python相当于clojure的分区 - 所有？

Question

在python的标准库或语法技巧中寻找东西。

非Clojure的程序员，分区都应该有这些语义：

partition_all(16, lst) == [lst[0:16], lst[16:32], lst[32:48], lst[48:60]]

假设LEN（LST）== 60

Answer 1

Python中没有这样的函数。你可以这样做：

from itertools import islice 
def chunkwise(n, iterable): 
    it = iter(iterable) 
    while True: 
     chunk = list(islice(it, n)) 
     if not chunk: 
      break 
     yield chunk 

print list(chunkwise(3, range(10))) 
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] 

Answer 2

添加第三个 “步长” 参数到range内置函数让你非常接近：

>>> range(0,60,16) 
[0, 16, 32, 48] 

你可以从那里为上下界创建元组：

>>> [(i, i+16) for i in range(0, 60, 16)] 
[(0, 16), (16, 32), (32, 48), (48, 64)] 

或者，如果你需要他们创建实际的范围：

>>> [range(i, i+16) for i in range(0, 60, 16)] 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31], [32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47], [48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63]] 

当然可以参数0，60，和16到你自己的功能，如果你需要。

Answer 3

我不认为有一个像标准库，所以我怕你编写自己的功能partition_all。但是看着https://github.com/clojure/clojure/blob/b578c69d7480f621841ebcafdfa98e33fcb765f6/src/clj/clojure/core.clj#L5599我想你可以实现它在Python这样的：

>>> from itertools import islice 
>>> lst = range(60) 
>>> def partition_all(n, lst, step=None, start=0): 
...  step = step if step is not None else n 
...  yield islice(lst, start, n) 
...  while n < len(lst): 
...    start, n = start + step, n + step 
...    yield islice(lst, start, n) 
... 
>>> 
>>> for partition in partition_all(16, lst): 
...  l = list(partition) 
...  print len(l), l 
... 
16 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15] 
16 [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31] 
16 [32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47] 
12 [48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59] 
>>> 

Answer 4

我相信我有一个较短的答案不需要导入任何库。

这里是一个班轮为您提供：

>>> lst = list(range(60)) 
>>> [lst[i * 16: (i + 1) * 16] for i in range(len(lst)/size + int((len(lst) % 16) > 0))) 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31], [32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47], [48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59]] 

现在，作为一个函数（假设你的工作与列表）：

def partition(lst, size): 
    assert(size >= 0) 
    if not lst: return() 
    return (lst[i * size: (i + 1) * size] for i in range(len(lst)/size + int((len(lst) % size) > 0))) 
>>> partition(list(range(78)), 17) 
<generator object <genexpr> at 0x7f284e33d5a0> 
>>> list(partition(list(range(78)), 17)) 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33], [34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50], [51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67], [68, 69, 70, 71, 72, 73, 74, 75, 76, 77]] 
>>> list(partition(range(16), 4)) 
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]] 
>>> print list(partition(range(0), 4)) 
[] 

编辑：新的解决方案基于三联的回答：

def partition(lst, size): 
    assert(size >= 0) 
    if not lst: return() 
    return (lst[i: i + size] for i in range(0, len(lst), size)) 

>>> partition(list(range(78)), 17) 
<generator object <genexpr> at 0x025F5A58> 
>>> list(partition(list(range(78)), 17)) 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33], [34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50], [51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67], [68, 69, 70, 71, 72, 73, 74, 75, 76, 77]] 
>>> list(partition(list(range(16)), 17)) 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]] 
>>> list(partition(list(range(16)), 4)) 
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]] 
>>> list(partition([], 17)) 
[] 
>>>