Develop Reference

Replacing string between two characters in the file

I have a string in my proprieties file as below:
line = "variables=ORACLE_BASE_HOME=/u02/test/oracle/landscape/1/db_50,DB_UNIQUE_NAME=cdms,ORACLE_BASE=//u02/test,PDB_NAME=,DB_NAME=cdms,ORACLE_HOME=/u02/test/product/19/db_21,SID=ss"
I would like to replace the following string with a different value:
DB_NAME=cdms -> DB_NAME=abc
I have the code below, however, it seems not doing as expected:
f = fileinput.FileInput(rsp_file_path)
for line in f:
re.sub(",DB_NAME=(.*?),", "abc", line, flags=re.DOTALL)
f.close()

It should be:
re.sub("(,DB_NAME=)(.*?),", "\g<1>abc,", line, flags=re.DOTALL)
or using raw string:
re.sub(r"(,DB_NAME=)(.*?),", r"\1abc,", line, flags=re.DOTALL)
That's because the documentation for re.sub() states:
In string-type repl arguments, in addition to the character escapes
and backreferences described above, \g will use the substring
matched by the group named name, as defined by the (?P...)
syntax. \g uses the corresponding group number; \g<2> is
therefore equivalent to \2, but isn’t ambiguous in a replacement such
as \g<2>0. \20 would be interpreted as a reference to group 20, not a
reference to group 2 followed by the literal character '0'. The
backreference \g<0> substitutes in the entire substring matched by the
RE.
In your case (,DB_NAME=) is the first captured group which you refer to with \g<1>.

you can use use string.replace()
s.replace('DB_NAME', 'cdms', 1).replace('DB_NAME', 'abc', 1)

Use python 3 regex to match a string in double quotes

I want to match a string contained in a pair of either single or double quotes. I wrote a regex pattern as so:
pattern = r"([\"\'])[^\1]*\1"
mytext = '"bbb"ccc"ddd'
re.match(pattern, mytext).group()
The expected output would be:
"bbb"
However, this is the output:
"bbb"ccc"
Can someone explain what's wrong with the pattern above? I googled and found the correct pattern to be:
pattern = r"([\"\'])[^\1]*?\1"
However, I don't understand why I must use ?.

In your regex
([\"'])[^\1]*\1
Character class is meant for matching only one character. So your use of [^\1] is incorrect. Think, what would have have happened if there were more than one characters in the first capturing group.
You can use negative lookahead like this
(["'])((?!\1).)*\1
or simply with alternation
(["'])(?:[^"'\\]+|\\.)*\1
or
(?<!\\)(["'])(?:[^"'\\]+|\\.)*\1
if you want to make sure "b\"ccc" does not matches in string bb\"b\"ccc"

You should use a negative lookahead assertion. And I assume there won't be any escaped quotes in your input string.
>>> pattern = r"([\"'])(?:(?!\1).)*\1"
>>> mytext = '"bbb"ccc"ddd'
>>> re.search(pattern, mytext).group()
'"bbb"'

You can use:
pattern = r"[\"'][^\"']*[\"']"
https://regex101.com/r/dO0cA8/1
[^\"']* will match everything that isn't " or '

Add [] around numbers in strings

I like to add [] around any sequence of numbers in a string e.g
"pixel1blue pin10off output2high foo9182bar"
should convert to
"pixel[1]blue pin[10]off output[2]high foo[9182]bar"
I feel there must be a simple way but its eluding me :(

Yes, there is a simple way, using re.sub():
result = re.sub(r'(\d+)', r'[\1]', inputstring)
Here \d matches a digit, \d+ matches 1 or more digits. The (...) around that pattern groups the match so we can refer to it in the second argument, the replacement pattern. That pattern simply replaces the matched digits with [...] around the group.
Note that I used r'..' raw string literals; if you don't you'd have to double all the \ backslashes; see the Backslash Plague section of the Python Regex HOWTO.
Demo:
>>> import re
>>> inputstring = "pixel1blue pin10off output2high foo9182bar"
>>> re.sub(r'(\d+)', r'[\1]', inputstring)
'pixel[1]blue pin[10]off output[2]high foo[9182]bar'

You can use re.sub :
>>> s="pixel1blue pin10off output2high foo9182bar"
>>> import re
>>> re.sub(r'(\d+)',r'[\1]',s)
'pixel[1]blue pin[10]off output[2]high foo[9182]bar
Here the (\d+) will match any combinations of digits and re.sub function will replace it with the first group match within brackets r'[\1]'.
You can start here to learn regular expression http://www.regular-expressions.info/

python. re.findall and re.sub with '^'

