I would like to create a single regular expression in Python that extracts two interleaved portions of text from a filename as named groups. An example filename is given below:
CM00626141_H12.d4_T0001F003L01A02Z03C02.tif
The part of the filename I'd like to extract is contained between the underscores, and consists of the following:
An uppercase letter: [A-H]
A zero-padded two-digit number: 01 to 12
A period
A lowercase letter: [a-d]
A single digit: 1 to 4
For the example above, I would like one group ('Row') to contain H.d, and the other group ('Column') to contain 12.4. However, I don't know how to do this this when the text is separated as it is here.
EDIT: A constraint which I omitted: it needs to be a single regex to handle the string. I've updated the text/title to reflect this point.
Regexp capturing groups (whether numbered or named) do not actually capture text - they capture starting/ending indices within the original text. Thus, it is impossible for them to capture non-contiguous text. Probably the best thing to do here is have four separate groups, and combine them into your two desired values manually.
You may do it in two steps using re.findall() as:
Step 1: Extract substring from the main string following your pattern as:
>>> import re
>>> my_file = 'CM00626141_H12.d4_T0001F003L01A02Z03C02.tif'
>>> my_content = re.findall(r'_([A-H])(0[0-9]|1[0-2])\.([a-d])([1-4])_', my_file)
# where content of my_content is: [('H', '12', 'd', '4')]
Step 2: Join tuples to get the value of row and column:
>>> row = ".".join(my_content[0][::2])
>>> row
'H.d'
>>> column = ".".join(my_content[0][1::2])
>>> column
'12.4'
I do not believe there is any way to capture everything you want in exactly two named capture groups and one regex call. The most straightforward way I see is to do the following:
>>> import re
>>> source = 'CM00626141_H12.d4_T0001F003L01A02Z03C02.tif'
>>> match = re.search(r'_([A-H])(0[0-9]|1[0-2])\.([a-d])([1-4])_', source)
>>> row, column = '.'.join(match.groups()[0::2]), '.'.join(match.groups()[1::2])
>>> row
'H.d'
>>> column
'12.4'
Alternatively, you might find it more appealing to handle the parsing almost completely in the regex:
>>> row, column = re.sub(
r'^.*_([A-H])(0[0-9]|1[0-2])\.([a-d])([1-4])_.*$',
r'\1.\3,\2.\4',
source).split(',')
>>> row, column
('H.d', '12.4')
Related
long time ago I wrote a tool for parsing text files, line by line, and do some stuff, depending on commands and conditions in the file.
I used regex for this, however, I was never good in regex.
A line holding a condition looks like this:
[type==STRING]
And the regex I use is:
re.compile(r'^[^\[\]]*\[([^\]\[=]*)==([^\]\[=]*)\][^\]\[]*$', re.MULTILINE)
This regex would result me the keyword "type" and the value "STRING".
However, now I need to update my tool to have more conditions in one line, e.g.
[type==STRING][amount==0]
I need to update my regex to get me two pairs of results, one pair type/STRING and one pair amount/0.
But I'm lost on this. My regex above gets me zero results with this line.
Any ideas how to do this?
You could either match a second pair of groups:
^[^\[\]]*\[([^\]\[=]*)==([^\]\[=]*)\][^\]\[]*(?:\[([^\]\[=]*)==([^\]\[=]*)\][^\]\[]*)?$
Regex demo
Or you can omit the anchors and the [^\[\]]* part to get the group1 and group 2 values multiple times:
\[([^\]\[=]*)==([^\]\[=]*)\]
Regex demo
Is it a requirement that you use regex? You can alternatively accomplish this pretty easily using the split function twice and stripping the first opening and last closing bracket.
line_to_parse = "[type==STRING]"
# omit the first and last char before splitting
pairs = line_to_parse[1:-1].split("][")
for pair in pairs:
x, y = pair.split("==")
Rather depends on the precise "rules" that describe your data. However, for your given data why not:
import re
text = '[type==STRING][amount==0]'
words = re.findall('\w+', text)
lst = []
for i in range(0, len(words), 2):
lst.append((words[i], words[i+1]))
print(lst)
Output:
[('type', 'STRING'), ('amount', '0')]
I have a String from which I want to take the values within the parenthesis. Then, get the values that are separated from a comma.
