从输入生成第n个值的列表

Question

I had the following code:

return [p.to_dict() for p in points]

I changed it to only print every nth row:

n = 100
count = 0

output = []

for p in points:
    if (count % n == 0):
        output.append(p.to_dict())
    count += 1
return output

Is there a more pythonic way to write this, to acheive the same result?

Answer 1

use enumerate and modulo on the index in a modified list comprehension to filter the ones dividable by n :

return [p.to_dict() for i,p in enumerate(points) if i % n == 0]

List comprehension filtering is good, but in that case, eduffy answer which suggests to use slicing with a step is better since the indices are directly computed. Use the filter part only when you cannot predict the indices.

Improving this answer even more: It's even better to use itertools.islice so not temporary list is generated:

import itertools
return [p.to_dict() for p in itertools.islice(points,None,None,n)]

itertools.islice(points,None,None,n) is equivalent to points[::n] but performing lazy evaluation.

Answer 2

The list slicing syntax takes an optional third argument to define the "step". This take every 3rd in a list:

 >>> range(10)
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 >>> range(10)[::3]
 [0, 3, 6, 9]

Answer 3

您可以对列表理解使用enumerate 。

[p.to_dict()  for i, p in enumerate(points) if i %100 == 0]