This is my string:
"Somestring8/9/0"
I need to get something like this:
['Somestring','8/9/0']
The moment I find a numeric char, I need to split the string to get:
'8/9/0'
This my code:
stringSample = "GigabitEthernet8/9/0"
print re.findall(r'(\w+?)(\d+)', stringSample)[0]
('GigabitEthernet', '8')
But I'm getting this result
What am I doing wrong?
I appreciate your help!!
Your second regex group accepts only digits. Allow it to include forward slashes too.
stringSample = "GigabitEthernet8/9/0"
print re.findall(r'(\w+?)([\d/]+)', stringSample)[0]
# ('GigabitEthernet', '8/9/0')
Try Using the re.split method to split your string in two, passing the maxsplit parameter
re.split('(\w+?)([\d/]+)', stringSample, 1)
Related
So I have this string:
'm(1,2),m(4,3)'
How can I split it to get list that contains only 2 elements:
['m(1,2)', 'm(4,3)']
I can't use str.split function because it will split the whole string and would give me something like this:
['m(1', '2)', 'm(4', '3)']
Can you help me?
You can try regex:
import re
regex = re.compile("m\([0-9],[0-9]\)")
s = "m(1,2),m(4,3)"
regex.findall(s)
should yield:
['m(1,2)', 'm(4,3)']
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 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'm a novice in python programming and i'm trying to split full name to first name and last name, can someone assist me on this ? so my example file is:
Sarah Simpson
I expect the output like this : Sarah,Simpson
You can use the split() function like so:
fullname=" Sarah Simpson"
fullname.split()
which will give you: ['Sarah', 'Simpson']
Building on that, you can do:
first=fullname.split()[0]
last=fullname.split()[-1]
print(first + ',' + last)
which would give you Sarah,Simpson with no spaces
This comes handly : nameparser 1.0.6 - https://pypi.org/project/nameparser/
>>> from nameparser import HumanName
>>> name = "Sarah Simpson"
>>> name = HumanName(name)
>>> name.last
'Simpson'
>>> name.first
'Sarah'
>>> name.last+', '+name.first
'Simpson, Sarah'
you can try the .split() function which returns a list of strings after splitting by a separator. In this case the separator is a space char.
first remove leading and trailing spaces using .strip() then split by the separator.
first_name, last_name=fullname.strip().split()
Strings in Python are immutable. Create a new String to get the desired output.
You can use split() method of string class.
name = "Sarah Simpson"
name.split()
split() by default splits on whitespace, and takes separator as parameter. It returns a list
["Sarah", "Simpson"]
Just concatenate the strings. For more reference https://docs.python.org/3.7/library/stdtypes.html?highlight=split#str.split
Output = "Sarah", "Simpson"
name = "Thomas Winter"
LastName = name.split()[1]
(note the parantheses on the function call split.)
split() creates a list where each element is from your original string, delimited by whitespace. You can now grab the second element using name.split()[1] or the last element using name.split()[-1]
split() is obviously the function to go for-
which can take a parameter or 0 parameter
fullname="Sarah Simpson"
ls=fullname.split()
ls=fullname.split(" ") #this will split by specified space
Extra Optional
And if you want the split name to be shown as a string delimited by coma, then you can use join() or replace
print(",".join(ls)) #outputs Sarah,Simpson
print(st.replace(" ",","))
Input: Sarah Simpson => suppose it is a string.
Then, to output: Sarah, Simpson. Do the following:
name_surname = "Sarah Simpson".split(" ")
to_output = name_surname[0] + ", " + name_surname[-1]
print(to_output)
The function split is executed on a string to split it by a specified argument passed to it. Then it outputs a list of all chars or words that were split.
In your case: the string is "Sarah Simpson", so, when you execute split with the argument " " -empty space- the output will be: ["Sarah", "Simpson"].
Now, to combine the names or to access any of them, you can right the name of the list with a square brackets containing the index of the desired word to return. For example: name_surname[0] will output "Sarah" since its index is 0 in the list.