The lambda I am working on gets triggered through API gateway.
I want to extract the a specific value from the path in the URL.
Sample URL : {id}/contacts
or
{id-0}/{id}/contacts
In order to extract the path variable I am using event.pathParamters
which gives me the value, but I need to only extract {id} from the path.
I am using the following code to split the path param and extract the {id}, but this is not a feasible option:
arr = path.split("/");
id = arr[arr.length-2];
Are there better ways to extract {id}? The position of this id will be always last right before api name (in his case <<contacts>>).
This would extract the string which is located between the last two occurrences of / or the first occurrence if two / do not exist
([^\/]+)\/[^\/]+$
https://regex101.com/r/tZNhrk/1
Would you please try the following;
import re
str = '{id-0}/{id}/contacts' # example
api_name = 'contacts' # api name
m = re.search(r'[^/]+(?=/%s)' % api_name, str)
if m:
id = m.group()
The regex [^/]+(?=/%s) matches a string of non-slash characters which is followed by a slash and the specified api_name. If the regex matches, m.group() is assigned to it.
Related
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('/').
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 have a list of IDs that I wish to extract, specifically for ID the intermediate values.
For example, the 1st ID is : garden/trade.FX.fwd/nyk12523adn
I wish to remove the values 'garden/', and '/nky12523adn', perhaps using string.replace()...
How might I do that efficiently? The starting logic would be to
string.find('/'), to get the location
then remove the prefix for the first '/', and suffix for the 2nd '/'
Is there an efficient way to do this?
you can also use .split() method
The split() method splits a string into a list.
You can specify the separator, default separator is any whitespace.
Syntax
string.split(separator, maxsplit)
id = "garden/trade.FX.fwd/nyk12523adn"
print (id.split('/')[1])
output:
trade.FX.fwd
if you want to use .find() function:
id = "garden/trade.FX.fwd/nyk12523adn"
idx_start = id.find('/') # find index of first '/'
idx_end = id.find('/',idx_start + 1) # find index of second '/'
print (id[idx_start+1:idx_end]) # use list slicing
output :
trade.FX.fwd
I have tried to remove prefix from the name. Now I use re.sup method to remove the prefix but some of the name are contain a character that included in the prefix.
Data example
-MisterClarkKent
-Mrs.Carol
-missjanedoemiss
I tried re.sub(r'(^\w{2,5}\ ?)', r'', name) to remove prefix with fix the position but it won't work because I have more than 10 prefix and each prefix has different size.
import re
name = 'mrjasontoddmr'
filter_name = re.sub(r'mr', r'', name)
print(filter_name)
#The result of filer_name is jasontodd but what I want is jasontoddmr
I expect the output of "jasontoddmr"
You can specify the count and ignore the case with the provided arguments in re.sub().
import re
names = ['MisterClarkKent','Mrs.Carol','missjanedoemiss', 'mrjasontoddmr']
filter_names = [re.sub(r'mrs?\.?|mister\s?|miss\s?', r'',name, count=1, flags=re.IGNORECASE) for name in names]
filter_names
Out[99]: ['ClarkKent', 'Carol', 'janedoemiss', 'jasontoddmr']
The ? means the character is optional so in mrs?\.? bother s and . are optional so it can capture both mr or mr. and mrs or mrs..