I am aware .capitalize() capitalizes the first letter of a string but what if the first character is a integer?
this
1bob
5sandy
to this
1Bob
5Sandy
Only because no one else has mentioned it:
>>> 'bob'.title()
'Bob'
>>> 'sandy'.title()
'Sandy'
>>> '1bob'.title()
'1Bob'
>>> '1sandy'.title()
'1Sandy'
However, this would also give
>>> '1bob sandy'.title()
'1Bob Sandy'
>>> '1JoeBob'.title()
'1Joebob'
i.e. it doesn't just capitalize the first alphabetic character. But then .capitalize() has the same issue, at least in that 'joe Bob'.capitalize() == 'Joe bob', so meh.
If the first character is an integer, it will not capitalize the first letter.
>>> '2s'.capitalize()
'2s'
If you want the functionality, strip off the digits, you can use '2'.isdigit() to check for each character.
>>> s = '123sa'
>>> for i, c in enumerate(s):
... if not c.isdigit():
... break
...
>>> s[:i] + s[i:].capitalize()
'123Sa'
This is similar to #Anon's answer in that it keeps the rest of the string's case intact, without the need for the re module.
def sliceindex(x):
i = 0
for c in x:
if c.isalpha():
i = i + 1
return i
i = i + 1
def upperfirst(x):
i = sliceindex(x)
return x[:i].upper() + x[i:]
x = '0thisIsCamelCase'
y = upperfirst(x)
print(y)
# 0ThisIsCamelCase
As #Xan pointed out, the function could use more error checking (such as checking that x is a sequence - however I'm omitting edge cases to illustrate the technique)
Updated per #normanius comment (thanks!)
Thanks to #GeoStoneMarten in pointing out I didn't answer the question! -fixed that
Here is a one-liner that will uppercase the first letter and leave the case of all subsequent letters:
import re
key = 'wordsWithOtherUppercaseLetters'
key = re.sub('([a-zA-Z])', lambda x: x.groups()[0].upper(), key, 1)
print key
This will result in WordsWithOtherUppercaseLetters
As seeing here answered by Chen Houwu, it's possible to use string package:
import string
string.capwords("they're bill's friends from the UK")
>>>"They're Bill's Friends From The Uk"
a one-liner: ' '.join(sub[:1].upper() + sub[1:] for sub in text.split(' '))
You can replace the first letter (preceded by a digit) of each word using regex:
re.sub(r'(\d\w)', lambda w: w.group().upper(), '1bob 5sandy')
output:
1Bob 5Sandy
def solve(s):
for i in s[:].split():
s = s.replace(i, i.capitalize())
return s
This is the actual code for work. .title() will not work at '12name' case
I came up with this:
import re
regex = re.compile("[A-Za-z]") # find a alpha
str = "1st str"
s = regex.search(str).group() # find the first alpha
str = str.replace(s, s.upper(), 1) # replace only 1 instance
print str
def solve(s):
names = list(s.split(" "))
return " ".join([i.capitalize() for i in names])
Takes a input like your name: john doe
Returns the first letter capitalized.(if first character is a number, then no capitalization occurs)
works for any name length
Related
I am new to Python and I have a String, I want to extract the numbers from the string. For example:
str1 = "3158 reviews"
print (re.findall('\d+', str1 ))
Output is ['4', '3']
I want to get 3158 only, as an Integer preferably, not as List.
You can filter the string by digits using str.isdigit method,
>>> int(filter(str.isdigit, str1))
3158
For Python3:
int(list(filter(str.isdigit, my_str))[0])
This code works fine. There is definitely some other problem:
>>> import re
>>> str1 = "3158 reviews"
>>> print (re.findall('\d+', str1 ))
['3158']
Your regex looks correct. Are you sure you haven't made a mistake with the variable names? In your code above you mixup total_hotel_reviews_string and str.
>>> import re
>>> s = "3158 reviews"
>>>
>>> print(re.findall("\d+", s))
['3158']
IntVar = int("".join(filter(str.isdigit, StringVar)))
You were quite close to the final answer. Your re.finadall expression was only missing the enclosing parenthesis to catch all detected numbers:
re.findall( '(\d+)', str1 )
For a more general string like str1 = "3158 reviews, 432 users", this code would yield:
Output: ['3158', '432']
Now to obtain integers, you can map the int function to convert strings into integers:
A = list(map(int,re.findall('(\d+)',str1)))
Alternatively, you can use this one-liner loop:
A = [ int(x) for x in re.findall('(\d+)',str1) ]
Both methods are equally correct. They yield A = [3158, 432].
