can anyone please explain what is wrong with this code?
str1='"xxx"'
print str1
if str1[:1].startswith('"'):
if str1[:-1].endswith('"'):
print "hi"
else:
print "condition fails"
else:
print "bye"
The output I got is:
Condition fails
but I expected it to print hi instead.
When you say [:-1] you are stripping the last element. Instead of slicing the string, you can apply startswith and endswith on the string object itself like this
if str1.startswith('"') and str1.endswith('"'):
So the whole program becomes like this
>>> str1 = '"xxx"'
>>> if str1.startswith('"') and str1.endswith('"'):
... print "hi"
>>> else:
... print "condition fails"
...
hi
Even simpler, with a conditional expression, like this
>>> print("hi" if str1.startswith('"') and str1.endswith('"') else "fails")
hi
You should either use
if str1[0] == '"' and str1[-1] == '"'
or
if str1.startswith('"') and str1.endswith('"')
but not slice and check startswith/endswith together, otherwise you'll slice off what you're looking for...
You are testing against the string minus the last character:
>>> '"xxx"'[:-1]
'"xxx'
Note how the last character, the ", is not part of the output of the slice.
I think you wanted just to test against the last character; use [-1:] to slice for just the last element.
However, there is no need to slice here; just use str.startswith() and str.endswith() directly.
When you set a string variable, it doesn't save quotes of it, they are a part of its definition.
so you don't need to use :1
Related
This question already has answers here:
Understanding slicing
(38 answers)
Closed 29 days ago.
I want to get a new string from the third character to the end of the string, e.g. myString[2:end]. If omitting the second part means 'to the end', and if you omit the first part, does it start from the start?
>>> x = "Hello World!"
>>> x[2:]
'llo World!'
>>> x[:2]
'He'
>>> x[:-2]
'Hello Worl'
>>> x[-2:]
'd!'
>>> x[2:-2]
'llo Worl'
Python calls this concept "slicing" and it works on more than just strings. Take a look here for a comprehensive introduction.
Just for completeness as nobody else has mentioned it. The third parameter to an array slice is a step. So reversing a string is as simple as:
some_string[::-1]
Or selecting alternate characters would be:
"H-e-l-l-o- -W-o-r-l-d"[::2] # outputs "Hello World"
The ability to step forwards and backwards through the string maintains consistency with being able to array slice from the start or end.
Substr() normally (i.e. PHP and Perl) works this way:
s = Substr(s, beginning, LENGTH)
So the parameters are beginning and LENGTH.
But Python's behaviour is different; it expects beginning and one after END (!). This is difficult to spot by beginners. So the correct replacement for Substr(s, beginning, LENGTH) is
s = s[ beginning : beginning + LENGTH]
A common way to achieve this is by string slicing.
MyString[a:b] gives you a substring from index a to (b - 1).
One example seems to be missing here: full (shallow) copy.
>>> x = "Hello World!"
>>> x
'Hello World!'
>>> x[:]
'Hello World!'
>>> x==x[:]
True
>>>
This is a common idiom for creating a copy of sequence types (not of interned strings), [:]. Shallow copies a list, see Python list slice syntax used for no obvious reason.
Is there a way to substring a string in Python, to get a new string from the 3rd character to the end of the string?
Maybe like myString[2:end]?
Yes, this actually works if you assign, or bind, the name,end, to constant singleton, None:
>>> end = None
>>> myString = '1234567890'
>>> myString[2:end]
'34567890'
Slice notation has 3 important arguments:
start
stop
step
Their defaults when not given are None - but we can pass them explicitly:
>>> stop = step = None
>>> start = 2
>>> myString[start:stop:step]
'34567890'
If leaving the second part means 'till the end', if you leave the first part, does it start from the start?
Yes, for example:
>>> start = None
>>> stop = 2
>>> myString[start:stop:step]
'12'
Note that we include start in the slice, but we only go up to, and not including, stop.
When step is None, by default the slice uses 1 for the step. If you step with a negative integer, Python is smart enough to go from the end to the beginning.
>>> myString[::-1]
'0987654321'
I explain slice notation in great detail in my answer to Explain slice notation Question.
I would like to add two points to the discussion:
You can use None instead on an empty space to specify "from the start" or "to the end":
'abcde'[2:None] == 'abcde'[2:] == 'cde'
This is particularly helpful in functions, where you can't provide an empty space as an argument:
def substring(s, start, end):
"""Remove `start` characters from the beginning and `end`
characters from the end of string `s`.