I try to change string like s='2.3^2+3^3-√0.04*2+√4',
where 2.3^2 has to change to pow(2.3,2), 3^3 - pow(3,3), √0.04 - sqrt(0.04) and
√4 - sqrt(4).
s='2.3^2+3^3-√0.04*2+√4'
patt1='[0-9]+\.[0-9]+\^[0-9]+|[0-9]+\^[0-9]'
patt2='√[0-9]+\.[0-9]+|√[0-9]+'
idx1=re.findall(patt1, s)
idx2=re.findall(patt2, s)
idx11=[]
idx22=[]
for i in range(len(idx1)):
idx11.append('pow('+idx1[i][:idx1[i].find('^')]+','+idx1[i][idx1[i].find('^')+1:]+')')
for i in range(len(idx2)):
idx22.append('sqrt('+idx2[i][idx2[i].find('√')+1:]+')')
for i in range(len(idx11)):
s=re.sub(idx1[i], idx11[i], s)
for i in range(len(idx22)):
s=re.sub(idx2[i], idx22[i], s)
print(s)
Temp results:
idx1=['2.3^2', '3^3']
idx2=['√0.04', '√4']
idx11=['pow(2.3,2)', 'pow(3,3)']
idx22=['sqrt(0.04)', 'sqrt(4)']
but string result:
2.3^2+3^3-sqrt(0.04)*2+sqrt(4)
Why calculating 'idx1' is right, but re.sub don't insert this value into string ?
(sorry for my english:)

Try this using only re.sub()
Input string:
s='2.3^2+3^3-√0.04*2+√4'
Replacing for pow()
s = re.sub("(\d+(?:\.\d+)?)\^(\d+)", "pow(\\1,\\2)", s)
Replacing for sqrt()
s = re.sub("√(\d+(?:\.\d+)?)", "sqrt(\\1)", s)
Output:
pow(2.3,2)+pow(3,3)-sqrt(0.04)*2+sqrt(4)
() means group capture and \\1 means first captured group from regex match. Using this link you can get the detail explanation for the regex.

I've only got python 2.7.5 but this works for me, using str.replace rather than re.sub. Once you've gone to the effort of finding the matches and constructing their replacements, this is a simple find and replace job:
for i in range(len(idx11)):
s = s.replace(idx1[i], idx11[i])
for i in range(len(idx22)):
s = s.replace(idx2[i], idx22[i])
edit
I think you're going about this in quite a long-winded way. You can use re.sub in one go to make these changes:
s = re.sub('(\d+(\.\d+)?)\^(\d+)', r'pow(\1,\3)', s)
Will substitute 2.3^2+3^3 for pow(2.3,2)+pow(3,3) and:
s = re.sub('√(\d+(\.\d+)?)', r'sqrt(\1)', s)
Will substitute √0.04*2+√4 to sqrt(0.04)*2+sqrt(4)
There's a few things going on here that are different. Firstly, \d, which matches a digit, the same as [0-9]. Secondly, the ( ) capture whatever is inside them. In the replacement, you can refer to these captured groups by the order in which they appear. In the pow example I'm using the first and third group that I have captured.
The prefix r before the replacement string means that the string is to be treated as "raw", so characters are interpreted literally. The groups are accessed by \1, \2 etc. but because the backslash \ is an escape character, I would have to escape it each time (\\1, \\2, etc.) without the r.

Python - why doesn't this simple regex work?

This code below should be self explanatory. The regular expression is simple. Why doesn't it match?
>>> import re
>>> digit_regex = re.compile('\d')
>>> string = 'this is a string with a 4 digit in it'
>>> result = digit_regex.match(string)
>>> print result
None
Alternatively, this works:
>>> char_regex = re.compile('\w')
>>> result = char_regex.match(string)
>>> print result
<_sre.SRE_Match object at 0x10044e780>
Why does the second regex work, but not the first?

Here is what re.match() says If zero or more characters at the beginning of string match the regular expression pattern ...
In your case the string doesn't have any digit \d at the beginning. But for the \w it has t at the beginning at your string.
If you want to check for digit in your string using same mechanism, then add .* with your regex:
digit_regex = re.compile('.*\d')

The second finds a match because string starts with a word character. If you want to find matches within the string, use the search or findall methods (I see this was suggested in a comment too). Or change your regex (e.g. .*(\d).*) and use the .groups() method on the result.