I need some help with a regular expression in python.
I have a string like this:
>>> s = '[i1]scale=-2:givenHeight_1[o1];'
How can I remove givenHeight_1 and turn the string to this?
>>> '[i1]scale=-2:360[o1];'
Is there an efficient one-liner regex for such a job?
UPDATE 1:
my regex so far is something like this but currently not working:
re.sub('givenHeight_1[o1]', '360[o1]', s)
You can use positive look around with re.sub :
>>> s = '[i1]scale=-2:givenHeight_1[o1];'
>>> re.sub(r'(?<=:).*(?=\[)','360',s)
'[i1]scale=-2:360[o1];'
The preceding regex will replace any thing that came after : and before [ with an '360'.
Or based on your need you can use str.replace directly :
>>> s.replace('givenHeight_1','360')
'[i1]scale=-2:360[o1];'
Related
I want to get String before last occurrence of my given sub string.
My String was,
path =
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov
my substring, 1001-1010 which will occurred twice. all i want is get string before its last occurrence.
Note: My substring is dynamic with different padding but only number.
I want,
D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v
I have done using regex and slicing,
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> q = re.findall("\d*-\d*",p)
>>> q[-1].join(p.split(q[-1])[:-1])
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v'
>>>
Is their any better way to do by purely using regex?
Please Note I have tried so many eg:
regular expression to match everything until the last occurrence of /
Regex Last occurrence?
I got answer by using regex with slicing but i want to achieve by using regex alone..
Why use regex. Just use built in string methods:
path = "D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov"
index = path.rfind("1001-1010")
print(path[:index])
You can use a simple greedy match and a capture group:
(.*)1001-1010
Your match is in capture group #1
Since .* is greedy by nature, it will match longest match before matching your keyword 1001-1010.
RegEx Demo
As per comments below if keyword is not a static string then you may use this regex:
r'(.*\D)\d+-\d+'
Python Code:
>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> print (re.findall(r'(.*\D)\d+-\d+', p))
['D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v']
Thanks #anubhava,
My first regex was,
.*(\d*-\d*)\/
Now i have corrected mine..
.*(\d*-\d*)
or
(.*)(\d*-\d*)
which gives me,
>>> q = re.search('.+(\d*-\d*)', p)
>>> q.group()
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v0001-1001'
>>>
(.*\D)\d+-\d+
this gives me exactly what i want...
>>> q = re.search('(.*\D)\d+-\d+', p)
>>> q.groups()
('D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v',)
>>>
How can i get word example from such string:
str = "http://test-example:123/wd/hub"
I write something like that
print(str[10:str.rfind(':')])
but it doesn't work right, if string will be like
"http://tests-example:123/wd/hub"
You can use this regex to capture the value preceded by - and followed by : using lookarounds
(?<=-).+(?=:)
Regex Demo
Python code,
import re
str = "http://test-example:123/wd/hub"
print(re.search(r'(?<=-).+(?=:)', str).group())
Outputs,
example
Non-regex way to get the same is using these two splits,
str = "http://test-example:123/wd/hub"
print(str.split(':')[1].split('-')[1])
Prints,
example
You can use following non-regex because you know example is a 7 letter word:
s.split('-')[1][:7]
For any arbitrary word, that would change to:
s.split('-')[1].split(':')[0]
many ways
using splitting:
example_str = str.split('-')[-1].split(':')[0]
This is fragile, and could break if there are more hyphens or colons in the string.
using regex:
import re
pattern = re.compile(r'-(.*):')
example_str = pattern.search(str).group(1)
This still expects a particular format, but is more easily adaptable (if you know how to write regexes).
