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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
a="I Am Siva". print(a[2:]) OUTPUT: >>> Am Siva. In the above code a [2:] declares to print a from index 2 till the last element. Remember that if you set the maximum limit to print a string, as (x) then it will print the string till (x-1) and also remember that the index of a list or string will always start from 0.
The Python documentation notes the difference between the three methods. str.isdigit. Return true if all characters in the string are digits and there is at least one character, false otherwise. Digits include decimal characters and digits that need special handling, such as the compatibility superscript digits.
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 ...
You may be interested in this: An optimization anecdote by Guido.Although it is worth remembering also that this is an old article and it predates the existence of things like ''.join (although I guess string.joinfields is more-or-less the same)
The find() method should be used only if you need to know the position of sub. To check if sub is a substring or not, use the in operator. (c) Python reference. +1 for highlighting the gotchas involved in substring searches. the obvious solution is if ' is ' in s: which will return False as is (probably) expected.
67. If you want to find out whether a whole word is in a space-separated list of words, simply use: def contains_word(s, w): return (' ' + w + ' ') in (' ' + s + ' ') contains_word('the quick brown fox', 'brown') # True. contains_word('the quick brown fox', 'row') # False. This elegant method is also the fastest.
A better approach for this job would be: from collections import defaultdict. text = 'Mary had a little lamb'. chars = defaultdict(int) for char in text: chars[char] += 1. So you'll have a dict that returns the number of occurrences of every letter in the string and 0 if it isn't present. >>>chars['a'] 4.
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...
2. You can do it with one variable you don't have to use multiple variable which can cause confusion. Also code will be much cleaner. You can do it in one line also. text = text.lower() text = text.replace(string.punctuation, ' ') text = text.split(' ') Also you can do it in one line.