We Keep Coding

Regex python - find match items on list that have the same digit between the second character "_" to character "."

I have the following list :
list_paths=imgs/foldeer/img_ABC_21389_1.tif.tif,
imgs/foldeer/img_ABC_15431_10.tif.tif,
imgs/foldeer/img_GHC_561321_2.tif.tif,
imgs_foldeer/img_BCL_871125_21.tif.tif,
...
I want to be able to run a for loop to match string with specific number,which is the number between the second occurance of "_" to the ".tif.tif", for example, when number is 1, the string to be matched is "imgs/foldeer/img_ABC_21389_1.tif.tif" , for number 2, the match string will be "imgs/foldeer/img_GHC_561321_2.tif.tif".
For that, I wanted to use regex expression. Based on this answer, I have tested this regex expression on Regex101:
[^\r\n_]+\.[^\r\n_]+\_([0-9])
But this doesn't match anything, and also doesn't make sure that it will take the exact number, so if number is 1, it might also select items with number 10 .
My end goal is to be able to match items in the list that have the request number between the 2nd occurrence of "_" to the first occirance of ".tif" , using regex expression, looking for help with the regex expression.
EDIT: The output should be the whole path and not only the number.

Your pattern [^\r\n_]+\.[^\r\n_]+\_([0-9]) does not match anything, because you are matching an underscore \_ (note that you don't have to escape it) after matching a dot, and that does not occur in the example data.
Then you want to match a digit, but the available digits only occur before any of the dots.
In your question, the numbers that you are referring to are after the 3rd occurrence of the _
What you could do to get the path(s) is to make the number a variable for the number you want to find:
^\S*?/(?:[^\s_/]+_){3}\d+\.tif\b[^\s/]*$
Explanation
\S*? Match optional non whitespace characters, as few as possible
/ Match literally
(?:[^\s_/]+_){3} Match 3 times (non consecutive) _
\d+ Match 1+ digits
\.tif\b[^\s/]* Match .tif followed by any char except /
$ End of string
See a regex demo and a Python demo.
Example using a list comprehension to return all paths for the given number:
import re
number = 10
pattern = rf"^\S*?/(?:[^\s_/]+_){{3}}{number}\.tif\b[^\s/]*$"
list_paths = [
"imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_15431_10.tif.tif",
"imgs/foldeer/img_GHC_561321_2.tif.tif",
"imgs_foldeer/img_BCL_871125_21.tif.tif",
"imgs_foldeer/img_BCL_871125_21.png.tif"
]
res = [lp for lp in list_paths if re.search(pattern, lp)]
print(res)
Output
['imgs/foldeer/img_ABC_15431_10.tif.tif']

I'll show you something working and equally ugly as regex which I hate:
data = ["imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_21389_1.tif.tif",
"imgs/foldeer/img_ABC_15431_10.tif.tif",
"imgs/foldeer/img_GHC_561321_2.tif.tif",
"imgs_foldeer/img_BCL_871125_21.tif.tif"]
numbers = [int(x.split("_",3)[-1].split(".")[0]) for x in data]
First split gives ".tif.tif"
extract the last element
split again by the dot this time, take the first element (thats your number as a string), cast it to int
Please keep in mind it's gonna work only for the format you provided, no flexibility at all in this solution (on the other hand regex doesn't give any neither)

without regex if allowed.
import re
s= 'imgs/foldeer/img_ABC_15431_10.tif.tif'
last =s[s.rindex('_')+1:]
print(re.findall(r'\d+', last)[0])
Gives #
10

[0-9]*(?=\.tif\.tif)
This regex expression uses a lookahead to capture the last set of numbers (what you're looking for)

Try this:
import re
s = '''imgs/foldeer/img_ABC_21389_1.tif.tif
imgs/foldeer/img_ABC_15431_10.tif.tif
imgs/foldeer/img_GHC_561321_2.tif.tif
imgs_foldeer/img_BCL_871125_21.tif.tif'''
number = 1
res1 = re.findall(f".*_{number}\.tif.*", s)
number = 21
res21 = re.findall(f".*_{number}\.tif.*", s)
print(res1)
print(res21)
Results
['imgs/foldeer/img_ABC_21389_1.tif.tif']
['imgs_foldeer/img_BCL_871125_21.tif.tif']