I am not sure why do you want to get a particular word from a string. I guess you wanted to see if this word is available in given string.
if that is the case, below code can be used.
import re
str1 = "http://tests-example:123/wd/hub"
matched = re.findall('example',str1)
Split on the -, and then on :
s = "http://test-example:123/wd/hub"
print(s.split('-')[1].split(':')[0])
#example
using re
import re
text = "http://test-example:123/wd/hub"
m = re.search('(?<=-).+(?=:)', text)
if m:
print(m.group())
Python strings has built-in function find:
a="http://test-example:123/wd/hub"
b="http://test-exaaaample:123/wd/hub"
print(a.find('example'))
print(b.find('example'))
will return:
12
-1
It is the index of found substring. If it equals to -1, the substring is not found in string. You can also use in keyword:
'example' in 'http://test-example:123/wd/hub'
True
I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']
I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)
Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')
If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']
Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.
>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']
I have a regex "value=4020a345-f646-4984-a848-3f7f5cb51f21"
if re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x ):
x = re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x )
m = x.group(1)
m only gives me 4020a345, not sure why it does not give me the entire "4020a345-f646-4984-a848-3f7f5cb51f21"
Can anyone tell me what i am doing wrong?
try out this regex, looks like you are trying to match a GUID
value=[0-9a-f]{8}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{12}
This should match what you want, if all the strings are of the form you've shown:
value=((\w*\d*\-?)*)
You can also use this website to validate your regular expressions:
http://regex101.com/
The below regex works as you expect.
value=([\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*]+)
You are trying to match on some hex numbers, that is why this regex is more correct than using [\w\d]
pattern = "value=([0-9a-fA-F]{8}-([0-9a-fA-F]{4}-){3}[0-9a-fA-F]{12})"
data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
res = re.search(pattern, data)
print(res.group(1))
If you dont care about the regex safety, aka checking that it is correct hex, there is no reason not to use simple string manipulation like shown below.
>>> data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
>>> print(data[7:])
020a345-f646-4984-a848-3f7f5cb51f21
>>> # or maybe
...
>>> print(data[7:].replace('-',''))
020a345f6464984a8483f7f5cb51f21
You can get the subparts of the value as a list
txt = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
parts = re.findall('\w+', txt)[1:]
parts is ['4020a345', 'f646', '4984', 'a848', '3f7f5cb51f21']
if you really want the entire string
full = "-".join(parts)
A simple way
full = re.findall("[\w-]+", txt)[-1]
full is 4020a345-f646-4984-a848-3f7f5cb51f21
value=([\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*)
Try this.Grab the capture.Your regex was not giving the whole as you had used | operator.So if regex on left side of | get satisfied it will not try the latter part.
See demo.
http://regex101.com/r/hQ1rP0/45
I have some strings that look like this:
東野 圭吾「夢幻花」「白夜行」他<br>
Now I want to extract the link and the strings inside the corner brackets ("「" and "」"), like this:
['higashino/index.html', '夢幻花', '白夜行']
I've tried:
import re
str = u'''東野 圭 吾「夢幻花」「白夜行」他<br>'''
myre = re.compile(ur'''\('(.*)'\)">.*「(.*?)」.*''', re.UNICODE)
myre.findall(str)
the result is:
['higashino/index.html', '白夜行']
then I tried to use the pattern\('(.*)'\)">.*「([^」]*)」.*, but the result was the same, only one element inside the corner brackets was found.
How can I get not just one, but all elements inside the corner brackets? Thanks.
Use re.findall() (or re.finditer) with the regex 「([^」]*?)」:
import re
str = '''東野 圭 吾「夢幻花」「白夜行」他<br>'''
match = re.findall(r'「([^」]*?)」', str)
print(match)
Giving:
['夢幻花', '白夜行']
Using python 3.
Also, if you're not using python 3 already I recommend doing so as it is much better with unicode strings than python 2
>>> myre = re.compile(ur'''(?<=\(').+?(?='\)">)|(?<=「)[^」]+''', re.UNICODE)
>>> myre.findall(str)
[u'higashino/index.html', u'\u5922\u5e7b\u82b1', u'\u767d\u591c\u884c']