I have a simple string as follows :
str = "id : c40d675f-19a9-4d40-9c6f-223eddafc81d"
Expected output :
c40d675f-19a9-4d40-9c6f-223eddafc81d
What I tried ?
print(str.strip('id : '))
What I get :
c40d675f-19a9-4d40-9c6f-223eddafc81
I am not able to understand, why is the last character 'd' from the end of string is stripped away? When I tried replacing the character 'd' with other alphabets, it works fine.
Because strip() takes a set of characters to remove from both sides of the string. If you have written: str.strip("i18d") it would remove 81d from the end and id from the beginning.
Maybe you wanted to do this :
str.split(":")[1].strip()
strip removes any possible combination of characters provided as its argument from both the start and end of the string. Here, the argument being 'id : ', it strips 'id : ' from the beginning and 'd' (one of the possible combinations of the string 'id : ') from the end of the string.
I think this link should help you.
https://www.tutorialspoint.com/python/string_strip.htm
look at the return value that strip gives
strip() strips the leading or trailing characters of the string off of every character in the sequence.
>>> 'abcadb'.strip('ab')
'cad'
>>> 'www.example.com'.strip('cmowz.')
'example'
strip([chars]):
The chars argument is not a prefix or suffix; rather, all combinations of its values are stripped
Be aware, it removes all combinations of the characters you provided!
>>> str.lstrip("id :")
'c40d675f-19a9-4d40-9c6f-223eddafc81d'
lstrip() returns a copy of the string in which all chars have been stripped from the beginning of the string
Alternatively, You can use str.replace("id : ", "", 1), specify that replace the 1st occurrence of "id : " to empty string
Related
For example i have the string "alba iuliara" and what i want to do is to convert all the "a" with "A" but not the first and the last "a". The result must be "albA iuliAra"
Any idea how can i do that using a statement like while, if and etc..
It's pretty easy if you use the replace method.
Code if you don't consider the first and last characters
your_string = "alba iuliara"
your_string = f"{your_string[0]}{your_string[1:-1].replace('a','A')}{your_string[-1]}"
print(your_string)
your_string = "Namaskara"
rep_char='a'
occ1=your_string.find(rep_char) #Searches for first occurence of replacing character
occn=your_string.rfind(rep_char) #Searches for last occurence of replacing character
#from startig poition of string to your first occurence of character and last occurence of character to end of the string nothing will be replaced. for remaining string character will be replaced with uppercase
your_string = f"{your_string[0:occ1+1]}{your_string[occ1+1:occn].replace(rep_char,rep_char.upper())}{your_string[occn:]}"
print(your_string)
output: NamAskAra
I have used #jock logic but modified to make it generic.
Here is the code i have until now :
dex = tree.xpath('//div[#class="cd-timeline-topic"]/text()')
names = filter(lambda n: n.strip(), dex)
table = str.maketrans(dict.fromkeys('?:,'))
for index, name in enumerate(dex, start = 0):
print('{}.{}'.format(index, name.strip().translate(table)))
The problem is that the output will print also strings with one special character "My name is/Richard". So what i need it's to replace that special character with a space and in the end the printing output will be "My name is Richard". Can anyone help me ?
Thanks!
Your call to dict.fromkeys() does not include the character / in its argument.
If you want to map all the special characters to None, just passing your list of special chars to dict.fromkeys() should be enough. If you want to replace them with a space, you could then iterate over the dict and set the value to for each key.
For example:
special_chars = "?:/"
special_char_dict = dict.fromkeys(special_chars)
for k in special_char_dict:
special_char_dict[k] = " "
You can do this by extending your translation table:
dex = ["My Name is/Richard????::,"]
table = str.maketrans({'?':None,':':None,',':None,'/':' '})
for index, name in enumerate(dex, start = 0):
print('{}.{}'.format(index, name.strip().translate(table)))
OUTPUT
0.My Name is Richard
You want to replace most special characters with None BUT forward slash with a space. You could use a different method to replace forward slashes as the other answers here do, or you could extend your translation table as above, mapping all the other special characters to None and forward slash to space. With this you could have a whole bunch of different replacements happen for different characters.
Alternatively you could use re.sub function following way:
import re
s = 'Te/st st?ri:ng,'
out = re.sub(r'\?|:|,|/',lambda x:' ' if x.group(0)=='/' else '',s)
print(out) #Te st string
Arguments meaning of re.sub is as follows: first one is pattern - it informs re.sub which substring to replace, ? needs to be escaped as otherwise it has special meaning there, | means: or, so re.sub will look for ? or : or , or /. Second argument is function which return character to be used in place of original substring: space for / and empty str for anything else. Third argument is string to be changed.
>>> a = "My name is/Richard"
>>> a.replace('/', ' ')
'My name is Richard'
To replace any character or sequence of characters from the string, you need to use `.replace()' method. So the solution to your answer is:
name.replace("/", " ")
here you can find details
After initializing a variable x with the content shown in below, I applied strip with a parameter. The result of strip is unexpected. As I'm trying to strip "ios_static_analyzer/", "rity/ios_static_analyzer/" is getting striped.
Kindly help me know why is it so.
>>> print x
/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/
>>> print x.strip()
/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/
>>> print x.strip('/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer
>>> print x.strip('ios_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/secu
>>> print x.strip('analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_
>>> print x.strip('_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static
>>> print x.strip('static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('s_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/security/io
>>> print x.strip('os_static_analyzer/')
Users/msecurity/Desktop/testspace/Hy5_Workspace/secu
Quoting from str.strip docs
Return a copy of the string with the leading and trailing characters
removed. The chars argument is a string specifying the set of
characters to be removed. If omitted or None, the chars argument
defaults to removing whitespace. The chars argument is not a prefix or
suffix; rather, all combinations of its values are stripped:
So, it removes all the characters in the parameter, from both the sides of the string.