Using Regex to move some letter of a string to a new location in the same string in a Series of strings in python

I have a list of 4000 strings. The naming convention needs to be changed for each string and I do not want to go through and edit each one individually.
The list looks like this:
data = list()
data = ['V2-FG2110-EMA-COMPRESSION',
'V2-FG2110-SA-COMPRESSION',
'V2-FG2110-UMA-COMPRESSION',
'V2-FG2120-EMA-DISTRIBUTION',
'V2-FG2120-SA-DISTRIBUTION',
'V2-FG2120-UMA-DISTRIBUTION',
'V2-FG2140-EMA-HEATING',
'V2-FG2140-SA-HEATING',
'V2-FG2140-UMA-HEATING',
'V2-FG2150-EMA-COOLING',
'V2-FG2150-SA-COOLING',
'V2-FG2150-UMA-COOLING',
'V2-FG2160-EMA-TEMPERATURE CONTROL']
I need all each 'SA' 'UMA' and 'EMA' to be moved to before the -FG.
Desired output is:
V2-EMA-FG2110-Compression
V2-SA-FG2110-Compression
V2-UMA-FG2110-Compression
...
The V2-FG2 does not change throughout the list so I have started there and I tried re.sub and re.search but I am pretty new to python so I have gotten a mess of different results. Any help is appreciated.

You can rearrange the strings.
new_list = []
for word in data:
arr = word.split('-')
new_word = '%s-%s-%s-%s'% (arr[0], arr[2], arr[1], arr[3])
new_list.append(new_word)

You can replace matches of the following regular expression with the contents of capture group 1:
(?<=^[A-Z]\d)(?=.*(-(?:EMA|SA|UMA))(?=-))|-(?:EMA|SA|UMA)(?=-)
Demo
The regular expression can be broken down as follows.
(?<=^[A-Z]\d) # current string position must be preceded by a capital
# letter followed by a digit at the start of the string
(?= # begin a positive lookahead
.* # match >= 0 chars other than a line terminator
(-(?:EMA|SA|UMA)) # match a hyphen followed by one of the three strings
# and save to capture group 1
(?=-) # the next char must be a hyphen
) # end positive lookahead
| # or
-(?:EMA|SA|UMA) # match a hyphen followed by one of the three strings
(?=-) # the next character must be a hyphen
(?=-) is a positive lookahead.
Evidently this may not work for versions of Python prior to 3.5, because the match in the second part of the alternation does not assign a value to capture group 1: "Before Python 3.5, backreferences to failed capture groups in Python re.sub were not populated with an empty string.. This quote is from
#WiktorStribiżew 's answer at the link. For what it's worth I confirmed that Ruby has the same behaviour ("V2-FG2110-EMA-COMPRESSION".gsub(rgx,'\1') #=> "V2-EMA-FG2110-COMPRESSION").
One could of course instead replace matches of (?<=^[A-Z]\d)(-[A-Z]{2}\d{4})(-(?:EMA|SA|UMA))(?=-)) with $2 + $1. That's probably more sensible even if it's less interesting.

python regex unexpected match groups

I am trying to find all occurrences of either "_"+digit or "^"+digit, using the regex ((_\^)[1-9])
The groups I'd expect back eg for "X_2ZZZY^5" would be [('_2'), ('^5')] but instead I am getting [('_2', '_'), ('^5', '^')]
Is my regex incorrect? Or is my expectation of what gets returned incorrect?
Many thanks
** my original re used (_|\^) this was incorrect, and should have been (_\^) -- question has been amended accordingly

You have 2 groups in your regex - so you're getting 2 groups. And you need to match atleast 1 number that follows.
try this:
([_\^][1-9]+)
See it in action here

Demand at least 1 digit (1-9) following the special characters _ or ^, placed inside a single capture group:
import re
text = "X_2ZZZY^5"
pattern = r"([_\^][1-9]{1,})"
regex = re.compile(pattern)
res = re.findall(regex, text)
print(res)
Returning:
['_2', '^5']

Extract date from inside a string with Python

I have the following string, while the first letters can differ and can also be sometimes two, sometimes three or four.
PR191030.213101.ABD
I want to extract the 191030 and convert that to a valid date.
filename_without_ending.split(".")[0][-6:]
PZA191030_392001_USB
Sometimes it looks liket his
This solution is not valid since this is also might differ from time to time. The only REAL pattern is really the first six numbers.
How do I do this?
Thank you!

