We Keep Coding

Regex to ignore Semicolon

I have one column in a dataframe with key value pairs I would like to extract.
'AF_ESP=0.00546;AF_EXAC=0.00165;AF_TGP=0.00619'
I would like to parse key value pairs like so
('AF_ESP', '0.00546')
('AF_EXAC', '0.00165')
('AF_TGP', '0.00619')
Here is my regex.
([^=]+)=([^;]+)
This gets me most of way there:
('AF_ESP', '0.00546')
(';AF_EXAC', '0.00165')
(';AF_TGP', '0.00619')
How can I adjust it so semicolons are not captured in the result?

You can consume the semi-colon or start of string in front:
(?:;|^)([^=]+)=([^;]+)
See the regex demo. Details:
(?:;|^) - a non-capturing group matching ; or start of string
([^=]+) - Group 1: one or more chars other than =
= - a = char
([^;]+) - Group 2: one or more chars other than ;.
See the Python demo:
import re
text = "AF_ESP=0.00546;AF_EXAC=0.00165;AF_TGP=0.00619"
print( re.findall(r'(?:;|^)([^=]+)=([^;]+)', text) )
# => [('AF_ESP', '0.00546'), ('AF_EXAC', '0.00165'), ('AF_TGP', '0.00619')]
A non-regex solution is also possible:
text = "AF_ESP=0.00546;AF_EXAC=0.00165;AF_TGP=0.00619"
print( [x.split('=') for x in text.split(';')] )
# => [['AF_ESP', '0.00546'], ['AF_EXAC', '0.00165'], ['AF_TGP', '0.00619']]
See this Python demo.

This can be also solved with a split method:
text = "AF_ESP=0.00546;AF_EXAC=0.00165;AF_TGP=0.00619"
print([tuple(i.split('=')) for i in text.split(';')])
output:
[('AF_ESP', '0.00546'), ('AF_EXAC', '0.00165'), ('AF_TGP', '0.00619')]

An alternate and somewhat simpler approach to #Wiktor's solution is, in steps:
Capture everything until the =.
Get the = but don't capture that.
Get everything after the = up until an optional ; if that exists.
This would translate to the following regex:
([^=]+)=([^;]+);?
And in python:
>>> re.findall(r'([^=]+)=([^;]+);?', "AF_ESP=0.00546;AF_EXAC=0.00165;AF_TGP=0.00619")
[('AF_ESP', '0.00546'), ('AF_EXAC', '0.00165'), ('AF_TGP', '0.00619')]

get all occurence of a regex in string python

I am trying to find in the following string TreeModel/Node/Node[1]/Node[4]/Node[1] this :
TreeModel/Node
TreeModel/Node/Node[1]
TreeModel/Node/Node[1]/Node[4]
TreeModel/Node/Node[1]/Node[4]/Node[1]
Using regular expression in python. Here is the code I tried:
string = 'TreeModel/Node/Node[1]/Node[4]/Node[1]'
pattern = r'.+?Node\[[1-9]\]'
print re.findall(pattern=pattern,string=string)
#result : ['TreeModel/Node/Node[1]', '/Node[4]', '/Node[1]']
#expected result : ['TreeModel/Node', 'TreeModel/Node/Node[1]', 'TreeModel/Node/Node[1]/Node[4]', 'TreeModel/Node/Node[1]/Node[4]/Node[1]']

You can use split here:
>>> s = 'TreeModel/Node/Node[1]/Node[4]/Node[1]'
>>> split_s = s.split('/')
>>> ['/'.join(split_s[:i]) for i in range(2, len(split_s)+1)]
['TreeModel/Node',
'TreeModel/Node/Node[1]',
'TreeModel/Node/Node[1]/Node[4]',
'TreeModel/Node/Node[1]/Node[4]/Node[1]']
You can also use regex:
for i in range(2, s.count('/')+2):
s_ = '[^/]+/*'
regex = re.search(r'('+s_*i+')', s).group(0)
print(regex)
TreeModel/Node/
TreeModel/Node/Node[1]/
TreeModel/Node/Node[1]/Node[4]/
TreeModel/Node/Node[1]/Node[4]/Node[1]

I'm not good in Python at all but for regex part with your specific structure of string below regex matches each segment:
/?(?:{[^{}]*})?[^/]+
Where braces and preceding / is optional. It matches a slash mark (if any) then braces with their content (if any) then the rest up to next slash mark.
Python code (see live demo here):
matches = re.findall(r'/?(?:{[^{}]*})?[^/]+', string)
output = ''
for i in range(len(matches)):
output += matches[i];
print(output)

regex python + variable

guys i hope you can give me a hand with this:
Im trying to find a match on a variable value:
net_card is a string
net_card = salida.read()
regex = re.compile('([a-z])\w+' % re.escape(net_card))
if i run this code it show me this error:
regex = re.compile('([a-z])\w+' % re.escape(net_card))
TypeError: not all arguments converted during string formatting
I haven't found a way to solve this, even with scape characters.
now if i do this:
net_card = salida.read()
match = re.search('([a-z])\w+', net_card)
whatIWant = match.group(1) if match else None
print whatIWant
it shows me just (e) in the output even when the value of net_card is NAME=ens32.

Your regex, ([a-z])\w+, will match a single character in the range a-z as the first group, and match the rest of the string as [a-zA-Z0-9_]+. Instead, match the two groups of \w+ (which is [a-zA-Z0-9_]+ in evaluation), separated by an equal sign. Here's an expression:
(\w+)=(\w+)
In practice (if you don't care about "NAME"), you can remove the first group and use:
net_card = salida.read()
match = re.match('\w+=(\w+)', net_card)
print(match.group(1) if match else None)
Which will output ens32.

