I have a string, I have to get digits only from that string.
url = "www.mylocalurl.com/edit/1987"
Now from that string, I need to get 1987 only.
I have been trying this approach,
id = [int(i) for i in url.split() if i.isdigit()]
But I am getting [] list only.
You can use regex and get the digit alone in the list.
import re
url = "www.mylocalurl.com/edit/1987"
digit = re.findall(r'\d+', url)
output:
['1987']
Replace all non-digits with blank (effectively "deleting" them):
import re
num = re.sub('\D', '', url)
See live demo.
You aren't getting anything because by default the .split() method splits a sentence up where there are spaces. Since you are trying to split a hyperlink that has no spaces, it is not splitting anything up. What you can do is called a capture using regex. For example:
import re
url = "www.mylocalurl.com/edit/1987"
regex = r'(\d+)'
numbers = re.search(regex, url)
captured = numbers.groups()[0]
If you do not what what regular expressions are, the code is basically saying. Using the regex string defined as r'(\d+)' which basically means capture any digits, search through the url. Then in the captured we have the first captured group which is 1987.
If you don't want to use this, then you can use your .split() method but this time provide a split using / as the separator. For example `url.split('/').
Related
I have list of demangled-function names like _Z6__comp7StudentS_
_Z4SortiSt6vectorI7StudentSaIS0_EE. I read wiki and found out that it follows some sort of defined structure. _Z is mangled Symbol followed by a number and then the function name of that length.
So I wanted to retrieve that function name using regex. I only come close to _Z(?:\d)(?<function_name>[a-z_A-Z]){\1}. But referring \1 won't work because its string, right? Is there a single regex pattern solution to this.
You can use 2 capture groups, and get the part of the string using the position of capture group 2
import re
pattern = r"_Z(\d+)([a-z_A-Z]+)"
s = "_Z4SortiSt6vectorI7StudentSaIS0_EE"
m = re.search(pattern, s)
if m:
print(m.group(2)[0: int(m.group(1))])
Output
Sort
Using _Z6__comp7StudentS_ will return __comp
I have list of websites unfortunately which looks like "rs--google.com--plain" how to remove 'rs--' and '--plain' from the url? I tried strip() but it didn't remove anything.
The way to remove "rs--" and "--plain" from that url (which is a string most likely) is to use some basic regex on it:
import re
url = 'rs--google.com--plain'
cleaned_url = re.search('rs--(.*)--plain', url).group(1)
print(cleaned_url)
Which prints out:
google.com
What is done here is use re's search module to check if anything exists between "rs--" and "--plain" and if it does match it to group 1, we then check for group 1 by doing .group(1) and set our entire "cleaned url" to it:
cleaned_url = re.search('rs--(.*)--plain', url).group(1)
And now we only "google.com" in our cleaned_url.
This assumes "rs--" and "--plain" are always in the url.
Updated to handle any letters on either side of --:
import re
url = 'po--google.com--plain'
cleaned_url = re.search('[A-z]+--(.*)--[A-z]+', url).group(1)
print(cleaned_url)
This will handle anything that has letters before -- and after -- and get only the url in the middle. What that does is check any letters on either side of -- regardless of how many letters are there. This will allow queries with letters that match that regular expression so long as --myurl.com-- letters exist before the first "--" and after the second "--"
A great resource for working on regex is regex101
You can use replace function in python.
>>> val = "rs--google.com--plain"
>>> newval =val.replace("rs--","").replace("--plain","")
>>> newval
'google.com'
I am having a string as follows:
A5697[2:10] = {ravi, rageev, raghav, smith};
I want the content after "A5697[2:10] =". So, my output should be:
{ravi, rageev, raghav, smith};
This is my code:
print(re.search(r'(?<=A\d+\[.*\] =\s).*', line).group())
But, this is giving error:
sre_constants.error: look-behind requires fixed-width pattern
Can anyone help to solve this issue? I would prefer to use regex.
You can try re.sub , like below, Since you have given only one data point. I am assuming all the other data points are following the similar pattern.
import re
text = "A5697[2:10] = {ravi, rageev, raghav, smith}"
re.sub(r'(A\d+\[\d+:\d+\]\s+=\s+)(.+)', r'\2', text)
returns,
'{ravi, rageev, raghav, smith}'
re.sub : substitutes the entire match as given as regex with the 2nd capturing group. The second capturing group captures every thing after '= '.
Simply replace the bits you don't want:
print re.sub(r'A\d[^=]*= *','',line)
See demo here: https://rextester.com/NSG17655
I am a newbee in python. I am trying to pull data (XXXX) out from a text with a pattern PDB:XXXX. The XXXX varies, but it is exactly what I want.
Since the data all contain PDB:, I use re.findall() to search and get this pattern. But this only gave me a list of PDB:. How can I get it to include the XXXX???
this is my code:
text = 'blah...........
PDB:AAAA
blah...........
blah...........
PDB:BBBB'
etc.
r = re.findall("PDB:",text)
and the output gave me:
['PDB:', 'PDB:']
My desired output should be something like
['AAAA', 'BBBB']
You need to use """ to quote multi-line strings in Python. Also, to get a specific subset of the matched pattern, you need to use capture groups (the parentheses in my regular expression below).
import re
text = """blah...........
PDB:AAAA
blah...........
blah...........
PDB:BBBB"""
results = re.findall(r"PDB:(.*)", text)
print results #['AAAA', 'BBBB']
Are there any alternatives that are similar to .replace() but that allow you to pass more than one old substring to be replaced?
I have a function with which I pass video titles so that specific characters can be removed (because the API I'm passing the videos too has bugs that don't allow certain characters):
def videoNameExists(vidName):
vidName = vidName.encode("utf-8")
bugFixVidName = vidName.replace(":", "")
search_url ='https://api.brightcove.com/services/library?command=search_videos&video_fields=name&page_number=0&get_item_count=true&token=kwSt2FKpMowoIdoOAvKj&any=%22{}%22'.format(bugFixVidName)
Right now, it's eliminating ":" from any video titles with vidName.replace(":", "") but I also would like to replace "|" when that occurs in the name string sorted in the vidName variable. Is there an alternative to .replace() that would allow me to replace more than one substring at a time?
>>> s = "a:b|c"
>>> s.translate(None, ":|")
'abc'
You may use re.sub
import re
re.sub(r'[:|]', "", vidName)