You could get the first 6 digits using a pattern an a capturing group
^[A-Z]{2,4}(\d{6})\.
^ Start of string
[A-Z]{2,4} Match 2, 3 or 4 uppercase chars
( Capture group 1
\d{6} Match 6 digits
)\. Close group and match trailing dot
Regex demo | Python demo
For example
import re
regex = r"^[A-Z]{2,4}(\d{6})\."
test_str = "PR191030.213101.ABD"
matches = re.search(regex, test_str)
if matches:
print(matches.group(1))
Output
191030

You can do:
a = 'PR191030.213101.ABD'
int(''.join([c for c in a if c.isdigit()][:6]))
Output:
191030

This can also be done by:
filename_without_ending.split(".")[0][2::]
This splits the string from the 3rd letter to the end.

Since first letters can differ we have to ignore alphabets and extract digits.
So using re module (for regular expressions) apply regex pattern on string. It will give matching pattern out of string.
'\d' is used to match [0-9]digits and + operator used for matching 1 digit atleast(1/more).
findall() will find all the occurences of matching pattern in a given string while #search() is used to find matching 1st occurence only.
import re
str="PR191030.213101.ABD"
print(re.findall(r"\d+",str)[0])
print(re.search(r"\d+",str).group())

python regex: get end digits from a string

I am quite new to python and regex (regex newbie here), and I have the following simple string:
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
I would like to extract only the last digits in the above string i.e 767980716 and I was wondering how I could achieve this using python regex.
I wanted to do something similar along the lines of:
re.compile(r"""-(.*?)""").search(str(s)).group(1)
indicating that I want to find the stuff in between (.*?) which starts with a "-" and ends at the end of string - but this returns nothing..
I was wondering if anyone could point me in the right direction..
Thanks.

You can use re.match to find only the characters:
>>> import re
>>> s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
>>> re.match('.*?([0-9]+)$', s).group(1)
'767980716'
Alternatively, re.finditer works just as well:
>>> next(re.finditer(r'\d+$', s)).group(0)
'767980716'
Explanation of all regexp components:
.*? is a non-greedy match and consumes only as much as possible (a greedy match would consume everything except for the last digit).
[0-9] and \d are two different ways of capturing digits. Note that the latter also matches digits in other writing schemes, like ୪ or ൨.
Parentheses (()) make the content of the expression a group, which can be retrieved with group(1) (or 2 for the second group, 0 for the whole match).
+ means multiple entries (at least one number at the end).
$ matches only the end of the input.

Nice and simple with findall:
import re
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
print re.findall('^.*-([0-9]+)$',s)
>>> ['767980716']
Regex Explanation:
^ # Match the start of the string
.* # Followed by anthing
- # Upto the last hyphen
([0-9]+) # Capture the digits after the hyphen
$ # Upto the end of the string
Or more simply just match the digits followed at the end of the string '([0-9]+)$'

Your Regex should be (\d+)$.
\d+ is used to match digit (one or more)
$ is used to match at the end of string.
So, your code should be: -
>>> s = "99-my-name-is-John-Smith-6376827-%^-1-2-767980716"
>>> import re
>>> re.compile(r'(\d+)$').search(s).group(1)
'767980716'
And you don't need to use str function here, as s is already a string.

Use the below regex
\d+$
$ depicts the end of string..
\d is a digit
+ matches the preceding character 1 to many times

Save the regular expressions for something that requires more heavy lifting.
>>> def parse_last_digits(line): return line.split('-')[-1]
>>> s = parse_last_digits(r"99-my-name-is-John-Smith-6376827-%^-1-2-767980716")
>>> s
'767980716'

I have been playing around with several of these solutions, but many seem to fail if there are no numeric digits at the end of the string. The following code should work.
import re
W = input("Enter a string:")
if re.match('.*?([0-9]+)$', W)== None:
last_digits = "None"
else:
last_digits = re.match('.*?([0-9]+)$', W).group(1)
print("Last digits of "+W+" are "+last_digits)

Try using \d+$ instead. That matches one or more numeric characters followed by the end of the string.