Fixed length lists from .split()?

[Bob Greschke]

I'm reading a file that has lines like

bcsn; 1000000; 1223
bcsn; 1000001; 1456
bcsn; 1000003
bcsn; 1000010; 4567

The problem is the line with only the one semi-colon.
Is there a fancy way to get Parts=Line.spli t(";") to make Parts always
have three items in it, or do I just have to check the length of Parts
and loop to add the required missing items (this one would just take
Parts+=[""], but there are other types of lines in the file that have
about 10 "fields" that also have this problem)?

Thanks!

Bob

[Duncan Booth]

Bob Greschke <bo*@passcal.nm t.eduwrote:

Is there a fancy way to get Parts=Line.spli t(";") to make Parts always
have three items in it, or do I just have to check the length of Parts
and loop to add the required missing items (this one would just take
Parts+=[""], but there are other types of lines in the file that have
about 10 "fields" that also have this problem)?

>>def nsplit(s, sep, n):

return (s.split(sep) + [""]*n)[:n]

>>nsplit("bcs n; 1000001; 1456", ";", 3)

['bcsn', ' 1000001', ' 1456']

>>nsplit("bcs n; 1000001", ";", 3)

['bcsn', ' 1000001', '']

>>>

[Bob Greschke]

On 2007-01-26 11:13:56 -0700, Duncan Booth <du**********@i nvalid.invalids aid:

Bob Greschke <bo*@passcal.nm t.eduwrote:

>Is there a fancy way to get Parts=Line.spli t(";") to make Parts always
have three items in it, or do I just have to check the length of Parts
and loop to add the required missing items (this one would just take
Parts+=[""], but there are other types of lines in the file that have
about 10 "fields" that also have this problem)?

>>>def nsplit(s, sep, n):

return (s.split(sep) + [""]*n)[:n]

>>>nsplit("bcsn ; 1000001; 1456", ";", 3)

['bcsn', ' 1000001', ' 1456']

>>>nsplit("bcsn ; 1000001", ";", 3)

['bcsn', ' 1000001', '']

That's fancy enough. :) I didn't know you could do [""]*n. I never
thought about it before.

Thanks!

Bob

[bearophileHUGS]

Duncan Booth:

def nsplit(s, sep, n):
return (s.split(sep) + [""]*n)[:n]

Another version, longer:

from itertools import repeat

def nsplit(text, sep, n):
"""

>>nsplit("bcs n; 1000001; 1456", ";", 3)

['bcsn', ' 1000001', ' 1456']

>>nsplit("bcs n; 1000001", ";", 3)

['bcsn', ' 1000001', '']

>>nsplit("bcsn" , ";", 3)

['bcsn', '', '']

>>nsplit("", ".", 4)

['', '', '', '']

>>nsplit("ab.ac .ad.ae", ".", 2)

['ab', 'ac', 'ad', 'ae']
"""
result = text.split(sep)
nparts = len(result)
result.extend(r epeat("", n-nparts))
return result

if __name__ == "__main__":
import doctest
doctest.testmod ()

Bye,
bearophile

[Steven Bethard]

On Jan 26, 11:07 am, Bob Greschke <b...@passcal.n mt.eduwrote:

I'm reading a file that has lines like

bcsn; 1000000; 1223
bcsn; 1000001; 1456
bcsn; 1000003
bcsn; 1000010; 4567

The problem is the line with only the one semi-colon.
Is there a fancy way to get Parts=Line.spli t(";") to make Parts always
have three items in it

In Python 2.5 you can use the .partition() method which always returns
a three item tuple:

>>text = '''\

.... bcsn; 1000000; 1223
.... bcsn; 1000001; 1456
.... bcsn; 1000003
.... bcsn; 1000010; 4567
.... '''

>>for line in text.splitlines ():

.... bcsn, _, rest = line.partition( ';')
.... num1, _, num2 = rest.partition( ';')
.... print (bcsn, num1, num2)
....
(' bcsn', ' 1000000', ' 1223')
(' bcsn', ' 1000001', ' 1456')
(' bcsn', ' 1000003', '')
(' bcsn', ' 1000010', ' 4567')

>>help(str.part ition)

Help on method_descript or:

partition(...)
S.partition(sep ) -(head, sep, tail)

Searches for the separator sep in S, and returns the part before
it,
the separator itself, and the part after it. If the separator is
not
found, returns S and two empty strings.
STeVe

[Bob Greschke]

This idiom is what I ended up using (a lot it turns out!):

Parts = Line.split(";")
Parts += (x-len(Parts))*[""]

where x knows how long the line should be. If the line already has
more parts than x (i.e. [""] gets multiplied by a negative number)
nothing seems to happen which is just fine in this program's case.

Bob

[George Sakkis]

On Feb 1, 2:40 pm, Bob Greschke <b...@passcal.n mt.eduwrote:

This idiom is what I ended up using (a lot it turns out!):

Parts = Line.split(";")
Parts += (x-len(Parts))*[""]

where x knows how long the line should be. If the line already has
more parts than x (i.e. [""] gets multiplied by a negative number)
nothing seems to happen which is just fine in this program's case.

Bob

Here's a more generic padding one liner:

from itertools import chain,repeat

def ipad(seq, minlen, fill=None):
return chain(seq, repeat(fill, minlen-len(seq)))

>>list(ipad('on e;two;three;fou r'.split(";"), 7, ''))

['one', 'two', 'three', 'four', '', '', '']

>>tuple(ipad(xr ange(1,5), 7))

(1, 2, 3, 4, None, None, None)
George