Example: x(142,1,23ERWA31)
I would like to get:
142
1
23ERWA31
Is it possible to get everything with one regex?
I have found a method to do so, but it is ugly.
This is how I did it in python:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
secondResult = re.search("(?<=\()(.*?)(?=\))", firstResult.group(0))
finalResult = [x.strip() for x in secondResult.group(0).split(',')]
for i in finalResult:
print(i)
142
1
23ERWA31
This works for your example string:
import re
string = "x(142,1,23ERWA31)"
l = re.findall (r'([^(,)]+)(?!.*\()', string)
print (l)
Result: a plain list
['142', '1', '23ERWA31']
The expression matches a sequence of characters not in (,,,) and – to prevent the first x being picked up – may not be followed by a ( anywhere further in the string. This makes it also work if your preamble x consists of more than a single character.
findall rather than search makes sure all items are found, and as a bonus it returns a plain list of the results.
You can make this a lot simpler. You are running your first Regex but then not taking the result. You want .group(1) (inside the brackets), not .group(0) (the whole match). Once you have that you can just split it on ,:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
for e in firstResult.group(1).split(','):
print(e)
A little wonky looking, and also assuming there's always going to be a grouping of 3 values in the parenthesis - but try this regex
\((.*?),(.*?),(.*?)\)
To extract all the group matches to a single object - your code would then look like
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?),(.*?),(.*?)\)", string).groups()
You can then call the firstResult object like a list
>> print(firstResult[2])
23ERWA31
I am trying to extract some groups of data from a text and validate if the input text is correct. In the simplified form my input text looks like this:
Sample=A,B;C,D;E,F;G,H;I&other_text
In which A-I are groups I am interested in extracting them.
In the generic form, Sample looks like this:
val11,val12;val21,val22;...;valn1,valn2;final_val
arbitrary number of comma separated pairs which are separated by semicolon, and one single value at the very end.
There must be at least two pairs before the final value.
The regular expression I came up with is something like this:
r'Sample=(\w),(\w);(\w),(\w);((\w),(\w);)*(\w)'
Assuming my desired groups are simply words (in reality they are more complex but this is out of the scope of the question).
It actually captures the whole text but fails to group the values correctly.
I am just assuming that your "values" are any composed of any characters other than , and ;, i.e. [^,;]+. This clearly needs to be modified in the re.match and re.finditer calls to meet your actual requirements.
import re
s = 'Sample=val11,val12;val21,val22;val31,val32;valn1,valn2;final_val'
# verify if there is a match:
m = re.match(r'^Sample=([^,;]+),+([^,;]+)(;([^,;]+),+([^,;]+))+;([^,;]+)$', s)
if m:
final_val = m.group(6)
other_vals = [(m.group(1), m.group(2)) for m in re.finditer(r'([^,;]+),+([^,;]+)', s[7:])]
print(final_val)
print(other_vals)
Prints:
final_val
[('val11', 'val12'), ('val21', 'val22'), ('val31', 'val32'), ('valn1', 'valn2')]
You can do this with a regex that has an OR in it to decide which kind of data you are parsing. I spaced out the regex for commenting and clarity.
data = 'val11,val12;val21,val22;valn1,valn2;final_val'
pat = re.compile(r'''
(?P<pair> # either comma separated ending in semicolon
(?P<entry_1>[^,;]+) , (?P<entry_2>[^,;]+) ;
)
| # OR
(?P<end_part> # the ending token which contains no comma or semicolon
[^;,]+
)''', re.VERBOSE)
results = []
for match in pat.finditer(data):
if match.group('pair'):
results.append(match.group('entry_1', 'entry_2'))
elif match.group('end_part'):
results.append(match.group('end_part'))
print(results)
This results in:
[('val11', 'val12'), ('val21', 'val22'), ('valn1', 'valn2'), 'final_val']
You can do this without using regex, by using string.split.