Your final result for the original question would be first entry in the array A, so we arrive at any of these expressions:
result = list(map(int,re.findall( '(\d+)' , str1 )))[0]
result = int(re.findall( '(\d+)' , str1 )[0])
Even if there is only one number present in str1, re.findall will still return a list, so you need to retrieve the first element A[0] manually.
To extract a single number from a string you can use re.search(), which returns the first match (or None):
>>> import re
>>> string = '3158 reviews'
>>> int(re.search(r'\d+', string).group(0))
3158
In Python 3.6+ you can also index into a match object instead of using group():
>>> int(re.search(r'\d+', string)[0])
3158
If the format is that simple (a space separates the number from the rest) then
int(str1.split()[0])
would do it
Best for every complex types
str1 = "sg-23.0 300sdf343fc -34rrf-3.4r" #All kinds of occurrence of numbers between strings
num = [float(s) for s in re.findall(r'-?\d+\.?\d*', str1)]
print(num)
Output:
[-23.0, 300.0, 343.0, -34.0, -3.4]
Above solutions seem to assume integers. Here's a minor modification to allow decimals:
num = float("".join(filter(lambda d: str.isdigit(d) or d == '.', inputString)
(Doesn't account for - sign, and assumes any period is properly placed in digit string, not just some english-language period lying around. It's not built to be indestructible, but worked for my data case.)
Python 2.7:
>>> str1 = "3158 reviews"
>>> int(filter(str.isdigit, str1))
3158
Python 3:
>>> str1 = "3158 reviews"
>>> int(''.join(filter(str.isdigit, str1)))
3158
There may be a little problem with code from Vishnu's answer. If there is no digits in the string it will return ValueError. Here is my suggestion avoid this:
>>> digit = lambda x: int(filter(str.isdigit, x) or 0)
>>> digit('3158 reviews')
3158
>>> digit('reviews')
0
For python3
input_str = '21ddd3322'
int(''.join(filter(str.isdigit, input_str)))
> 213322
a = []
line = "abcd 3455 ijkl 56.78 ij"
for word in line.split():
try:
a.append(float(word))
except ValueError:
pass
print(a)
OUTPUT
3455.0 56.78
I am a beginner in coding. This is my attempt to answer the questions. Used Python3.7 version without importing any libraries.
This code extracts and returns a decimal number from a string made of sets of characters separated by blanks (words).
Attention: In case there are more than one number, it returns the last value.
line = input ('Please enter your string ')
for word in line.split():
try:
a=float(word)
print (a)
except ValueError:
pass
My answer does not require any additional libraries, and it's easy to understand. But you have to notice that if there's more than one number inside a string, my code will concatenate them together.
def search_number_string(string):
index_list = []
del index_list[:]
for i, x in enumerate(string):
if x.isdigit() == True:
index_list.append(i)
start = index_list[0]
end = index_list[-1] + 1
number = string[start:end]
return number
#Use this, THIS IS FOR EXTRACTING NUMBER FROM STRING IN GENERAL.
#To get all the numeric occurences.
*split function to convert string to list and then the list comprehension
which can help us iterating through the list
and is digit function helps to get the digit out of a string.
getting number from string
use list comprehension+isdigit()
test_string = "i have four ballons for 2 kids"
print("The original string : "+ test_string)
# list comprehension + isdigit() +split()
res = [int(i) for i in test_string.split() if i.isdigit()]
print("The numbers list is : "+ str(res))
#To extract numeric values from a string in python
*Find list of all integer numbers in string separated by lower case characters using re.findall(expression,string) method.
*Convert each number in form of string into decimal number and then find max of it.
import re
def extractMax(input):
# get a list of all numbers separated by lower case characters
numbers = re.findall('\d+',input)
# \d+ is a regular expression which means one or more digit
number = map(int,numbers)
print max(numbers)
if __name__=="__main__":
input = 'sting'
extractMax(input)
you can use the below method to extract all numbers from a string.
def extract_numbers_from_string(string):
number = ''
for i in string:
try:
number += str(int(i))
except:
pass
return number
(OR) you could use i.isdigit() or i.isnumeric(in Python 3.6.5 or above)
def extract_numbers_from_string(string):
number = ''
for i in string:
if i.isnumeric():
number += str(int(i))
return number
a = '343fdfd3'
print (extract_numbers_from_string(a))
# 3433
Using a list comprehension and Python 3:
>>> int("".join([c for c in str1 if str.isdigit(c)]))
3158
I have just started to learn python and would like help using string.replace(x,y).
