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

[Paul Rubin]

"unexpected " <su***********@ gmail.comwrites :

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

The most direct way would be something like (untested):

for i in xrange(0, 100, 5):
print ' '.join(str(j) for j in a[i:i+5])

where a is the array. If you prefer a "coordinate-free" style, you
could use something like:

for x,y in itertools.group by(a, lambda n: n//5):
print ' '.join(str(k) for k in y)

I hope I got that right.

[Paul Rubin]

Paul Rubin <http://ph****@NOSPAM.i nvalidwrites:

for x,y in itertools.group by(a, lambda n: n//5):
print ' '.join(str(k) for k in y)

I hope I got that right.

Of course I meant to say

for x,y in itertools.group by(enumerate(a) , lambda (n,x): n//5):
print ' '.join(str(k) for n,k in y)

This is still way ugly and I've never found a really good solution.

[Dan Sommers]

On 15 Aug 2006 16:51:29 -0700,
"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

Perhaps not the prettiest, but I can't think of anything simpler to read
six months from now:

counter = 0
for an_element in the_list:
print an_element,
counter = counter + 1
if counter == n:
print
counter = 0

Regards,
Dan

--
Dan Sommers
<http://www.tombstoneze ro.net/dan/>
"I wish people would die in alphabetical order." -- My wife, the genealogist

[Simon Forman]

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

>From http://docs.python.org/lib/itertools-recipes.html there's the

grouper() function:

from itertools import izip, chain, repeat

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

R = range(1, 101)

for N in grouper(5, R, ''):
print ' '.join(str(n) for n in N)

If your iterable is not a multiple of n (of course not the case for 100
and 5), and you don't want the extra spaces at the end of your last
line, you could join the lines with '\n' and stick a call to rstrip()
in there:

G = grouper(5, R, '')
print '\n'.join(' '.join(str(n) for n in N) for N in G).rstrip()

but then you're back to ugly. lol.

Peace,
~Simon

[John Machin]

Dan Sommers wrote:

>
counter = 0
for an_element in the_list:
print an_element,
counter = counter + 1
if counter == n:
print
counter = 0

Yes, often simple old-fashioned ways are the best. A little verbose,
though. And I'd do some old-fashioned testing -- the above needs "if
counter: print" after the loop has finished, to get the final '\n' for
cases where there are not an even multiple of n elements.

>>def wrapn(alist, n):

.... ctr = 0
.... for item in alist:
.... print item,
.... ctr = (ctr + 1) % n
.... if not ctr:
.... print
.... if ctr:
.... print
....

>>for k in range(8):

.... print '--- %d ---' % k
.... wrapn(range(k), 3)
....
--- 0 ---
--- 1 ---
0
--- 2 ---
0 1
--- 3 ---
0 1 2
--- 4 ---
0 1 2
3
--- 5 ---
0 1 2
3 4
--- 6 ---
0 1 2
3 4 5
--- 7 ---
0 1 2
3 4 5
6

>>>

Cheers,
John

[Steven D'Aprano]

On Tue, 15 Aug 2006 16:51:29 -0700, 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

I don't see why you think that's an ugly hack.

def printitems(sequ ence, count=5):
"""Print count items of sequence per line."""
numrows = len(sequence)//count
if len(sequence)%c ount != 0: numrows += 1
for start in range(0, numrows):
items = sequence[start*count:(st art+1)*count]
for item in items:
print item,
print

--
Steven D'Aprano

[John Machin]

Steven D'Aprano wrote:

I think it's possible to

hack it up using while loops and some ugly slicing, but hopefully I'm
missing something

I don't see why you think that's an ugly hack.

def printitems(sequ ence, count=5):
"""Print count items of sequence per line."""
numrows = len(sequence)//count
if len(sequence)%c ount != 0: numrows += 1
for start in range(0, numrows):
items = sequence[start*count:(st art+1)*count]
for item in items:
print item,
print

Ugliness is in the eye of the beholder.
It is more blessed to add than to multiply.
Gaze upon this alternative:

def printitems2(seq uence, count=5):
"""Print count items of sequence per line."""
for pos in range(0, len(sequence), count):
for item in sequence[pos:pos+count]:
print item,
print

Cheers,
John

[Simon Forman]

John Machin wrote:

Steven D'Aprano wrote:

I think it's possible to

hack it up using while loops and some ugly slicing, but hopefully I'm
missing something

I don't see why you think that's an ugly hack.

def printitems(sequ ence, count=5):
"""Print count items of sequence per line."""
numrows = len(sequence)//count
if len(sequence)%c ount != 0: numrows += 1
for start in range(0, numrows):
items = sequence[start*count:(st art+1)*count]
for item in items:
print item,
print

Ugliness is in the eye of the beholder.
It is more blessed to add than to multiply.
Gaze upon this alternative:

def printitems2(seq uence, count=5):
"""Print count items of sequence per line."""
for pos in range(0, len(sequence), count):
for item in sequence[pos:pos+count]:
print item,
print

Cheers,
John

Very nice.

~Simon

[Dan Sommers]

On 15 Aug 2006 18:33:51 -0700,
"John Machin" <sj******@lexic on.netwrote:

Dan Sommers wrote:

>>
counter = 0
for an_element in the_list:
print an_element,
counter = counter + 1
if counter == n:
print
counter = 0

Yes, often simple old-fashioned ways are the best. A little verbose,
though. And I'd do some old-fashioned testing -- the above needs "if
counter: print" after the loop has finished, to get the final '\n' for
cases where there are not an even multiple of n elements.

I know I *thought* "untested"; I could have sworn that I typed it, too.
Sorry.

OTOH, pasted directly from my interactive python session (I changed
sys.ps2 to four spaces to make pasting from these sessions into an
editor that much easier):

Python 2.4.2 (#1, Jun 17 2006, 00:09:19)
[GCC 4.0.1 (Apple Computer, Inc. build 5247)] on darwin
Type "help", "copyright" , "credits" or "license" for more information.

>>c = 0
for e in range(10):

print e,
c = c + 1
if c == 3:
print
c = 0

0 1 2
3 4 5
6 7 8
9

>>>

And pasted right from an xterm:

$ python -c 'print 3,'
3
$

(All of which only explains your more complex test harness, and I still
should have looked more carefully before I posted.)

... ctr = (ctr + 1) % n

I'm old enough to remember the days when we avoided division like the
plague. Remember those zippy 1MHz (yes, that's an M and not a G) CPUs
with less than a handful of 8-bit registers? Old habits die hard. ;-)

Regards,
Dan

--
Dan Sommers
<http://www.tombstoneze ro.net/dan/>
"I wish people would die in alphabetical order." -- My wife, the genealogist