CODEWITHSUNDEEP
pythonlistprintingformat
user1692261
user1692261

Reputation: 1217

python print in format way unknown list length

If I have a 6 length list like this:

l = ["AA","BB","CC","DD"]

I can print it with:

print "%-2s %-2s %-2s %-2s" % tuple(l)

The output will be:

AA BB CC DD

But what if the list l could be in any length? Is there a way to print the list in the same format with unknown number of elements?

Upvotes: 5

Views: 7995

Answers (3)

Martijn Pieters
Martijn Pieters

Reputation: 1121962

Generate separate snippets and join them:

print ' '.join(['%-2s' % (i,) for i in l])

Or you can use string multiplication:

print ('%-2s ' * len(l))[:-1] % tuple(l)

The [:-1] removes the extraneous space at the end; you could use .rstrip() as well.

Demo:

>>> print ' '.join(['%-2s' % (i,) for i in l])
AA BB CC DD
>>> print ' '.join(['%-2s' % (i,) for i in (l + l)])
AA BB CC DD AA BB CC DD
>>> print ('%-2s ' * len(l))[:-1] % tuple(l)
AA BB CC DD
>>> print ('%-2s ' * len(l))[:-1] % tuple(l + l)
AA BB CC DD AA BB CC DD

Timing stats:

>>> def joined_snippets(l):
...     ' '.join(['%-2s' % (i,) for i in l])
... 
>>> def joined_template(l):
...     ' '.join(['%-2s' for i in l])%tuple(l)
... 
>>> def multiplied_template(l):
...     ('%-2s ' * len(l))[:-1] % tuple(l)
... 
>>> from timeit import timeit
>>> l = ["AA","BB","CC","DD"]
>>> timeit('f(l)', 'from __main__ import l, joined_snippets as f')
1.3180170059204102
>>> timeit('f(l)', 'from __main__ import l, joined_template as f')
1.080280065536499
>>> timeit('f(l)', 'from __main__ import l, multiplied_template as f')
0.7333378791809082
>>> l *= 10
>>> timeit('f(l)', 'from __main__ import l, joined_snippets as f')
10.041708946228027
>>> timeit('f(l)', 'from __main__ import l, joined_template as f')
5.52706503868103
>>> timeit('f(l)', 'from __main__ import l, multiplied_template as f')
2.8013129234313965

The multiplied template option leaves the other options in the dust.

Upvotes: 11

John La Rooy
John La Rooy

Reputation: 304175

Another approach

' '.join(['%-2s' for i in l])%tuple(l)

I found this to be more than twice as fast as using a generator expression

' '.join('%-2s' for i in l)%tuple(l)

This is faster still

'%-2s '*len(l)%tuple(l) # leaves an extra trailing space though

Upvotes: 1

7stud
7stud

Reputation: 48599

tests = [
    ["AA"],
    ["AA", "BB"],
    ["AA", "BBB", "CCC"]
]

for test in tests:
    format_str = "%-2s " * len(test)
    print format_str

--output:--
%-2s 
%-2s %-2s 
%-2s %-2s %-2s

Upvotes: 0

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