Specifically, replacing all to X's and x's depending whether the letter was originally capitalized or not.
e.g.
John S. Smith -> Xxxx X. Xxxxx
What I have created currently is below.
print("Enter text to translate:", end ="")
sentence = input ()
replaced = sentence.replace("", "x")
print(replaced)
However when I input text like "John Smith". I am returned with "xJxoxhxnx xSx.x xSxmxixtxhx".
Thank you in advance!
Edit: Although string.replace(x,y) may be longer to perform, I'd like to slowly build on my knowledge before finding faster and shorter ways to perform the same operation. I'd highly appreciate it if it was explained in terms of string.replace(x, y) instead of re.sub
Edit2: I have been notified that string.replace is the wrong tool to use. Thank you for your help! I will be reading into re.sub instead.
If you insist on using replace even though it's the wrong tool for the job (because it can only replace one letter at a time and has to go through the whole string every time), here's a way:
>>> s = 'John S. Smith'
>>> for c in s:
if c.islower():
s = s.replace(c, 'x')
if c.isupper():
s = s.replace(c, 'X')
>>> s
'Xxxx X. Xxxxx'
And a somewhat neat more efficient way:
>>> ''.join('x' * c.islower() or 'X' * c.isupper() or c for c in s)
'Xxxx X. Xxxxx'
And a regex way:
>>> re.sub('[A-Z]', 'X', re.sub('[a-z]', 'x', s))
'Xxxx X. Xxxxx'
import re
print("Enter text to translate:", end ="")
sentence = input()
replaced = re.sub("[A-Z]", 'X', re.sub("[a-z]", 'x', sentence))
print replaced
Use re.sub to replace individual character of string or iterate through the string.
Not the correct usecase for string replace.
There are 2 things that you can do:
Loop through the string and perform the operation
Use re.sub to replace using regex. (How to input a regex in string.replace?)
Plain way:
>>> my_string = "John S. Smith"
>>> replaced = ''
>>> for character in my_string:
if character.isupper():
replaced += 'X'
elif character.islower():
replaced += 'x'
else:
replaced += character
>>> replaced
'Xxxx X. Xxxxx'
One liner:
>>> ''.join('x' if c.islower() else 'X' if c.isupper() else c for c in my_string)
'Xxxx X. Xxxxx'
print("Enter text to translate:", end ="")
sentence = input ()
replaced = ''.join(['x' if (i>='a' and i<='z') else 'X' if (i>='A' and i<='Z') else i for i in sentence])
print(replaced)
You can use the re module, and provide re.sub with a callback function to achieve this.
import re
def get_capital(ch):
if ch.isupper():
return "X"
return "x"
def get_capitals(s):
return re.sub("[a-zA-Z]", lambda ch: get_capital(ch.group(0)), s)
print(get_capitals("John S. Smith"))
Has the output:
Xxxx X. Xxxxx
As a matter of style, I've not condensed any of this into lambdas or one-liners, although you probably could.
This will be significantly faster than just repeated concatenation.
** edit fixed code **
Try this instead, it is a bit of a simple and easily breakable example but this is what you would need to achieve what you want:
s = "Testing This"
r = ""
for c in s:
if c.isalpha():
if c.islower():
r += "x"
else:
r += "X"
else:
r += c
print(r)
I am new to Python and I have a String, I want to extract the numbers from the string. For example:
str1 = "3158 reviews"
print (re.findall('\d+', str1 ))
Output is ['4', '3']
I want to get 3158 only, as an Integer preferably, not as List.
You can filter the string by digits using str.isdigit method,
>>> int(filter(str.isdigit, str1))
3158
For Python3:
int(list(filter(str.isdigit, my_str))[0])
This code works fine. There is definitely some other problem:
>>> import re
>>> str1 = "3158 reviews"
>>> print (re.findall('\d+', str1 ))
['3158']
Your regex looks correct. Are you sure you haven't made a mistake with the variable names? In your code above you mixup total_hotel_reviews_string and str.