An example:
words = map(lambda x : x.split(','), 'val11,val12;val21,val22;valn1,valn2;final_val'.split(';'))
This will result in the following list:
[
['val11', 'val12'],
['val21', 'val22'],
['valn1', 'valn2'],
['final_val']
]
Using Regular Expression, I want to find all the match words in a sentence and extract the wanted part in the matches words at the same time.
I use the API "findall" from "re" module to find the match words and plus the brackets to extract the parts I want.
For example I have a string "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C".
I only want the remaining two words after "0xQQ" or "0xWW", which will result in a list ["1A", "2B, "4C"].
Here is my code:
import re
MyString = "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C"
MySearch = re.compile("0xQQ(\w{2})|0xWW(\w{2})")
MyList = MySearch.findall(MyString)
print MyList
So my expected result is ["1A", "2B, "4C"].
But the actual result is [('1A', ''), ('', '2B'), ('4C', '')]
I think I might have used the combination of "()" and "|" in the wrong way.
Thx for the help!
Two different capturing groups will result in two items in the output (whatever matched each).
Instead, use a single capturing group and put your | (OR) earlier:
re.compile("0x(?:QQ|WW)(\w{2})")
((?:...) is a non-capturing group that matches ... - used to limit the effects of the | to only the QQ/WW split, without adding another capture to the output.)
You can try this:
import re
string = "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C"
pattern = re.compile(r"(0xQQ|0xWW)(\w{2})")
result = [match[2] for match in pattern.finditer(string)]
result will be:
['1A', '2B', '4C']
I am trying to do something which I thought would be simple (and probably is), however I am hitting a wall. I have a string that contains document numbers. In most cases the format is ######-#-### however in some cases, where the single digit should be, there are multiple single digits separated by a comma (i.e. ######-#,#,#-###). The number of single digits separated by a comma is variable. Below is an example:
For the string below:
('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
I need to return:
['030421-1-001', '030421-2-001' '030421-1-002', '030421-1-002', '030421-2-002', '030421-3-002' '030421-1-003']
I have only gotten as far as returning the strings that match the ######-#-### pattern:
import re
p = re.compile('\d{6}-\d{1}-\d{3}')
m = p.findall('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
print m
Thanks in advance for any help!
Matt
Perhaps something like this:
>>> import re
>>> s = '030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003'
>>> it = re.finditer(r'(\b\d{6}-)(\d(?:,\d)*)(-\d{3})\b', s)
>>> for m in it:
a, b, c = m.groups()
for x in b.split(','):
print a + x + c
...
030421-1-001
030421-2-001
030421-1-002
030421-1-002
030421-2-002
030421-3-002
030421-1-003
Or using a list comprehension
>>> [a+x+c for a, b, c in (m.groups() for m in it) for x in b.split(',')]
['030421-1-001', '030421-2-001', '030421-1-002', '030421-1-002', '030421-2-002', '030421-3-002', '030421-1-003']
Use '\d{6}-\d(,\d)*-\d{3}'.
* means "as many as you want (0 included)".
It is applied to the previous element, here '(,\d)'.
I wouldn't use a single regular expression to try and parse this. Since it is essentially a list of strings, you might find it easier to replace the "&" with a comma globally in the string and then use split() to put the elements into a list.
Doing a loop of the list will allow you to write a single function to parse and fix the string and then you can push it onto a new list and the display your string.
replace(string, '&', ',')
initialList = string.split(',')
for item in initialList:
newItem = myfunction(item)
newList.append(newItem)
newstring = newlist(join(','))
(\d{6}-)((?:\d,?)+)(-\d{3})
We take 3 capturing groups. We match the first part and last part the easy way. The center part is optionally repeated and optionally contains a ','. Regex will however only match the last one, so ?: won't store it at all. What where left with is the following result:
>>> p = re.compile('(\d{6}-)((?:\d,?)+)(-\d{3})')
>>> m = p.findall('030421-1,2-001 & 030421-1-002,030421-1,2,3-002, 030421-1-003')
>>> m
[('030421-', '1,2', '-001'), ('030421-', '1', '-002'), ('030421-', '1,2,3', '-002'), ('030421-', '1', '-003')]
You'll have to manually process the 2nd term to split them up and join them, but a list comprehension should be able to do that.