import re
path = "xyz pqr /all_it_abc type = cell"
s = re.findall(r'\/([^]*)\ ', path)
is giving me error "return _compile(pattern, flags).findall(string"
s= re.findall(r'\/([^]]*)\ ', path)
is giving me output as ['all_it_abc type =']
why? I have to use "]]"
print(s)
Desired output is all_it_abc
Thanks in advance to all and stackoverflow
The regex your are trying to find is is capturing a group [^]] This represents any character except ']'. That's why you are getting undesired output.
Your expected output can be obtained by using this simple regex instead.
s = re.findall(r'\/([a-z_]*)', path)
where s gives you this list as output
['all_it_abc']
Explanation - it captures everything after the slash that has an underscore or alphabet in it. You can add more characters to include them in the captured group.
r'\/([a-z_]*)'
In regex the pattern
pattern = r'[^abcd]'
means that this position can be matched by anything BUT 'a', 'b' ,'c' or 'd'.
pattern = r'[^]'
is an incomplete pattern that could match anything BUT ']' - if you complete it to a valid r[^]] pattern by closing the bracket.
You need r'/([^ ]+):
import re
path = "xyz pqr /all_it_abc type = cell"
s = re.findall(r'/([^ ]+)', path) # find all after / thats not a space
print(type(s), s) # this is a list of all matches
To get a list of all matches:
(<type 'list'>, ['all_it_abc'])
Related
I have a verilog file in which some inputs and ouputs are named as 133GAT(123).For example
nand2 g679(.a(n752), .b(n750), .O(1355GAT(558) ));
Here, I have to only replace 1355GAT(558) with 1355GAT_588 and not for .a(n752) There are multiple such instance.
I tried with python3.
re.sub(r'GAT*\((\w+)\)',r'_\1',"nand2 g679(.a(n752), .b(n750), .O(1355GAT(558) ) ")
It is giving output as
'nand2 g679(.a(n752), .b(n750), .O(1355_558 ) '
My expectation is to get the output as
'nand2 g679(.a(n752), .b(n750), .O(1355GAT_558 ) '
Why your code is not giving you expected results
Your regex GAT*\((\w+)\) matches GA, GAT, GATT, etc., and while it matches GAT in your string, you're effectively replacing it with your substitution since you never capture it and include it again in the substitution.
Regex 1
This works and gives you the option to check for digits before GAT.
See this regex in use here
# regex
(\d+GAT)\((\d+)\)
# replacement
\1_\2
Code 1
See code in use here
import re
s = "nand2 g679(.a(n752), .b(n750), .O(1355GAT(558) ));"
r = r'(\d+GAT)\((\d+)\)'
x = re.sub(r,r'\1_\2',s)
print(x)
Regex 2
This works too, but uses one capture group rather than two.
See this regex in use here
# regex
(?<=\dGAT)\((\d+)\)
# replacement
_\1
Code 2
See code in use here
import re
s = "nand2 g679(.a(n752), .b(n750), .O(1355GAT(558) ));"
r = r'(?<=\dGAT)\((\d+)\)'
x = re.sub(r,r'_\1',s)
print(x)
I have some sample strings:
s = 'neg(able-23, never-21) s2-1/3'
i = 'amod(Market-8, magical-5) s1'
I've got the problem where I can figure out if the string has 's1' or 's3' using:
word = re.search(r's\d$', s)
But if I want to know if the contains 's2-1/3' in it, it won't work.
Is there a regex expression that can be used so that it works for both cases of 's#' and 's#+?
Thanks!
You can allow the characters "-" and "/" to be captured as well, in addition to just digits. It's hard to tell the exact pattern you're going for here, but something like this would capture "s2-1/3" from your example:
import re
s = "neg(able-23, never-21) s2-1/3"
word = re.search(r"s\d[-/\d]*$", s)
I'm guessing that maybe you would want to extract that with some expression, such as:
(s\d+)-?(.*)$
Demo 1
or:
(s\d+)-?([0-9]+)?\/?([0-9]+)?$
Demo 2
Test
import re
expression = r"(s\d+)-?(.*)$"
string = """
neg(able-23, never-21) s211-12/31
neg(able-23, never-21) s2-1/3
amod(Market-8, magical-5) s1
"""
print(re.findall(expression, string, re.M))
Output
[('s211', '12/31'), ('s2', '1/3'), ('s1', '')]
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).
First off, I know this may seem like a duplicate question, however, I could find no working solution to my problem.
I have string that looks like the following:
string = "api('randomkey123xyz987', 'key', 'text')"
I need to extract randomkey123xyz987 which will always be between api(' and ',
I was planning on using Regex for this, however, I seem to be having some trouble.
This is the only progress that I have made:
import re
string = "api('randomkey123xyz987', 'key', 'text')"
match = re.findall("\((.*?)\)", string)[0]
print match
The following code returns 'randomkey123xyz987', 'key', 'text'
I have tried to use [^'], but my guess is that I am not properly inserting it into the re.findall function.
Everything that I am trying is failing.
Update: My current workaround is using [2:-4], but I would still like to avoid using match[2:-4].
If the string contains only one instance, use re.search() instead:
>>> import re
>>> s = "api('randomkey123xyz987', 'key', 'text')"
>>> match = re.search(r"api\('([^']*)'", s).group(1)
>>> print match
randomkey123xyz987
You want the string between the ( and ,, you are catching everything between the parens:
match = re.findall("api\((.*?),", string)
print match
["'randomkey123xyz987'"]
Or match between the '':
match = re.findall("api\('(.*?)'", string)
print match
['randomkey123xyz987']
If that is how your strings actually look you can split:
string = "api('randomkey123xyz987', 'key', 'text')"
print(string.split(",",1)[0][4:])
You should use the following regex:
api\('(.*?)'
Assuming that api( is fixed prefix
It matches api(, then captures what appears next, until ' token.
>>> re.findall(r"api\('(.*?)'", "api('randomkey123xyz987', 'key', 'text')")
['randomkey123xyz987']
If you are certain that randomkey123xyz987 will always be between "api('" and "',", then using the split() method can get it done in one line. This way you will not have to use regex matching. It will match the pattern between the starting and ending delimiter which is "api('" and "',
".
>>> string = "api('randomkey123xyz987', 'key', 'text')"
>>> value = (string.split("api('")[1]).split("',")[0]
>>> print value
randomkey123xyz987
This is one of those things where I'm sure I'm missing something simple, but... In the sample program below, I'm trying to use Python's RE library to parse the string "line" to get the floating-point number just before the percent sign, i.e. "90.31". But the code always prints "no match".
I've tried a couple other regular expressions as well, all with the same result. What am I missing?
#!/usr/bin/python
import re
line = ' 0 repaired, 90.31% done'
pct_re = re.compile(' (\d+\.\d+)% done$')
#pct_re = re.compile(', (.+)% done$')
#pct_re = re.compile(' (\d+.*)% done$')
match = pct_re.match(line)
if match: print 'got match, pct=' + match.group(1)
else: print 'no match'
match only matches from the beginning of the string. Your code works fine if you do pct_re.search(line) instead.
You should use re.findall instead:
>>> line = ' 0 repaired, 90.31% done'
>>>
>>> pattern = re.compile("\d+[.]\d+(?=%)")
>>> re.findall(pattern, line)
['90.31']
re.match will match at the start of the string. So you would need to build the regex for complete string.
try this if you really want to use match:
re.match(r'.*(\d+\.\d+)% done$', line)
r'...' is a "raw" string ignoring some escape sequences, which is a good practice to use with regexp in python. – kratenko (see comment below)