matches an optional &Ns=<digits>, and is anchored to the end of the string with $.

If you just want to allow strings 999-123-222-...-22 and 999-123-222-...-22&Ns=12 you better use a string function.
If you want to allow any numbers between - you can use the regex:
^(\d+-){3}[.]{3}-\d+(&Ns=\d+)?$
If the numbers must be of only 3 digits and the last number of only 2 digits you can use:
^(\d{3}-){3}[.]{3}-\d{2}(&Ns=\d{2})?$

This looks like a phone number and extension information..
Why not make things simpler for yourself (and anyone who has to read this later) and split the input rather than use a complicated regex?
s = '999-123-222-...-22&Ns=12'
parts = s.split('&Ns=') # splits on Ns and removes it
If the piece before the "&" is a phone number, you could do another split and get the area code etc into separate fields, like so:
phone_parts = parts[0].split('-') # breaks up the digit string and removes the '-'
area_code = phone_parts[0]
The portion found after the the optional '&Ns=' can be checked to see if it is numeric with the string method isdigit, which will return true if all characters in the string are digits and there is at least one character, false otherwise.
if len(parts) > 1:
extra_digits_ok = parts[1].isdigit()