将清单大致等份的最佳方法是什么?例如，如果列表有7个元素，并将其分为2部分，我们希望其中一部分有3个元素，而另一部分应该有4个元素。
我正在寻找类似even_split(L, n)的东西，它将L分解为n部分。
def chunks(L, n):
""" Yield successive n-sized chunks from L.
"""
for i in range(0, len(L), n):
yield L[i:i+n]
上面的代码给出了3个块，而不是3个块。我可以简单地转置(遍历这个，取每列的第一个元素，称之为第一部分，然后取第二个元素，把它放在第二部分，等等)，但这破坏了项目的顺序。
当前回答
在这种情况下，我自己编写了代码:
def chunk_ports(port_start, port_end, portions):
if port_end < port_start:
return None
total = port_end - port_start + 1
fractions = int(math.floor(float(total) / portions))
results = []
# No enough to chuck.
if fractions < 1:
return None
# Reverse, so any additional items would be in the first range.
_e = port_end
for i in range(portions, 0, -1):
print "i", i
if i == 1:
_s = port_start
else:
_s = _e - fractions + 1
results.append((_s, _e))
_e = _s - 1
results.reverse()
return results
Divide_ports(1,10,9)将返回
[(1, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10)]
其他回答
将代码更改为产生n个块，而不是n个块:
def chunks(l, n):
""" Yield n successive chunks from l.
"""
newn = int(len(l) / n)
for i in xrange(0, n-1):
yield l[i*newn:i*newn+newn]
yield l[n*newn-newn:]
l = range(56)
three_chunks = chunks (l, 3)
print three_chunks.next()
print three_chunks.next()
print three_chunks.next()
这使:
[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]
这将把额外的元素分配给最后一组，这并不完美，但在你的“大致N相等部分”的规范范围内:-)我的意思是，56个元素作为(19,19,18)会更好，而这给出(18,18,20)。
你可以用下面的代码得到更均衡的输出:
#!/usr/bin/python
def chunks(l, n):
""" Yield n successive chunks from l.
"""
newn = int(1.0 * len(l) / n + 0.5)
for i in xrange(0, n-1):
yield l[i*newn:i*newn+newn]
yield l[n*newn-newn:]
l = range(56)
three_chunks = chunks (l, 3)
print three_chunks.next()
print three_chunks.next()
print three_chunks.next()
输出:
[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]
由于舍入错误，此代码被破坏。不要使用它!!
assert len(chunkIt([1,2,3], 10)) == 10 # fails
这里有一个可行的方法:
def chunkIt(seq, num):
avg = len(seq) / float(num)
out = []
last = 0.0
while last < len(seq):
out.append(seq[int(last):int(last + avg)])
last += avg
return out
测试:
>>> chunkIt(range(10), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> chunkIt(range(11), 3)
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]
>>> chunkIt(range(12), 3)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]
假设你想分成5个部分:
p1, p2, p3, p4, p5 = np.split(df, 5)
def chunk_array(array : List, n: int) -> List[List]:
chunk_size = len(array) // n
chunks = []
i = 0
while i < len(array):
# if less than chunk_size left add the remainder to last element
if len(array) - (i + chunk_size + 1) < 0:
chunks[-1].append(*array[i:i + chunk_size])
break
else:
chunks.append(array[i:i + chunk_size])
i += chunk_size
return chunks
这是我的版本(灵感来自Max)
使用列表推导式:
def divide_list_to_chunks(list_, n):
return [list_[start::n] for start in range(n)]
