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Suppose, I want to define a two-line (or a multi-line) string.
I can do this in two ways:
Using escape sequence for the newline character.\n
Ex: "This is the first sentence. \n This is the second sentence."
Using triple-quoted strings.
Ex: """ This is the first sentence.
This is the second sentence."""
Which is the more efficient or conventional ? Why ?
I'm tempted to say it doesn't matter since each one still scans inside for escaped characters while parsing the text.
>>> print "a\n\tb"
a
b
>>> print """a\n\tb"""
a
b
Related
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How can we reduce a string like haaaaaaapppppyyyyyy to haappyy
Such that repetition is allowed to a maximum of twice in a row for a character in a string?
including any character ( special characters also )
converting --------------------- to --
We can use a regex replacement:
inp = "haaaaaaapppppyyyyyy"
output = re.sub(r'(\w)\1{2,}', r'\1\1', inp)
print(output) # haappyy
The above logic matches any one character which is followed by itself two or more times. It then replaces with just two of the character.
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I am looking for a way to automatically extract dates from a string, but following each other without a delimiter
For example my string is: \n-\n24-04-201923-04-201922-04-201921-04-201920-04-201919-04-201918-04-2019
How can I get this output:
24-04-2019
23-04-2019
22-04-2019
21-04-2019
20-04-2019
19-04-2019
18-04-2019
Any help would be appreciated!
Given that they're all of equal length, you can just clear the \n's then use textwrap:
import textwrap
print(textwrap.wrap(my_string, 10))
You can remove \n's using strip():
my_string = my_string.strip()
You can use this code also.
string='\n-\n24-04-201923-04-201922-04-201921-04-201920-04-201919-04-201918-04-2019'
newStr=string[3:]
for char in range(0,len(newStr),10):
print newStr[char:char+10]
Here's the output
24-04-2019
23-04-2019
22-04-2019
21-04-2019
20-04-2019
19-04-2019
18-04-2019
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I have a list output:
['Go497f9te(40RAAC34)\n','G0THDU433(40RAAC33)\n']
and I want to clean it up in order to output:
[40RAAC34,40RAAC33]
If you have a string:
'hello (world)'
and want the text between the brackets, you can either use a regex:
import re
re.findall('\((.*?)\)', s)[0]
#'world'
or, if you are sure that there is only one set of brackets (i.e. no leading ) chars) then you can just use slicing:
s[s.index('(')+1:s.index(')')]
#'world'
So then you just need to throw this into a list-comprehension or similar.
l = ['Go497f9te(40RAAC34)\n','G0THDU433(40RAAC33)\n']
[s[s.index('(')+1:s.index(')')] for s in l]
#['40RAAC34', '40RAAC33']
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I took a python course back when i was in high school but now I barely remember anything about it. I'm bored today and though I should try some python exercises.
Example:
string = '3dc8uo8c33a v8c08oizl6ga'
The code needs to remove 3d 8u 8c ... ect
so that the
answer = 'coca cola'
Assuming the rule is "split the string along whitespace, then take every third letter of the words, and add them back together", you can use
>>> string = '3dc8uo8cc33a v8c08oizl6ga'
>>> " ".join("".join(s[2::3]) for s in string.split())
'coca cola'
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How can I replace all the " with the raw string \" in a string such as "Hello", said he. in Python?
s = '"Hello", said he.'
print s.replace('"', r'\"')
# output
\"Hello\", said he.
It helps to use the r'' notation to indicate that the string should be raw and not interpreted. Helps with backslashes.
Use replace().
>>> print '"Hello"'.replace('"', '\\"')
\"Hello\"