Shortest way to slice even/odd lines from a python array?

Or, a more general question would be, how to slice an array to get every n-th line, so for even/odd you’d want to skip one line, but in the general case you’d want to get every n-th lines, skipping n-1 lines.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Assuming you are talking about a list, you specify the step in the slice (and start index). The syntax is list[start:end:step].

You probably know the normal list access to get an item, e.g. l[2] to get the third item. Giving two numbers and a colon in between, you can specify a range that you want to get from the list. The return value is another list. E.g. l[2:5] gives you the third to sixth item. You can also pass an optional third number, which specifies the step size. The default step size is one, which just means take every item (between start and end index).

Example:

>>> l = range(10)
>>> l[::2]         # even  - start at the beginning at take every second item
[0, 2, 4, 6, 8]
>>> l[1::2]        # odd - start at second item and take every second item
[1, 3, 5, 7, 9]

See lists in the Python tutorial.

If you want to get every n-th element of a list (i.e. excluding the first element), you would have to slice like l[(n-1)::n].

Example:

>>> l = range(20)
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

Now, getting every third element would be:

>>> l[2::3]
[2, 5, 8, 11, 14, 17]

If you want to include the first element, you just do l[::n].

Method 2

This is more for me as a complete example 😉

>>> import itertools
>>> ret = [[1,2], [3,4,5,6], [7], [8,9]]
>>> itertools.izip_longest(*ret)
>>> [x for x in itertools.chain.from_iterable(tmp) if x is not None]
[1, 3, 7, 8, 2, 4, 9, 5, 6]

Method 3

example for indices 0,2,4… of myArr

myArr[list(range(0,len(myArr),2))]

example for indices 1,3,5… of myArr

myArr[list(range(1,len(myArr)+1,2))]

you can manipulate it anyway you want with the step parameter, in this case it is equal to 2.
hope this helped

Method 4

> map(lambda index: arr[index],filter(lambda x: x%n == 0,range(len(arr))))

where arr is a list, and n slices are required.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0