>>> import re
>>> s = "3158 reviews"
>>>
>>> print(re.findall("\d+", s))
['3158']
IntVar = int("".join(filter(str.isdigit, StringVar)))
You were quite close to the final answer. Your re.finadall expression was only missing the enclosing parenthesis to catch all detected numbers:
re.findall( '(\d+)', str1 )
For a more general string like str1 = "3158 reviews, 432 users", this code would yield:
Output: ['3158', '432']
Now to obtain integers, you can map the int function to convert strings into integers:
A = list(map(int,re.findall('(\d+)',str1)))
Alternatively, you can use this one-liner loop:
A = [ int(x) for x in re.findall('(\d+)',str1) ]
Both methods are equally correct. They yield A = [3158, 432].
Your final result for the original question would be first entry in the array A, so we arrive at any of these expressions:
result = list(map(int,re.findall( '(\d+)' , str1 )))[0]
result = int(re.findall( '(\d+)' , str1 )[0])
Even if there is only one number present in str1, re.findall will still return a list, so you need to retrieve the first element A[0] manually.
To extract a single number from a string you can use re.search(), which returns the first match (or None):
>>> import re
>>> string = '3158 reviews'
>>> int(re.search(r'\d+', string).group(0))
3158
In Python 3.6+ you can also index into a match object instead of using group():
>>> int(re.search(r'\d+', string)[0])
3158
If the format is that simple (a space separates the number from the rest) then
int(str1.split()[0])
would do it
Best for every complex types
str1 = "sg-23.0 300sdf343fc -34rrf-3.4r" #All kinds of occurrence of numbers between strings
num = [float(s) for s in re.findall(r'-?\d+\.?\d*', str1)]
print(num)
Output:
[-23.0, 300.0, 343.0, -34.0, -3.4]
Above solutions seem to assume integers. Here's a minor modification to allow decimals:
num = float("".join(filter(lambda d: str.isdigit(d) or d == '.', inputString)
(Doesn't account for - sign, and assumes any period is properly placed in digit string, not just some english-language period lying around. It's not built to be indestructible, but worked for my data case.)
Python 2.7:
>>> str1 = "3158 reviews"
>>> int(filter(str.isdigit, str1))
3158
Python 3:
>>> str1 = "3158 reviews"
>>> int(''.join(filter(str.isdigit, str1)))
3158
There may be a little problem with code from Vishnu's answer. If there is no digits in the string it will return ValueError. Here is my suggestion avoid this:
>>> digit = lambda x: int(filter(str.isdigit, x) or 0)
>>> digit('3158 reviews')
3158
>>> digit('reviews')
0
For python3
input_str = '21ddd3322'
int(''.join(filter(str.isdigit, input_str)))
> 213322
a = []
line = "abcd 3455 ijkl 56.78 ij"
for word in line.split():
try:
a.append(float(word))
except ValueError:
pass
print(a)
OUTPUT
3455.0 56.78
I am a beginner in coding. This is my attempt to answer the questions. Used Python3.7 version without importing any libraries.
This code extracts and returns a decimal number from a string made of sets of characters separated by blanks (words).
Attention: In case there are more than one number, it returns the last value.
line = input ('Please enter your string ')
for word in line.split():
try:
a=float(word)
print (a)
except ValueError:
pass
My answer does not require any additional libraries, and it's easy to understand. But you have to notice that if there's more than one number inside a string, my code will concatenate them together.
def search_number_string(string):
index_list = []
del index_list[:]
for i, x in enumerate(string):
if x.isdigit() == True:
index_list.append(i)
start = index_list[0]
end = index_list[-1] + 1
number = string[start:end]
return number
#Use this, THIS IS FOR EXTRACTING NUMBER FROM STRING IN GENERAL.
#To get all the numeric occurences.
*split function to convert string to list and then the list comprehension
which can help us iterating through the list
and is digit function helps to get the digit out of a string.
getting number from string
use list comprehension+isdigit()
test_string = "i have four ballons for 2 kids"
print("The original string : "+ test_string)
# list comprehension + isdigit() +split()
res = [int(i) for i in test_string.split() if i.isdigit()]
print("The numbers list is : "+ str(res))
#To extract numeric values from a string in python
*Find list of all integer numbers in string separated by lower case characters using re.findall(expression,string) method.
