Only create list in python with every 5th item? [duplicate]

问题

This question already has answers here:

Pythonic way to return list of every nth item in a larger list (9 answers)

Closed 3 years ago.

I have a list of 20 items. I want to create a new list than only contains every 5th item of the list.

Here is was I have:

    listA=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
    for i in len(range(listA)):
      print i

How can I create listB from listA using Python? listB should be equal to [5, 10, 15, 20].

回答1:

You can use this idiom

listB = listA[4::5]

This means starting from index 4, pick every 5th element.

回答2:

Another method is using the range() function to create a list of integers with increments (steps) of 5 between them, then using them to access the indexes of the original list to append them to a new one:

listB=[]
for i in range(4,len(listA),5):
    listB.append(listA[i])

The range(start, stop[, step]) function returns a list of integers, based on the arguments used:

1. start - the first integer in the list (optional, defaults to 0).
2. stop - the last integer generated in the list will be the greatest, but less than the value of stop (stop will not be in the list).
3. step - the increment for values in the list (defaults to 1, can be made negative - to produce decreasing lists).

来源：https://stackoverflow.com/questions/40335148/only-create-list-in-python-with-every-5th-item