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for making lowercase from uppercase string just use "string".lower() where "string" is your string that you want to convert lowercase. for this question concern it will like this: s.lower() If you want to make your whole string variable use. s="sadf" # sadf s=s.upper() # SADF
After Python 2.5, string concatenation with the + operator is pretty fast. ... over methods to do string ...
On Python 3.9 and newer you can use the removeprefix and removesuffix methods to remove an entire substring from either side of the string: url = 'abcdc.com' url.removesuffix('.com') # Returns 'abcdc' url.removeprefix('abcdc.') # Returns 'com' The relevant Python Enhancement Proposal is PEP-616. On Python 3.8 and older you can use endswith and ...
All string characters are unicode literal in Python 3; as a consequence, since str.split() splits on all white space characters, that means it splits on unicode white space characters. So split + join syntax (as in 1 , 2 , 3 ) will produce the same output as re.sub with the UNICODE flag (as in 4 ); in fact, the UNICODE flag is redundant here ...
@gimel: Actually, [:] on an immutable type doesn't make a copy at all. While mysequence[:] is mostly harmless when mysequence is an immutable type like str, tuple, bytes (Py3) or unicode (Py2), a = b[:] is equivalent to a = b, it just wastes a little time dispatching the slicing byte codes which the object responds to by returning itself since it's pointless to shallow copy when, aside from ...
If you only have one reference to a string and you concatenate another string to the end, CPython now special cases this and tries to extend the string in place. The end result is that the operation is amortized O(n). e.g. s = "" for i in range(n): s += str(i) used to be O(n^2), but now it is O(n). More information. From the source (bytesobject.c):
What surprises me is that the seemingly impractical string.rpartition('hello')[0] == '' method always finds a way to be listed first, before the string.startswith('hello') method, every now and then. The results show that using str.partition to determine if a string starts with another string is more efficient then using both rfind and rindex.
When you want to split a string by a specific delimiter like: __ or | or , etc. it's much easier and faster to split using .split() method as in the top answer because Python string methods are intuitive and optimized. However, if you need to split a string using a pattern (e.g. " __ "and "__"), then using the built-in re module might be useful.
I'm working with Python, and I'm trying to find out if you can tell if a word is in a string. I have found some information about identifying if the word is in the string - using .find , but is there a way to do an if statement.
How do I get the probability of a string being similar to another string in Python? I want to get a decimal value like 0.9 (meaning 90%) etc. Preferably with standard Python and library. e.g. si...