For example,
my_str = "abcd"
print my_str.strip("da") # bc
Note: You can think of it like this, it stops removing the characters from the string when it finds a character which is not found in the input parameter string.
To actually, remove the particular string, you should use str.replace
x = "/Users/Desktop/testspace/Hy5_Workspace/security/ios_static_analyzer/"
print x.replace('analyzer/', '')
# /Users/msecurity/Desktop/testspace/Hy5_Workspace/security/ios_static_
But replace will remove the matches everywhere,
x = "abcd1abcd2abcd"
print x.replace('abcd', '') # 12
But if you want to remove words only at the beginning and ending of the string, you can use RegEx, like this
import re
pattern = re.compile("^{0}|{0}$".format("abcd"))
x = "abcd1abcd2abcd"
print pattern.sub("", x) # 1abcd2
What you need, I think, is replace:
>>> x.replace('ios_static_analyzer/','')
'/Users/msecurity/Desktop/testspace/Hy5_Workspace/security/'
string.replace(s, old, new[, maxreplace])
Return a copy of string s with all occurrences of substring old replaced by new.
So you can replace your string with nothing and get the desired output.
Python x.strip(s) remove from the begginning or the end of the string x any character appearing in s ! So s is just a set of characters, not a string being matched for substring.
string.strip removes a set of characters given as an argument. The chars argument is not a prefix or suffix; rather, all combinations of its values are stripped.
strip does not remove the string given as argument from the object; it removes the characters in the argument.
In this case, strip sees the string s_static_analyzer/ as an iterable of characters that needs to be stripped.
i am trying to delete certain portion of a string if a match found in the string as below
string = 'Newyork, NY'
I want to delete all the characters after the comma from the string including comma, if comma is present in the string
Can anyone let me now how to do this .
Use .split():
string = string.split(',', 1)[0]
We split the string on the comma once, to save python the work of splitting on more commas.
Alternatively, you can use .partition():
string = string.partition(',')[0]
Demo:
>>> 'Newyork, NY'.split(',', 1)[0]
'Newyork'
>>> 'Newyork, NY'.partition(',')[0]
'Newyork'
.partition() is the faster method:
>>> import timeit
>>> timeit.timeit("'one, two'.split(',', 1)[0]")
0.52929401397705078
>>> timeit.timeit("'one, two'.partition(',')[0]")
0.26499605178833008
You can split the string with the delimiter ",":
string.split(",")[0]
Example:
'Newyork, NY'.split(",") # ['Newyork', ' NY']
'Newyork, NY'.split(",")[0] # 'Newyork'
Try this :
s = "this, is"
m = s.index(',')
l = s[:m]
A fwe options:
string[:string.index(",")]
This will raise a ValueError if , cannot be found in the string. Here, we find the position of the character with .index then use slicing.
string.split(",")[0]
The split function will give you a list of the substrings that were separated by ,, and you just take the first element of the list. This will work even if , is not present in the string (as there'd be nothing to split in that case, we'd have string.split(...) == [string])
This question already has answers here:
How do I remove a substring from the end of a string?
(24 answers)
Closed 4 years ago.
>>> path = "/Volumes/Users"
>>> path.lstrip('/Volume')
's/Users'
>>> path.lstrip('/Volumes')
'Users'
>>>
I expected the output of path.lstrip('/Volumes') to be '/Users'
lstrip is character-based, it removes all characters from the left end that are in that string.
To verify this, try this:
"/Volumes/Users".lstrip("semuloV/") # also returns "Users"
Since / is part of the string, it is removed.
You need to use slicing instead:
if s.startswith("/Volumes"):
s = s[8:]
Or, on Python 3.9+ you can use removeprefix:
s = s.removeprefix("/Volumes")
Strip is character-based. If you are trying to do path manipulation you should have a look at os.path
>>> os.path.split("/Volumes/Users")
('/Volumes', 'Users')
The argument passed to lstrip is taken as a set of characters!
>>> ' spacious '.lstrip()
'spacious '
>>> 'www.example.com'.lstrip('cmowz.')
'example.com'
See also the documentation
You might want to use str.replace()
str.replace(old, new[, count])
# e.g.
'/Volumes/Home'.replace('/Volumes', '' ,1)
Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.
For paths, you may want to use os.path.split(). It returns a list of the paths elements.
>>> os.path.split('/home/user')
('/home', '/user')
To your problem:
>>> path = "/vol/volume"
>>> path.lstrip('/vol')
'ume'
The example above shows, how lstrip() works. It removes '/vol' starting form left. Then, is starts again...
So, in your example, it fully removed '/Volumes' and started removing '/'. It only removed the '/' as there was no 'V' following this slash.
HTH
lstrip doc says:
Return a copy of the string S with leading whitespace removed.
If chars is given and not None, remove characters in chars instead.
If chars is unicode, S will be converted to unicode before stripping
So you are removing every character that is contained in the given string, including both 's' and '/' characters.
Here is a primitive version of lstrip (that I wrote) that might help clear things up for you:
def lstrip(s, chars):
for i in range len(s):
char = s[i]
if not char in chars:
return s[i:]
else:
return lstrip(s[i:], chars)
Thus, you can see that every occurrence of a character in chars is is removed until a character that is not in chars is encountered. Once that happens, the deletion stops and the rest of the string is simply returned