Examples
--------
>>> substring('abcde', 0, 3)
'abc'
>>> substring('abcde', 1, None)
'bcde'
"""
return s[start:end]
Python has slice objects:
idx = slice(2, None)
'abcde'[idx] == 'abcde'[2:] == 'cde'
You've got it right there except for "end". It's called slice notation. Your example should read:
new_sub_string = myString[2:]
If you leave out the second parameter it is implicitly the end of the string.
text = "StackOverflow"
#using python slicing, you can get different subsets of the above string
#reverse of the string
text[::-1] # 'wolfrevOkcatS'
#fist five characters
text[:5] # Stack'
#last five characters
text[-5:] # 'rflow'
#3rd character to the fifth character
text[2:5] # 'rflow'
#characters at even positions
text[1::2] # 'tcOefo'
If myString contains an account number that begins at offset 6 and has length 9, then you can extract the account number this way: acct = myString[6:][:9].
If the OP accepts that, they might want to try, in an experimental fashion,
myString[2:][:999999]
It works - no error is raised, and no default 'string padding' occurs.
Well, I got a situation where I needed to translate a PHP script to Python, and it had many usages of substr(string, beginning, LENGTH).
If I chose Python's string[beginning:end] I'd have to calculate a lot of end indexes, so the easier way was to use string[beginning:][:length], it saved me a lot of trouble.
str1='There you are'
>>> str1[:]
'There you are'
>>> str1[1:]
'here you are'
#To print alternate characters skipping one element in between
>>> str1[::2]
'Teeyuae'
#To print last element of last two elements
>>> str1[:-2:-1]
'e'
#Similarly
>>> str1[:-2:-1]
'e'
#Using slice datatype
>>> str1='There you are'
>>> s1=slice(2,6)
>>> str1[s1]
'ere '
Maybe I missed it, but I couldn't find a complete answer on this page to the original question(s) because variables are not further discussed here. So I had to go on searching.
Since I'm not yet allowed to comment, let me add my conclusion here. I'm sure I was not the only one interested in it when accessing this page:
>>>myString = 'Hello World'
>>>end = 5
>>>myString[2:end]
'llo'
If you leave the first part, you get
>>>myString[:end]
'Hello'
And if you left the : in the middle as well you got the simplest substring, which would be the 5th character (count starting with 0, so it's the blank in this case):
>>>myString[end]
' '
Using hardcoded indexes itself can be a mess.
In order to avoid that, Python offers a built-in object slice().
string = "my company has 1000$ on profit, but I lost 500$ gambling."
If we want to know how many money I got left.
Normal solution:
final = int(string[15:19]) - int(string[43:46])
print(final)
>>>500
Using slices:
EARNINGS = slice(15, 19)
LOSSES = slice(43, 46)
final = int(string[EARNINGS]) - int(string[LOSSES])
print(final)
>>>500
Using slice you gain readability.
a="Helloo"
print(a[:-1])
In the above code, [:-1] declares to print from the starting till the maximum limit-1.
OUTPUT :
>>> Hello
Note: Here a [:-1] is also the same as a [0:-1] and a [0:len(a)-1]
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.
I have a simpler solution using for loop to find a given substring in a string.
Let's say we have two string variables,
main_string = "lullaby"
match_string = "ll"
If you want to check whether the given match string exists in the main string, you can do this,
match_string_len = len(match_string)
for index,value in enumerate(main_string):
sub_string = main_string[index:match_string_len+index]
if sub_string == match_string:
print("match string found in main string")
I need to write a function that returns the location of the first sub-string in a string in Python without using the string.find(sub)
def sublocation(string,sub):
i=0
while i<len(string):
o=0
while o<len(sub):
u=i
if string[i] == sub[o]:
o=o+1
u=u+1
result=True
i=len(string)
else:
i=i+1-u
result=False
You can use find().
Ex:
>>> s = "this is the string for finding sub string"
>>> s.find('string')
12
I tried this :
myString = "Hellow World!"
index = str.find(myString, "World")
print(index)
And the output was :
>>>
7
>>>
Is that what you want?
you can use the building-in function find,e.m.,"Hello world".find("world") will return 6.The fuction index can do it too,but it will raise exception when string does not have the substring you search.
I think it might help you to learn how to loop in python. If the sub-strings are separated you can use .split() to split it into a list of sub-strings over which you can iterate natively. 'enumerate' enumerate even gives you the index.
def sub_search_location(inp_str,sub_string):
for pos, sub in enumerate(inp_str.split(' ')):
if sub_string == sub:
return pos
else:
return False
>>> sub_search_location("this is the string for finding sub string", "string")
3
However this works only for sub strings without spaces
>>> sub_search_location("this is the string for finding sub string", "sub string")
False
Of course you can use str.find as others have recommended, but looking at your code I think you have more important things to learn than just a call to an arbitrary method.
Try using the method .find(). It will return an index of the substring that you're trying to find.
I have a code
print(re.split(r"[\s\?\!\,\;]+", "Holy moly, feferoni!"))
which results
['Holy', 'moly', 'feferoni', '']
How can I get rid of this last blank element, what caused it?