Splitting a string using re module of python

I have a string
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
#I have to capture only the field 'count_EVENT_GENRE'
field = re.split(r'[(==)(>=)(<=)(in)(like)]', s)[0].strip()
#o/p is 'cou'
# for s = 'sum_EVENT_GENRE in [1,2,3,4,5]' o/p = 'sum_EVENT_GENRE'
which is fine
My doubt is for any character in (in)(like) it is splitting the string s at that character and giving me first slice.(as after "cou" it finds one matching char i:e n). It's happening for any string that contains any character from (in)(like).
Ex : 'percentage_AMOUNT' o/p = 'p'
as it finds a matching char as 'e' after p.
So i want some advice how to treat (in)(like) as words not as characters , when splitting occurs/matters.
please suggest a syntax.

Answering your question, the [(==)(>=)(<=)(in)(like)] is a character class matching single characters you defined inside the class. To match sequences of characters, you need to remove [ and ] and use alternation:
r'==?|>=?|<=?|\b(?:in|like)\b'
or better:
r'[=><]=?|\b(?:in|like)\b'
You code would look like:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
field = re.split(r'[=><]=?|\b(?:in|like)\b', s)[0].strip()
print(field)
However, there might be other (easier, or safer - depending on the actual specifications) ways to get what you want (splitting with space and getting the first item, use re.match with r'\w+' or r'[a-z]+(?:_[A-Z]+)+', etc.)
If your value is at the start of the string and starts with lowercase ASCII letters, and then can have any amount of sequences of _ followed with uppercase ASCII letters, use:
re.match(r'[a-z]+(?:_[A-Z]+)*', s)
Full demo code:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
fieldObj = re.match(r'[a-z]+(?:_[A-Z]+)*', s)
if fieldObj:
print(fieldObj.group())

If you want only the first word of your string, then this should do the job:
import re
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
field = re.split(r'\W', s)[0]
# count_EVENT_GENRE

Is there anything wrong with using split?
>>> s = 'count_EVENT_GENRE in [1,2,3,4,5]'
>>> s.split(' ')[0]
'count_EVENT_GENRE'
>>> s = 'coint_EVENT_GENRE = "ROMANCE"'
>>> s.split(' ')[0]
'coint_EVENT_GENRE'
>>>

Need help extracting data from a file

I'm a newbie at python.
So my file has lines that look like this:
-1 1:-0.294118 2:0.487437 3:0.180328 4:-0.292929 5:-1 6:0.00149028 7:-0.53117 8:-0.0333333
I need help coming up with the correct python code to extract every float preceded by a colon and followed by a space (ex: [-0.294118, 0.487437,etc...])
I've tried dataList = re.findall(':(.\*) ', str(line)) and dataList = re.split(':(.\*) ', str(line)) but these come up with the whole line. I've been researching this problem for a while now so any help would be appreciated. Thanks!

try this one:
:(-?\d\.\d+)\s
In your code that will be
p = re.compile(':(-?\d\.\d+)\s')
m = p.match(str(line))
dataList = m.groups()
This is more specific on what you want.
In your case .* will match everything it can
Test on Regexr.com:
In this case last element wasn't captured because it doesnt have space to follow, if this is a problem just remove the \s from the regex

This will do it:
import re
line = "-1 1:-0.294118 2:0.487437 3:0.180328 4:-0.292929 5:-1 6:0.00149028 7:-0.53117 8:-0.0333333"
for match in re.finditer(r"(-?\d\.\d+)", line, re.DOTALL | re.MULTILINE):
print match.group(1)
Or:
match = re.search(r"(-?\d\.\d+)", line, re.DOTALL | re.MULTILINE)
if match:
datalist = match.group(1)
else:
datalist = ""
Output:
-0.294118
0.487437
0.180328
-0.292929
0.00149028
-0.53117
-0.0333333
Live Python Example:
http://ideone.com/DpiOBq
Regex Demo:
https://regex101.com/r/nR4wK9/3
Regex Explanation
(-?\d\.\d+)
Match the regex below and capture its match into backreference number 1 «(-?\d\.\d+)»
Match the character “-” literally «-?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match a single character that is a “digit” (ASCII 0–9 only) «\d»
Match the character “.” literally «\.»
Match a single character that is a “digit” (ASCII 0–9 only) «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»

Given:
>>> s='-1 1:-0.294118 2:0.487437 3:0.180328 4:-0.292929 5:-1 6:0.00149028 7:-0.53117 8:-0.0333.333'
With your particular data example, you can just grab the parts that would be part of a float with a regex:
>>> re.findall(r':([\d.-]+)', s)
['-0.294118', '0.487437', '0.180328', '-0.292929', '-1', '0.00149028', '-0.53117', '-0.0333.333']
You can also split and partition, which would be substantially faster:
>>> [e.partition(':')[2] for e in s.split() if ':' in e]
['-0.294118', '0.487437', '0.180328', '-0.292929', '-1', '0.00149028', '-0.53117', '-0.0333.333']
Then you can convert those to a float using try/except and map and filter:
>>> def conv(s):
... try:
... return float(s)
... except ValueError:
... return None
...
>>> filter(None, map(conv, [e.partition(':')[2] for e in s.split() if ':' in e]))
[-0.294118, 0.487437, 0.180328, -0.292929, -1.0, 0.00149028, -0.53117, -0.0333333]

A simple oneliner using list comprehension -
str = "-1 1:-0.294118 2:0.487437 3:0.180328 4:-0.292929 5:-1 6:0.00149028 7:-0.53117 8:-0.0333333"
[float(s.split()[0]) for s in str.split(':')]
Note: this is simplest to understand (and pobably fastest) as we are not doing any regex evaluation. But this would only work for the particular case above. (eg. if you've to get the second number - in the above not so correctly formatted string would need more work than a single one-liner above).