This question already has answers here:
Process escape sequences in a string in Python
(8 answers)
Closed 7 months ago.
I have an array of strings which looks like:
["U0001f308", "U0001F602"]
I need to add “\” in front of the first letter U so the output will be like:
["\U0001f308", "\U0001F602"]
This is the code I have tried so far:
matches = ["U0001f308", "U0001F602"]
emojis = [emoji.replace('U', r"\U") for emoji in matches]
print(emojis) #this prints ['\\U0001f308', '\\U0001F602'] which has two blacklashes
How can i add only one backslash in front of every string?
I guess what you want is the following code:
matches = ["U0001f308", "U0001F602"]
emojis = [emoji.replace('U', r"\U").encode().decode('unicode-escape') for emoji in matches]
print(emojis)
which prints
['🌈', '😂']
It's the same result as when we execute the following code:
print(["\U0001f308", "\U0001F602"])
Related
This question already has answers here:
Python split() without removing the delimiter [duplicate]
(4 answers)
Closed 11 months ago.
Is it possible to separate the string "a!b!" into two strings "a!" and "b!" and store that in a list? I have tried the split() function (and even with the delimiter "!"), but it doesn't seem to give me the right result that I want. Also, the character "!" could be any character.
How about :
string = 'a!ab!b!'
deliminator = '!'
word_list = [section+deliminator for section in string.split(deliminator) if section]
print(word_list)
Output :
['a!', 'ab!', 'b!']
split() is used when you need to seperate a string with particular character. If you want split a string into half, Try this
s = "a!b!"
l = [s[ : len(s)//2], s[len(s)//2 : ]]
# output : ["a!", "b!"]
This question already has answers here:
get index of character in python list
(4 answers)
Regular expression to match a dot
(7 answers)
Closed 3 years ago.
I want to find the position of '.', but when i run code below:
text = 'Hello world.'
pattern = '.'
search = re.search(pattern,text)
print(search.start())
print(search.end())
Output is:
0
1
Place of '.' isn't 0 1.
So why is it giving wrong output?
You can use find method for this task.
my_string = "test"
s_position = my_string.find('s')
print (s_position)
Output
2
If you really want to use RegEx be sure to escape the dot character or it will be interpreted as a special character.
The dot in RegEx matches any character except the newline symbol.
text = 'Hello world.'
pattern = '\.'
search = re.search(pattern,text)
print(search.start())
print(search.end())
This question already has answers here:
How do I split a string into a list of characters?
(15 answers)
Closed 5 years ago.
How would you go about separating each character in a given input and turn it into a list?
For example I have
import string
print ("Enter string")
x = input("")
Enter string
The quick brown
I want the end result to be
['T','h','e',' ','q','u','i','c','k',' ','b','r','o','w','n']
Y'know, to turn every character as a separate string in a list instead of every word as a separate string.
Thanks!
Simply use list(x) where x is the string.
This question already has answers here:
Split a string at uppercase letters
(22 answers)
Closed 6 years ago.
In general I have a string say
temp = "ProgramFields"
Now I want to split strings like these into two terms(I can identify tow strings based on uppercase character)
term1 = "Program"
term2 = "Field"
How to achieve this in python?
I tried regular expression and splitting terms but nothing gave me the result that I expected
Python code -
re.split("[A-Z][a-z]*","ProgramField")
Any suggestions?
You have to include groups:
re.split('([A-Z][a-z]*)', 'ProgramField)
This question already has answers here:
re.sub replace with matched content
(4 answers)
Closed 8 years ago.
I want to replace the digits in the middle of telephone with regex but failed. Here is my code:
temp= re.sub(r'1([0-9]{1}[0-9])[0-9]{4}([0-9]{4})', repl=r'$1****$2', tel_phone)
print temp
In the output, it always shows:
$1****$2
But I want to show like this: 131****1234. How to accomplish it ? Thanks
I think you're trying to replace four digits present in the middle (four digits present before the last four digits) with ****
>>> s = "13111111234"
>>> temp= re.sub(r'^(1[0-9]{2})[0-9]{4}([0-9]{4})$', r'\1****\2', s)
>>> print temp
131****1234
You might have seen $1 in replacement string in other languages. However, in Python, use \1 instead of $1. For correctness, you also need to include the starting 1 in the first capturing group, so that the output also include the starting 1; otherwise, the starting 1 will be lost.