Simple List division problem

[marcstuart]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ?
consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x


what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,
the overriding logic, is that I will get 3 lists and find a way for
python to try to break it evenly, if not one list can have a greater
amount of elements

Would I use itertools ? How would I do this ?

Thanks

 

[marcstuart]

I have gotten a little further, but not in the order of the original
list

def divide_list(lst, n):
return [lst[i::n] for i in range(n)]

x = [1,2,3,4,5,6,7,8,9,10]
y = 3
z = divide_list(x,y)

print z[0]
print z[1]
print z[2]


this prints:

[1, 4, 7, 10]
[2, 5, 8]
[3, 6, 9]



closer, but I would like to maintain the original order of the list.

 

[Gary Herron]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ?
consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x


what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,
the overriding logic, is that I will get 3 lists and find a way for
python to try to break it evenly, if not one list can have a greater
amount of elements

Would I use itertools ? How would I do this ?

Thanks

Calculate the size of a normal sublist, and the amount of extra that
goes with the last sublist.
Then extract y-1 normal sized sublists and one possibly larger sublist.

x = [1,2,3,4,5,6,7,8,9,10]
y = 3
s = len(x)/y
s # size of normal sublists 3
e = len(x) - y*s # extra elements for last sublist
e 1
z = [x[s*i:s*i+s] for i in range(y-1)] + [x[-s-e:]] #extract y-1 normal + 1 larger sublists
z

[[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]

Done!

Gary Herron

 

[Fredrik Lundh]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ? consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,

here's one way to do it:

# chop it up
n = len(x) / y
z = [x[i:i+n] for i in xrange(0, len(x), n)]

# if the last piece is too short, add it to one before it
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))

</F>

 

[mensanator]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ?
consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3      # number of lists I want to break x into
z = y/x

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,
the overriding logic, is that I will get 3 lists and find a way for
python to try to break it evenly, if not one list can have a greater
amount of elements

Would I use itertools ? How would I do this ?

Thanks

def list_split(x,y):
dm = divmod(len(x),y)
if dm[1] != 0:
z = [x[i*y:i*y+y] for i in xrange(len(x)/y) if len(x[i*y:])>=2*y]
z.append(x[(len(x)/y-1)*y:])
else:
z = [x[i*y:i*y+y] for i in xrange(len(x)/y)]
return z

list_split([1,2,3,4,5,6,7,8,9,10],3) [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]
list_split([1,2,3,4,5,6,7,8,9,10],5) [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
list_split([1,2,3,4,5,6,7,8,9,10],4) [[1, 2, 3, 4], [5, 6, 7, 8, 9, 10]]
list_split([1,2,3,4,5,6,7,8,9,10],2)

[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]

 

[Paul Rubin]

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

Are you SURE you want that? In almost every situation I've seen,

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9]
print z[4] = [10]

is preferable.

 

[Dustan]

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

Are you SURE you want that? In almost every situation I've seen,

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9]
print z[4] = [10]

is preferable.

Even more preferable is:

print z[0] = [1,2,3]
print z[1] = [4,5,6]
print z[2] = [7,8,9]
print z[3] = [10]

 

[thebjorn]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ? consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,

here's one way to do it:

# chop it up
n = len(x) / y
z = [x[i:i+n] for i in xrange(0, len(x), n)]

# if the last piece is too short, add it to one before it
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))

</F>

Eh...

def chop(lst, length):
n = len(lst) / length
z = [lst[i:i+n] for i in xrange(0, len(lst), n)]
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))
return z

gives

chop(range(1,9), 3) [[1, 2], [3, 4], [5, 6], [7, 8]]
chop(range(1,8), 3) [[1, 2], [3, 4], [5, 6, 7]]
chop(range(1,6), 3) [[1], [2], [3], [4], [5]]
chop([1], 3)

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "beforemeth.py", line 9, in chop
if len(z[-1]) < n and len(z) > 1:
ValueError: xrange() arg 3 must not be zero

Perhaps something like this?

def chop(lst, length):
from itertools import islice
it = iter(lst)
z = [list(islice(it, length)) for i in xrange(1 + len(lst) //
length)]
if len(z) > 1:
z[-2].extend(z.pop()) # the last item will be empty or contain
"overflow" elements.
return z

-- bjorn

 

[thebjorn]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ? consider this example:
x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x
what I would like to get is 3 sublists
print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]
obviously not even, one list will have 4 elements, the other 2 will
have 3.,

here's one way to do it:

# chop it up
n = len(x) / y
z = [x[i:i+n] for i in xrange(0, len(x), n)]

# if the last piece is too short, add it to one before it
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))

</F>

Eh...

def chop(lst, length):
n = len(lst) / length
z = [lst[i:i+n] for i in xrange(0, len(lst), n)]
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))
return z

gives

[[1, 2], [3, 4], [5, 6], [7, 8]]>>> chop(range(1,8), 3)

[[1, 2], [3, 4], [5, 6, 7]]>>> chop(range(1,6), 3)

[[1], [2], [3], [4], [5]]>>> chop([1], 3)

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "beforemeth.py", line 9, in chop
if len(z[-1]) < n and len(z) > 1:
ValueError: xrange() arg 3 must not be zero

Perhaps something like this?

def chop(lst, length):
from itertools import islice
it = iter(lst)
z = [list(islice(it, length)) for i in xrange(1 + len(lst) //
length)]
if len(z) > 1:
z[-2].extend(z.pop()) # the last item will be empty or contain
"overflow" elements.
return z

