Python: loop through list item x times?

Question

I am using Python2.7 and I would like to loop through a list x times.

a=['string1','string2','string3','string4','string5']
for item in a:
  print item

The above code will print all five items in the list, What if I just want to print the first 3 items? I searched over the internet but couldn't find an answer, it seems that xrange() will do the trick, but I can't figure out how.

Thanks for your help!

Answer 1

Sequence Slicing is what you are looking for. In this case, you need to slice the sequence to the first three elements to get them printed.

a=['string1','string2','string3','string4','string5']
for item in a[:3]:
      print item

Even, you don't need to loop over the sequence, just join it with a newline and print it

print '\n'.join(a[:3])

Answer 2

I think this would be considered pythonic:

for item in a[:3]:
    print item

Edit: since a matter of seconds made this answer redundant, I will try to provide some background information:

Array slicing allows for quick selection in sequences like Lists of Strings. A subsequence of a one-dimensional sequence can be specified by the indices of left and right endpoints:

>>> [1,2,3,4,5][:3] # every item with an index position < 3
[1, 2, 3]
>>> [1,2,3,4,5][3:] # every item with an index position >= 3
[4, 5]
>>> [1,2,3,4,5][2:3] # every item with an index position within the interval [2,3)
[3]

Note that the left endpoint is included, the right one is not. You can add a third argument to select only every nth element of a sequence:

>>> [1,2,3,4,5][::2] # select every second item from list
[1, 3, 5]
>>> [1,2,3,4,5][::-1] # select every single item in reverse order
[5,4,3,2,1]
>>> [1,2,3,4,5][1:4:2] # every second item from subsequence [1,4) = [2,3,4]
[2, 4]

By converting lists to numpy arrays, it is even possible to perform multi-dimensional slicing:

>>> numpy.array([[1,2,3,4,5], [1,2,3,4,5]])[:, ::2]
array([[1, 3, 5],
       [1, 3, 5]])