using regexp

[s99999999s2003]

hi
how can i use regexp to group these digits into groups of 3?

eg
line 123456789123456789

i have :

pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)

but this only gives the first 3. I also tried

"line\s+(\d{3})+"
but also not working.
I need output to be ['123' ,'456','789', '123','456','789', .....]
thanks.

[Shane Geiger]

You don't even need regex.

def
split_seq(seq,size):

# this is sort of the inverse of
flatten

# Source:
http://aspn.activestate.com/ASPN/Coo.../Recipe/425044

return [seq[i:i+size] for i in range(0, len(seq),
size)]

line =
'123456789123456789'

print
split_seq(line,3)

Will that work for you?

s9************@yahoo.com wrote:

hi
how can i use regexp to group these digits into groups of 3?

eg
line 123456789123456789

i have :

pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)

but this only gives the first 3. I also tried

"line\s+(\d{3})+"
but also not working.
I need output to be ['123' ,'456','789', '123','456','789', .....]
thanks.

--
Shane Geiger
IT Director
National Council on Economic Education
sg*****@ncee.net | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

[Shane Geiger]

import re
line = '123456789123456789'
print re.findall('([0-9]{3})', line)

Shane Geiger wrote:

You don't even need regex.

def
split_seq(seq,size):

# this is sort of the inverse of
flatten

# Source:
http://aspn.activestate.com/ASPN/Coo.../Recipe/425044

return [seq[i:i+size] for i in range(0, len(seq),
size)]

line =
'123456789123456789'

print
split_seq(line,3)

Will that work for you?

s9************@yahoo.com wrote:

>hi
how can i use regexp to group these digits into groups of 3?

eg
line 123456789123456789

i have :

pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)

but this only gives the first 3. I also tried

"line\s+(\d{3})+"
but also not working.
I need output to be ['123' ,'456','789', '123','456','789', .....]
thanks.

--
Shane Geiger
IT Director
National Council on Economic Education
sg*****@ncee.net | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

[s99999999s2003]

On Mar 20, 1:57 pm, Shane Geiger <[e-mail]:

import re
line = '123456789123456789'
print re.findall('([0-9]{3})', line)

Shane Geiger wrote:

You don't even need regex.

def
split_seq(seq,size):

# this is sort of the inverse of
flatten

# Source:
http://aspn.activestate.com/ASPN/Coo.../Recipe/425044

return [seq[i:i+size] for i in range(0, len(seq),
size)]

line =
'123456789123456789'

print
split_seq(line,3)

Will that work for you?

[e-mail] wrote:

hi
how can i use regexp to group these digits into groups of 3?

eg
line 123456789123456789

i have :

pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)

but this only gives the first 3. I also tried

"line\s+(\d{3})+"
but also not working.
I need output to be ['123' ,'456','789', '123','456','789', .....]
thanks.

--
Shane Geiger
IT Director
National Council on Economic Education
[e-mail] | 402-438-8958 | http://www.ncee.net

Leading the Campaign for Economic and Financial Literacy

sgeiger.vcf
1KDownload

hi
thanks.
I know it can be done without regexp, but its just for learning
purpose, i want to use regexp
eg
someword 123456789123456789 #this is a line of a file. 'someword' is a
word before the digits
I need to get the digits into groups of 3 based on whether i found
'someword'.
so i use this pattern :
"someword\s+(\d{3})"
but this will on give me the first 3. I need to group them all.
hope i explain it clearly.

[attn.steven.kuo]

On Mar 19, 10:33 pm, [e-mail] wrote:

hi
how can i use regexp to group these digits into groups of 3?

eg
line 123456789123456789

i have :

pat = re.compile("line\s+(\d{3})" , re.M|re.DOTALL)

but this only gives the first 3. I also tried

"line\s+(\d{3})+"
but also not working.
I need output to be ['123' ,'456','789', '123','456','789', .....]
thanks.

Try:

import re

target_string = """not 123456789 but
try this line 987654321"""

try:
digits = re.compile(r'line\s+(\d
+)').search(target_string).group(1)
except AttributeError:
digits = ''

three_at_a_time = re.compile(r'\d{3}').findall(digits)
print three_at_a_time

--
Hope this helps,
Steven