-- bjorn

Bad for to reply to myself, I know, but I just realized that the OP
wanted to control the _number_ of chunks, not the size of the
chunks... Building on the above would give something like

from itertools import islice
from operator import add

def chop(lst, nchunks):
chunksize, extra = divmod(len(lst), nchunks)
if not chunksize:
raise ValueError('More chunks than elements in list.')
it = iter(lst)
z = [list(islice(it, chunksize)) for i in xrange(nchunks + extra)]
z, extra = z[:nchunks], z[nchunks:]
z[-1].extend(reduce(add, extra, [])) # because sum ain't add :-(
return z

-- bjorn

 

[Fredrik Lundh]

> Eh...

oh, forgot that it was "pulling requirements out of thin air" week on
c.l.python.

def chop(lst, length):
n = len(lst) / length
z = [lst[i:i+n] for i in xrange(0, len(lst), n)]
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))
return z

gives

chop([1], 3)

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "beforemeth.py", line 9, in chop
if len(z[-1]) < n and len(z) > 1:
ValueError: xrange() arg 3 must not be zero

well, it doesn't. there's no xrange on that line.

> Perhaps something like this?

> from itertools import islice

or just use an if-statement, or the max function. but I guess those
tools are too old and boring for c.l.python these days...

</F>

 

[Paul Rubin]

Perhaps something like this?

def chop(lst, length):
from itertools import islice
it = iter(lst)
z = [list(islice(it, length)) for i in xrange(1 + len(lst) // length)]
if len(z) > 1:
z[-2].extend(z.pop()) # the last item will be empty or contain "overflow" elements.
return z

def chop(lst, length):
def chop1():
t = len(lst) // length - 1
for i in xrange(t):
yield lst[i*length: (i+1)*length]
yield lst[t*length:]
return list(chop1())

 

[thebjorn]

oh, forgot that it was "pulling requirements out of thin air" week on
c.l.python.

Well, the OP requirements were to control the number of chunks, not
the size of them, so I guess we both got it wrong initially.

def chop(lst, length):
n = len(lst) / length
z = [lst[i:i+n] for i in xrange(0, len(lst), n)]
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))
return z

gives
chop([1], 3)

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "beforemeth.py", line 9, in chop
if len(z[-1]) < n and len(z) > 1:
ValueError: xrange() arg 3 must not be zero

well, it doesn't. there's no xrange on that line.

It's from this line

z = [lst[i:i+n] for i in xrange(0, len(lst), n)]

(I executed the file with python -i beforemeth.py to get to an
interactive prompt, I'm sure you're familiar with the technique. You
could have just debugged your own program to find it though, or just
looked at the code -- not that many xrange calls in there, eh?)

or just use an if-statement, or the max function. but I guess those
tools are too old and boring for c.l.python these days...

I didn't realize correct code was too new-school. Perhaps you should
publish a list of modules you don't like, or perhaps I should just use
a sufficiently venerable version of Python? Ok, here you go:

C:\Python22>python
'import site' failed; use -v for traceback
Python 2.2.3 (#42, May 30 2003, 18:12:08) [MSC 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information..... chunksize = len(lst) // nchunks
.... if not chunksize:
.... raise ValueError('More chunks than elements in list.')
.... res = []
.... begin, end = 0, chunksize
.... for i in range(nchunks-1):
.... res.append(lst[begin:end])
.... begin, end = end, end+chunksize
.... res.append(lst[begin:])
.... return res
....

chop2(range(1,6), 2) [[1, 2], [3, 4, 5]]
chop2(range(1,6), 3) [[1], [2], [3, 4, 5]]
chop2(range(1,11), 3) [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]

Sufficiently old-school (or do I need to take out the // division
also?)

</F>

Shall we perhaps drop some of the attitude? (you used to be so much
nicer before you wrote sre ;-)

-- bjorn

 

[Pierre Quentel]

How do I divide a list into a set group of sublist's- if the list is
not evenly dividable ?
consider this example:

x = [1,2,3,4,5,6,7,8,9,10]
y = 3 # number of lists I want to break x into
z = y/x

what I would like to get is 3 sublists

print z[0] = [1,2,3]
print z[2] = [4,5,6]
print z[3] = [7,8,9,10]

obviously not even, one list will have 4 elements, the other 2 will
have 3.,
the overriding logic, is that I will get 3 lists and find a way for
python to try to break it evenly, if not one list can have a greater
amount of elements

Would I use itertools ? How would I do this ?

Thanks

Hi,

If you want to split the list in 4, do you want
[1,2],[3,4],[5,6],[7,8,9,10] : all extra items in the last sublist
or
[1,2],[3,4],[5,6,7],[8,9,10] : one extra item in each of the last
sublists ?

Assuming you want the second version :

=======================
def split_list(lst,nb):
# ln = length of smaller sublists
# extra = number of longer sublists (they have ln+1 items)
ln,extra = divmod(len(lst),nb)
pos = ln*(nb-extra) # position where larger sublists begin
return [ lst[i*lni+1)*ln] for i in xrange(nb-extra) ] \
+ [lst[pos+i*(ln+1)os+(i+1)*(ln+1)] for i in xrange(extra)]

======================

x = range(1,11)

print split_list(x,1)

[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]] print split_list(x,2)
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]] print split_list(x,3)
[[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]] print split_list(x,4)
[[1, 2], [3, 4], [5, 6, 7], [8, 9, 10]] print split_list(x,5)
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]] print split_list(x,10)
[[1], [2], [3], [4], [5], [6], [7], [8], [9], [10]]