In general case, we could use string .format():

>>> '{:<30}'.format('left aligned')
'left aligned                  '
>>> '{:>30}'.format('right aligned')
'                 right aligned'
>>> '{:^30}'.format('centered')
'           centered           '
>>> '{:*^30}'.format('centered')  # use '*' as a fill char
'***********centered***********'

We could also do that with f strings for python >3.6

>>> f"{'left aligned':<30}"
'left aligned                  '
>>> f"{'right aligned':>30}"
'                 right aligned'
>>> f"{'centered':^30}"
'           centered           '
>>> f"{'centered':*^30}"  # use '*' as a fill char
'***********centered***********'

If you need to separate certain elements with spaces you could do something like

print "hello", "there"

Notice the comma between "hello" and "there".

If you want to print a new line (i.e. \n) you could just use print without any arguments.

You can also use third-parties like PrettyTable or texttable. Example using texttable:

import texttable

l = [(4, 3, 7, 23),(17, 4021, 4, 92)]

table = texttable.Texttable()
# table.set_chars(["", "|", "", ""])
table.add_rows(l)

print(table.draw())

Would produce:

+----+------+---+----+
| 4  |  3   | 7 | 23 |
+====+======+===+====+
| 17 | 4021 | 4 | 92 |
+----+------+---+----+

Use string interpolation instead.

print '%d   %f' % (count,conv)

Answer of @user136036 is quite good, but unfortunately it does not take into account reality of Python bugs. Full answer could be following:

Variant A

If locale of your platform is working right, then just use locale:

import locale
locale.setlocale(locale.LC_ALL, '')
print("{:,d}".format(7123001))

Result is dependent on your locale and Python implementation working right.

But what if Python formatting according to locale is broken, e.g. Python 3.5 on Linux?

Variant B

If Python does not respect grouping=True parameter, you can use locale and a workaround (use monetary format):

locale.setlocale(locale.LC_ALL, '')
locale._override_localeconv = {'mon_thousands_sep': '.'}
print(locale.format('%.2f', 12345.678, grouping=True, monetary=True))

Above gives 12.345,68 on my platform. Setting monetary to False or omitting it - Python does not group thousands. Specifying locale._override_localeconv = {'thousands_sep': '.'} do nothing.

Variant C

If you don't have time to check what is working OK and what is broken with Python on your platform, you can just use regular string replace function (if you want to swap commas and dot to dots and comma):

print("{:,.2f}".format(7123001.345).replace(",", "X").replace(".", ",").replace("X", "."))

Replacing comma for space is trivial (point is assumed decimal separator):

print("{:,.2f}".format(7123001.345).replace(",", " ")

import re
x=[1,15,20,3]
print [re.sub(r"^0"," ",str(i).zfill(2)) for i  in x]

Output:[' 1', '15', '20', ' 3']

A tough way to do easy things :P

You can use string formatting:

print(" ".join("{:2d}".format(e) for e in mylist))

Also, " ".join can be replaced by unpacking due to the default sep=' ' of the print function:

print(*("{:2d}".format(e) for e in mylist))

Although the accepted answer is absolutely clear, I just wanted to check efficiency in terms of time.

The best way is to print joined string of numbers converted to strings.

print(" ".join(list(map(str,l))))

Note that I used map instead of loop. I wrote a little code of all 4 different ways to compare time:

import time as t

a, b = 10, 210000
l = list(range(a, b))
tic = t.time()
for i in l:
    print(i, end=" ")

print()
tac = t.time()
t1 = (tac - tic) * 1000
print(*l)
toe = t.time()
t2 = (toe - tac) * 1000
print(" ".join([str(i) for i in l]))
joe = t.time()
t3 = (joe - toe) * 1000
print(" ".join(list(map(str, l))))
toy = t.time()
t4 = (toy - joe) * 1000
print("Time",t1,t2,t3,t4)

Result:

Time 74344.76 71790.83 196.99 153.99

The output was quite surprising to me. Huge difference of time in cases of 'loop method' and 'joined-string method'.

Conclusion: Do not use loops for printing list if size is too large( in order of 10**5 or more).