*Convert each number in form of string into decimal number and then find max of it.
import re
def extractMax(input):
# get a list of all numbers separated by lower case characters
numbers = re.findall('\d+',input)
# \d+ is a regular expression which means one or more digit
number = map(int,numbers)
print max(numbers)
if __name__=="__main__":
input = 'sting'
extractMax(input)
you can use the below method to extract all numbers from a string.
def extract_numbers_from_string(string):
number = ''
for i in string:
try:
number += str(int(i))
except:
pass
return number
(OR) you could use i.isdigit() or i.isnumeric(in Python 3.6.5 or above)
def extract_numbers_from_string(string):
number = ''
for i in string:
if i.isnumeric():
number += str(int(i))
return number
a = '343fdfd3'
print (extract_numbers_from_string(a))
# 3433
Using a list comprehension and Python 3:
>>> int("".join([c for c in str1 if str.isdigit(c)]))
3158
I need help in trying to write a certain part of a program.
The idea is that a person would input a bunch of gibberish and the program will read it till it reaches an "!" (exclamation mark) so for example:
input("Type something: ")
Person types: wolfdo65gtornado!salmontiger223
If I ask the program to print the input it should only print wolfdo65gtornado and cut anything once it reaches the "!" The rest of the program is analyzing and counting the letters, but those part I already know how to do. I just need help with the first part. I been trying to look through the book but it seems I'm missing something.
I'm thinking, maybe utilizing a for loop and then placing restriction on it but I can't figure out how to make the random imputed string input be analyzed for a certain character and then get rid of the rest.
If you could help, I'll truly appreciate it. Thanks!
The built-in str.partition() method will do this for you. Unlike str.split() it won't bother to cut the rest of the str into different strs.
text = raw_input("Type something:")
left_text = text.partition("!")[0]
Explanation
str.partition() returns a 3-tuple containing the beginning, separator, and end of the string. The [0] gets the first item which is all you want in this case. Eg.:
"wolfdo65gtornado!salmontiger223".partition("!")
returns
('wolfdo65gtornado', '!', 'salmontiger223')
>>> s = "wolfdo65gtornado!salmontiger223"
>>> s.split('!')[0]
'wolfdo65gtornado'
>>> s = "wolfdo65gtornadosalmontiger223"
>>> s.split('!')[0]
'wolfdo65gtornadosalmontiger223'
if it doesnt encounter a "!" character, it will just grab the entire text though. if you would like to output an error if it doesn't match any "!" you can just do like this:
s = "something!something"
if "!" in s:
print "there is a '!' character in the context"
else:
print "blah, you aren't using it right :("
You want itertools.takewhile().
>>> s = "wolfdo65gtornado!salmontiger223"
>>> '-'.join(itertools.takewhile(lambda x: x != '!', s))
'w-o-l-f-d-o-6-5-g-t-o-r-n-a-d-o'
>>> s = "wolfdo65gtornado!salmontiger223!cvhegjkh54bgve8r7tg"
>>> i = iter(s)
>>> '-'.join(itertools.takewhile(lambda x: x != '!', i))
'w-o-l-f-d-o-6-5-g-t-o-r-n-a-d-o'
>>> '-'.join(itertools.takewhile(lambda x: x != '!', i))
's-a-l-m-o-n-t-i-g-e-r-2-2-3'
>>> '-'.join(itertools.takewhile(lambda x: x != '!', i))
'c-v-h-e-g-j-k-h-5-4-b-g-v-e-8-r-7-t-g'
Try this:
s = "wolfdo65gtornado!salmontiger223"
m = s.index('!')
l = s[:m]
To explain accepted answer.
Splitting
partition() function splits string in list with 3 elements:
mystring = "123splitABC"
x = mystring.partition("split")
print(x)
will give:
('123', 'split', 'ABC')
Access them like list elements:
print (x[0]) ==> 123
print (x[1]) ==> split
print (x[2]) ==> ABC
Suppose we have:
s = "wolfdo65gtornado!salmontiger223" + some_other_string
s.partition("!")[0] and s.split("!")[0] are both a problem if some_other_string contains a million strings, each a million characters long, separated by exclamation marks. I recommend the following instead. It's much more efficient.
import itertools as itts
get_start_of_string = lambda stryng, last, *, itts=itts:\
str(itts.takewhile(lambda ch: ch != last, stryng))
###########################################################
s = "wolfdo65gtornado!salmontiger223"
start_of_string = get_start_of_string(s, "!")
Why the itts=itts
Inside of the body of a function, such as get_start_of_string, itts is global.
itts is evaluated when the function is called, not when the function is defined.
