This question already has answers here:
Split a string to even sized chunks
(9 answers)
Closed 8 years ago.
how can I separate a string: "Blahblahblahblah" into "Blah" "blah" "blah" "blah" on python. I've tried the following:
str = "Blahblahblahblah"
for letter[0:3] on str
How can I do it?
If you do not mind to use re library. In this example the regex .{4} means any character except \n of length 4.
import re
str = "Blahblahblahblah"
print re.findall(".{4}", str)
output:
['Blah', 'blah', 'blah', 'blah']
Note: str is not a very good name for a variable name. Because there is a function named str() in python that converts the given variable into a string.
Try:
>>> SUBSTR_LEN = 4
>>> string = "bla1bla2bla3bla4"
>>> [string[n:n + SUBSTR_LEN] for n in range(0, len(string), SUBSTR_LEN)]
['bla1', 'bla2', 'bla3', 'bla4']
Related
This question already has answers here:
Do regular expressions from the re module support word boundaries (\b)?
(5 answers)
Closed 4 years ago.
I have the following a string, I need to check if
the string contains App2 and iPhone,
but not App and iPhone
I wrote the following:
campaign_keywords = "App2 iPhone"
my_string = "[Love]App2 iPhone Argentina"
pattern = re.compile("r'\b" + campaign_keywords + "\b")
print pattern.search(my_string)
It prints None. Why?
The raw string notation is wrong, the r should not be inside the the quotes. and the second \b should also be a raw string.
The match function tries to match at the start of the string. You need to use search or findall
Difference between re.search and re.match
Example
>>> pattern = re.compile(r"\b" + campaign_keywords + r"\b")
>>> pattern.findall(my_string)
['App2 iPhone']
>>> pattern.match(my_string)
>>> pattern.search(my_string)
<_sre.SRE_Match object at 0x10ca2fbf8>
>>> match = pattern.search(my_string)
>>> match.group()
'App2 iPhone'
This question already has answers here:
Tuple to List - Python / PostgreSQL return type of SETOF Record
(2 answers)
Closed 8 years ago.
so I got this code:
from dosql import *
import cgi
import simplejson as json
import re
def index(req, userID):
userID = cgi.escape(userID)
get = doSql()
rec = get.execqry("select get_progressrecord('" + userID + "');",False)[0][0]
result = str(rec)
stringed = re.split(',', result)
return json.dumps(stringed)
And it returns this:
But I want to exclude the parenthesis "(" ")" too. How could I put multiple delimiters in the regex?
Using str.strip, you can remove surround characters specified:
>>> row = ["(178.00", "65.00", "20.52", "normal", "18", "0.00)"]
>>> [x.strip('(),') for x in row]
['178.00', '65.00', '20.52', 'normal', '18', '0.00']
BTW, if get.execqry(..) returns a tuple, string manipulation is not necessary.
a_tuple = get.execqry(....)
# or (if you want a list)
a_list = list(get.execqry(....))
Put a | between them:
stringed = re.split(',|(|)', result)
You can use a simple regex like this:
[,()]
Working demo
The idea is to match the characters you want using a regex class [...]. So, this will match commas or parentheses.
On the other hand, if you want to capture the following content:
(250.00", "612.00", "55.55", "normal", "1811", "0.00)
You could use something like this:
([\w.]+)
Working demo
This question already has answers here:
How can I split and parse a string in Python? [duplicate]
(3 answers)
Closed 8 years ago.
I have a file which contains each line in the following format
"('-1259656819525938837', 598679497)\t0.036787946" # "\t" within the string is the tab sign
I need to get the components out
-1259656819525938837 #string, it is the content within ' '
598679497 # long
0.036787946 # float
Python 2.6
You can use regular expressions from re module:
import re
s = "('-1259656819525938837', 598679497)\t0.036787946"
re.findall(r'[-+]?[0-9]*\.?[0-9]+', s)
% gives: ['-1259656819525938837', '598679497', '0.036787946']
"2.7.0_bf4fda703454".split("_") gives a list of strings:
In [1]: "2.7.0_bf4fda703454".split("_")
Out[1]: ['2.7.0', 'bf4fda703454']
This splits the string at every underscore. If you want it to stop after the first split, use "2.7.0_bf4fda703454".split("_", 1).
If you know for a fact that the string contains an underscore, you can even unpack the LHS and RHS into separate variables:
In [8]: lhs, rhs = "2.7.0_bf4fda703454".split("_", 1)
In [9]: lhs
Out[9]: '2.7.0'
In [10]: rhs
Out[10]: 'bf4fda703454'
You can use a regex to extract number and float from string:
>>> import re
>>> a = "('-1259656819525938837', 598679497)\t0.036787946"
>>> re.findall(r'[-?\d\.\d]+', a)
['-1259656819525938837', '598679497', '0.036787946']
This question already has answers here:
How to convert string to Title Case in Python?
(10 answers)
Closed 9 years ago.
I'm having trouble trying to create a function that can do this job. The objective is to convert strings like
one to One
hello_world to HelloWorld
foo_bar_baz to FooBarBaz
I know that the proper way to do this is using re.sub, but I'm having trouble creating the right regular expressions to do the job.
You can try something like this:
>>> s = 'one'
>>> filter(str.isalnum, s.title())
'One'
>>>
>>> s = 'hello_world'
>>> filter(str.isalnum, s.title())
'HelloWorld'
>>>
>>> s = 'foo_bar_baz'
>>> filter(str.isalnum, s.title())
'FooBarBaz'
Relevant documentation:
str.title()
str.isalnum()
filter()
Found solution:
def uppercase(name):
return ''.join(x for x in name.title() if not x.isspace()).replace('_', '')
This question already has answers here:
Python Trailing L Problem
(5 answers)
Closed 9 years ago.
I receive from a module a string that is a representation of an long int
>>> str = hex(5L)
>>> str
'0x5L'
What I now want is to convert the string str back to a number (integer)
int(str,16) does not work because of the L.
Is there a way to do this without stripping the last L out of the string? Because it is also possible that the string contains a hex without the L ?
Use str.rstrip; It works for both cases:
>>> int('0x5L'.rstrip('L'),16)
5
>>> int('0x5'.rstrip('L'),16)
5
Or generate the string this way:
>>> s = '{:#x}'.format(5L) # remove '#' if you don' want '0x'
>>> s
'0x5'
>>> int(s, 16)
5
You could even just use:
>>> str = hex(5L)
>>> long(str,16)
5L
>>> int(long(str,16))
5
>>>