Printing n elements per line in a list

[unexpected]

If have a list from 1 to 100, what's the easiest, most elegant way to
print them out, so that there are only n elements per line.

So if n=5, the printed list would look like:

1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
etc.

My search through the previous posts yields methods to print all the
values of the list on a single line, but that's not what I want. I feel
like there is an easy, pretty way to do this. I think it's possible to
hack it up using while loops and some ugly slicing, but hopefully I'm
missing something

[johnzenger]

def perline(n):
count = 1
while 1:
yield (count == n) and "\n" or " "
count = count % n + 1

r = range(1,101)
p = perline(5)

print "".join("%d %s" % (x, p.next()) for x in r)

unexpected wrote:

If have a list from 1 to 100, what's the easiest, most elegant way to
print them out, so that there are only n elements per line.

So if n=5, the printed list would look like:

1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
etc.

My search through the previous posts yields methods to print all the
values of the list on a single line, but that's not what I want. I feel
like there is an easy, pretty way to do this. I think it's possible to
hack it up using while loops and some ugly slicing, but hopefully I'm
missing something

[John Machin]

Matimus wrote:

Well, I have another at bat, so I will try to redeem myself... using
recursion:

def printTable(l,c) :
print(("%d "*len(l[:c]))%tuple(l[:c]))
if( len(l[:c]) 0 ):
printTable(l[c:],c)

printTable(rang e(1,101),5)

Sorry. Recursion disqualified your batter before he got out of the
dugout. Besides the %d restricts it to use wirh integers, and using
"L".lower() as a variable name made the umpire barf through his
facemask all over the catcher. Fortunately the ump didn't notice the
weirdly-spaced and superfluous () in the if statement ...

[Neil Cerutti]

On 2006-08-15, unexpected <su***********@ gmail.comwrote:

If have a list from 1 to 100, what's the easiest, most elegant
way to print them out, so that there are only n elements per
line.

So if n=5, the printed list would look like:

1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
etc.

My search through the previous posts yields methods to print
all the values of the list on a single line, but that's not
what I want. I feel like there is an easy, pretty way to do
this. I think it's possible to hack it up using while loops and
some ugly slicing, but hopefully I'm missing something

I can't resist putting in my oar:

def print_per(seq, n, isep=" ", rsep="\n"):
"""Print the items in seq, splitting into records of length n.

Trailing records may be shorter than length n."""
t = len(seq)
for i in xrange(n, t+1, n):
print isep.join(map(s tr, seq[i-n:i]))+rsep,
t = t % n
if t 0:
print isep.join(map(s tr, seq[-t:]))+rsep,

That's probably similar to some of the other mostly
non-functional solutions posted.

--
Neil Cerutti

[Rhamphoryncus]

unexpected wrote:

If have a list from 1 to 100, what's the easiest, most elegant way to
print them out, so that there are only n elements per line.

I've run into this problem a few times, and although many solutions
have been presented specifically for printing I would like to present a
more general alternative.

from itertools import chain
def istepline(step, iterator):
i = 0
while i < step:
yield iterator.next()
i += 1

def istep(iterable, step):
iterator = iter(iterable) # Make sure we won't restart iteration
while True:
# We rely on istepline()'s side-effect of progressing the
# iterator.
start = iterator.next()
rest = istepline(step - 1, iterator)
yield chain((start,), rest)
for i in rest:
pass # Exhaust rest to make sure the iterator has
# progressed properly.

>>i = istep(range(12) , 5)
for x in i: print list(x)

....
[0, 1, 2, 3, 4]
[5, 6, 7, 8, 9]
[10, 11]

>>i = istep(range(12) , 5)
for x in i: print x

....
<itertools.chai n object at 0xa7d3268c>
<itertools.chai n object at 0xa7d3260c>
<itertools.chai n object at 0xa7d3266c>

>>from itertools import islice, chain, repeat
def pad(iterable, n, pad): return islice(chain(it erable, repeat(pad)), n)
i = istep(range(12) , 5)
for x in i: print list(pad(x, 5, None))

....
[0, 1, 2, 3, 4]
[5, 6, 7, 8, 9]
[10, 11, None, None, None]

Would anybody else find this useful? Maybe worth adding it to itertool?

[Justin Azoff]

Rhamphoryncus wrote:

I've run into this problem a few times, and although many solutions
have been presented specifically for printing I would like to present a
more general alternative.

[snip interesting istep function]

Would anybody else find this useful? Maybe worth adding it to itertool?

yeah, but why on earth did you make it so complicated?

def istep(iterable, step):
a=[]
for x in iterable:
if len(a) >= step:
yield a
a=[]
a.append(x)
if a:
yield a

--
- Justin

[Marc 'BlackJack' Rintsch]

In <11************ *********@p79g2 000cwp.googlegr oups.com>, Justin Azoff
wrote:

Rhamphoryncus wrote:
[snip interesting istep function]

>Would anybody else find this useful? Maybe worth adding it to itertool?

yeah, but why on earth did you make it so complicated?

def istep(iterable, step):
a=[]
for x in iterable:
if len(a) >= step:
yield a
a=[]
a.append(x)
if a:
yield a

This is not as "lazy" as Rhamphoryncus' function anymore. Lets say the
`iterable` is a stream of events, then your function always needs to
"receive" `step` events before the caller can do anything else with the
events. In Rhamphoryncus' function the caller can react on the event as
soon as it's "delivered" by `iterable`.

Ciao,
Marc 'BlackJack' Rintsch

[Neil Cerutti]

On 2006-08-19, Rhamphoryncus <rh****@gmail.c omwrote:

unexpected wrote:

>If have a list from 1 to 100, what's the easiest, most elegant
way to print them out, so that there are only n elements per
line.

I've run into this problem a few times, and although many
solutions have been presented specifically for printing I would
like to present a more general alternative.

from itertools import chain
def istepline(step, iterator):
i = 0
while i < step:
yield iterator.next()
i += 1

def istep(iterable, step):
iterator = iter(iterable) # Make sure we won't restart iteration
while True:
# We rely on istepline()'s side-effect of progressing the
# iterator.
start = iterator.next()
rest = istepline(step - 1, iterator)
yield chain((start,), rest)
for i in rest:
pass # Exhaust rest to make sure the iterator has
# progressed properly.

Would anybody else find this useful? Maybe worth adding it to
itertool?

Your note me curious enough to re-read the itertools
documentation, and I found the following in 5.16.3 Recipes:

def grouper(n, iterable, padvalue=None):
"grouper(3, 'abcdefg', 'x') --('a','b','c'), ('d','e','f'), ('g','x','x')"
return izip(*[chain(iterable, repeat(padvalue , n-1))]*n)

Wish I'd found that yesterday. ;)

--
Neil Cerutti