If this is a dirty way to get rid of punctuation and spaces from a string, how else can I write but in regex?
Expanding on what #HamZa said in his comment, you would use re.findall and a negative character set:
>>> from re import findall
>>> findall(r"[^\s?!,;]+", "Holy moly, feferoni!")
['Holy', 'moly', 'feferoni']
>>>
You get the empty string as the last element of you list, because the RegEx splits after the last !. It ends up giving you what's before the ! and what's after it, but after it, there's simply nothing, i.e. an empty string! You might have the same problem in the middle of the string if you didn't wisely add the + to your RegEx.
Add a call to list if you can't work with an iterator. If you want to elegantly get rid of the optional empty string, do:
filter(None, re.split(r"[\s?!,;]+", "Holy moly, feferoni!"))
This will result in:
['Holy', 'moly', 'feferoni']
What this does is remove every element that is not a True value. The filter function generally only returns elements that satisfy a requirement given as a function, but if you pass None it will check if the value itself is True. Because an empty string is False and every other string is True it will remove every empty string from the list.
Also note I removed the escaping of special characters in the character class, as it is simply not neccessary and just makes the RegEx harder to read.
the first thing which comes to my mind is something like this:
>>> mystring = re.split(r"[\s\?\!\,\;]+", "Holy moly, feferoni!")
['Holy', 'moly', 'feferoni', '']
>>> mystring.pop(len(mystring)-1)
>>> print mystring
['Holy', 'moly', 'feferoni']
__import__('re').findall('[^\s?!,;]+', 'Holy moly, feferoni!')
Input
str = 'test1,test2,test3,'
Ouput
str = 'test1,test2,test3'
Requirement to strip the last occurence of ','
Just use rstrip().
result = your_string.rstrip(',')
str = 'test1,test2,test3,'
str[:-1] # 'test1,test2,test3'
The question is very old but tries to give the better answer
str = 'test1,test2,test3,'
It will check the last character, if the last character is a comma it will remove otherwise will return the original string.
result = str[:-1] if str[-1]==',' else str
Though it is little bit over work for something like that. I think this statement will help you.
str = 'test1,test2,test3,'
result = ','.join([s for s in str.split(',') if s]) # 'test1,test2,test3'
If you have to remove the last comma (also as the last character?) you can do this via the function removesuffix()
Here is an example:
>>>'hello,'.removesuffix(',')
'hello'
Actually we have to consider the worst case also.
The worst case is,
str= 'test1,test2,test3, ,,,, '
for above code, please use following code,
result = ','.join([s.strip() for s in str.split(',') if s.strip()!=''])
It will work/remove the prefix 'comma' also. For example,
str= ' , ,test1,test2,test3, ,,,, '
I have a string, example:
s = "this is a string, a"
Where a ',' (comma) will always be the 3rd to the last character, aka s[-3].
I am thinking of ways to remove the ',' but can only think of converting the string into a list, deleting it, and converting it back to a string. This however seems a bit too much for simple task.
How can I accomplish this in a simpler way?
Normally, you would just do:
s = s[:-3] + s[-2:]
The s[:-3] gives you a string up to, but not including, the comma you want removed ("this is a string") and the s[-2:] gives you another string starting one character beyond that comma (" a").
Then, joining the two strings together gives you what you were after ("this is a string a").
A couple of variants, using the "delete the last comma" rather than "delete third last character" are:
s[::-1].replace(",","",1)[::-1]
or
''.join(s.rsplit(",", 1))
But these are pretty ugly. Slightly better is:
a, _, b = s.rpartition(",")
s = a + b
This may be the best approach if you don't know the comma's position (except for last comma in string) and effectively need a "replace from right". However Anurag's answer is more pythonic for the "delete third last character".
Python strings are immutable. This means that you must create at least 1 new string in order to remove the comma, as opposed to editing the string in place in a language like C.
For deleting every ',' character in the text, you can try
s = s.split(',')
>> ["this is a string", " a"]
s = "".join(s)
>> "this is a string a"
Or in one line:
s0 = "".join(s.split(','))
The best simple way is : You can use replace function as :-
>>> s = 'this is a string, a'
>>> s = s.replace(',','')
>>> s
'this is a string a'
Here, replace() function search the character ',' and replace it by '' i.e. empty character
Note that , the replace() function defaults all ',' but if you want only replace some ',' in some case you can use : s.replace(',' , '', 1)
To slice a string of arbitrary length into multiple equal length slices of arbitrary length you could do
def slicer(string, slice_length):
return [string[i:i + slice_length]
for i in xrange(0, len(string), slice_length)]
If slice_length does not divide exactly into len(string) then there will be a single slice at the end of the list that holds the remainder.