Consider the following example:
color = "white"
get_fleece_color = lambda shoop: shoop + ", whose fleece was as " + color + " as snow."
print(get_fleece_color("Igor"))
# [... many lines of code later...]
color = "pink polka-dotted"
print(get_fleece_color("Igor's cousin, 3 times removed"))
The output is:
Igor, whose fleece was white as snow.
Igor's cousin, 3 times removed Igor, whose fleece was as pink polka-dotted as snow.
You can extract the beginning of a string, up until the first delimiter is encountered, by using regular expressions.
import re
slash_if_special = lambda ch:\
"\\" if ch in "\\^$.|?*+()[{" else ""
prefix_slash_if_special = lambda ch, *, _slash=slash_if_special: \
_slash(ch) + ch
make_pattern_from_char = lambda ch, *, c=prefix_slash_if_special:\
"^([^" + c(ch) + "]*)"
def get_string_up_untill(x_stryng, x_ch):
i_stryng = str(x_stryng)
i_ch = str(x_ch)
assert(len(i_ch) == 1)
pattern = make_pattern_from_char(ch)
m = re.match(pattern, x_stryng)
return m.groups()[0]
An example of the code above being used:
s = "wolfdo65gtornado!salmontiger223"
result = get_string_up_untill(s, "!")
print(result)
# wolfdo65gtornado
We can use itertools
s = "wolfdo65gtornado!salmontiger223"
result = "".join(itertools.takewhile(lambda x : x!='!' , s))
>>"wolfdo65gtornado"
I have a string s with nested brackets: s = "AX(p>q)&E((-p)Ur)"
I want to remove all characters between all pairs of brackets and store in a new string like this: new_string = AX&E
i tried doing this:
p = re.compile("\(.*?\)", re.DOTALL)
new_string = p.sub("", s)
It gives output: AX&EUr)
Is there any way to correct this, rather than iterating each element in the string?
Another simple option is removing the innermost parentheses at every stage, until there are no more parentheses:
p = re.compile("\([^()]*\)")
count = 1
while count:
s, count = p.subn("", s)
Working example: http://ideone.com/WicDK
You can just use string manipulation without regular expression
>>> s = "AX(p>q)&E(qUr)"
>>> [ i.split("(")[0] for i in s.split(")") ]
['AX', '&E', '']
I leave it to you to join the strings up.
>>> import re
>>> s = "AX(p>q)&E(qUr)"
>>> re.compile("""\([^\)]*\)""").sub('', s)
'AX&E'
Yeah, it should be:
>>> import re
>>> s = "AX(p>q)&E(qUr)"
>>> p = re.compile("\(.*?\)", re.DOTALL)
>>> new_string = p.sub("", s)
>>> new_string
'AX&E'
Nested brackets (or tags, ...) are something that are not possible to handle in a general way using regex. See http://www.amazon.de/Mastering-Regular-Expressions-Jeffrey-Friedl/dp/0596528124/ref=sr_1_1?ie=UTF8&s=gateway&qid=1304230523&sr=8-1-spell for details why. You would need a real parser.
It's possible to construct a regex which can handle two levels of nesting, but they are already ugly, three levels will already be quite long. And you don't want to think about four levels. ;-)
You can use PyParsing to parse the string:
from pyparsing import nestedExpr
import sys
s = "AX(p>q)&E((-p)Ur)"
expr = nestedExpr('(', ')')
result = expr.parseString('(' + s + ')').asList()[0]
s = ''.join(filter(lambda x: isinstance(x, str), result))
print(s)
Most code is from: How can a recursive regexp be implemented in python?
You could use re.subn():
import re
s = 'AX(p>q)&E((-p)Ur)'
while True:
s, n = re.subn(r'\([^)(]*\)', '', s)
if n == 0:
break
print(s)
Output
AX&E
this is just how you do it:
# strings
# double and single quotes use in Python
"hey there! welcome to CIP"
'hey there! welcome to CIP'
"you'll understand python"
'i said, "python is awesome!"'
'i can\'t live without python'
# use of 'r' before string
print(r"\new code", "\n")
first = "code in"
last = "python"
first + last #concatenation
# slicing of strings
user = "code in python!"
print(user)
print(user[5]) # print an element
print(user[-3]) # print an element from rear end
print(user[2:6]) # slicing the string
print(user[:6])
print(user[2:])
print(len(user)) # length of the string
print(user.upper()) # convert to uppercase
print(user.lstrip())
print(user.rstrip())
print(max(user)) # max alphabet from user string
print(min(user)) # min alphabet from user string
print(user.join([1,2,3,4]))
input()