更快地将列表划分为块的方法
更快地将列表划分为块的方法
提问于 2018-07-17 00:34:23
我有一个需要分成块的dict列表：
input_size = len(input_rows) # num of dicts
slice_size = int(input_size / 4) # size of each chunk
remain = input_size % 4 # num of remaining dicts which cannot be divided into chunks
result = [] # initializes the list for containing lists of dicts
iterator = iter(input_rows) # gets a iterator on input
for i in range(4):
result.append([]) # creates an empty list as an element/block in result for containing rows for each core
for j in range(slice_size):
result[i].append(iterator.__next__()) # push in rows into the current list
if remain:
result[i].append(iterator.__next__()) # push in one remainder row into the current list
remain -= 1input_rows包含一个dict列表，将其划分为4个区块/切片；如果有任何剩余的dict不能平均划分为4个区块，则这些剩余的dict将被放入其中的一些区块中。一个列表(result)用于包含每个块，而每个块又包含一个dict列表。
我想知道如何以一种更有效的方式做这件事。
回答 3
Stack Overflow用户
回答已采纳
发布于 2018-07-18 23:24:07
看起来这不会很快作为副本关闭，所以我已经调整了tixxit’s great answer from 2010来解决这个问题，将他的生成器转换为列表理解，并希望使事情更容易理解。
chunks = 4
quot, rem = divmod(len(input_rows), chunks)
divpt = lambda i: i * quot + min(i, rem)
return [input_rows[divpt(i):divpt(i+1)] for i in range(chunks)]下面的测试框架显示，生成的代码生成的结果与OP的代码完全相同。
def main():
for top in range(1, 18):
print("{}:".format(top))
input_list = list(range(1, top + 1))
daiyue = chunkify_daiyue(input_list[:])
print('daiyue: {}'.format(daiyue))
zych = chunkify_zych(input_list[:])
match = 'Same' if (zych == daiyue) else 'Different'
print('Zych: {} {}'.format(zych, match))
print('')
def chunkify_daiyue(input_rows):
"Divide into chunks with daiyue's code"
input_size = len(input_rows) # num of dicts
slice_size = int(input_size / 4) # size of each chunk
remain = input_size % 4 # num of remaining dicts which cannot be divided into chunks
result = [] # initializes the list for containing lists of dicts
iterator = iter(input_rows) # gets a iterator on input
for i in range(4):
# creates an empty list as an element/block in result for
# containing rows for each core
result.append([])
for j in range(slice_size):
# push in rows into the current list
result[i].append(iterator.__next__())
if remain:
# push in one remainder row into the current list
result[i].append(iterator.__next__())
remain -= 1
return result
def chunkify_zych(input_rows):
"Divide into chunks with Tom Zych's code"
chunks = 4
quot, rem = divmod(len(input_rows), chunks)
divpt = lambda i: i * quot + min(i, rem)
return [input_rows[divpt(i):divpt(i+1)] for i in range(chunks)]
if __name__ == '__main__':
main()Stack Overflow用户
发布于 2018-07-17 04:12:16
使用标准库
R = list()
L = list(range(10))
remainder = int(len(L) % 4)
chunk_size = int(len(L) / 4)
position = 0
while position < len(L):
this_chunk = chunk_size
if remainder:
this_chunk += 1
remainder -= 1
R.append(L[position:this_chunk + position])
position += this_chunk
print(R)
[[0, 1, 2], [3, 4, 5], [6, 7], [8, 9]]这应该会更快，因为你的迭代和插入要少得多。在这种情况下，您实际上只需要抓取4个切片，然后根据列表元数据的计算插入4次……
此外，这是使用numpy.array_split*的具体原因:这应该更快……
>>> print(*np.array_split(range(10), 4))
[0 1 2] [3 4 5] [6 7] [8 9]编辑:由于评论部分中的反馈和上面答案中的潜在错误(在输入列表大小小于潜在组块数量的情况下)，下面是一个替代函数，它执行相同的操作，但将始终生成请求的组块数量
def array_split(input_list, chunks):
chunk_size = int(len(input_list) / chunks)
remainder = len(input_list) % chunks
new_list = list()
position = 0
while position < len(input_list):
this_chunk = chunk_size
if remainder:
this_chunk, remainder = this_chunk + 1, remainder - 1
new_list.append(input_list[position:this_chunk + position])
position += this_chunk
new_list.extend([[] for _ in range(chunks - len(new_list))])
return new_list
def unit_test():
L = [1, 2]
print(array_split(L, 4))
L = list(range(13))
print(array_split(L, 3))
L = list(range(22))
print(array_split(L, 5))
>>> unit_test()
[[1], [2], [], []]
[[0, 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13], [14, 15, 16, 17], [18, 19, 20, 21]]Stack Overflow用户
发布于 2018-07-17 02:50:52
myList = [1,2,3,4,5,6,9]
numOfChunks = 2
newList = []
for i in range(0, len(myList), numOfChunks):
newList.append(myList[i:i + numOfChunks])
print (newList) # [[1, 2], [3, 4], [5, 6], [9]]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51366397
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