[英]How to pick a sequence of numbers from a list?
I have a startnumber and an endnumber. 我有一个开始编号和一个结束编号。
From these numbers I need to pick a sequence of numbers. 从这些数字中,我需要选择一个数字序列。
The sequences is not always the same. 顺序并不总是相同的。
Example: 例:
startnumber = 1
endnumber = 32
I need to create a list of numbers with a certain sequence 我需要创建具有特定顺序的数字列表
pe pe
3 numbers yes, 2 numbers no, 3 numbers yes, 2 numbers no.. etc 3个数字,2个数字,3个数字,2个数字..等
Expected output: 预期产量:
[[1-3],[6-8],[11-13],[16-18],[21-23],[26-28],[31-32]]
(at the end there are only 2 numbers remaining (31 and 32)) (最后只剩下两个数字(31和32))
Is there a simple way in python to select sequences of line from a range of numbers? python中有没有一种简单的方法可以从一系列数字中选择行序列?
numbers = range(1,33)
take = 3
skip = 2
seq = [list(numbers[idx:idx+take]) for idx in range(0, len(numbers),take+skip)]
Extrapolating this out: 推断出来:
def get_data(data, filterfunc=None):
if filterfunc is None:
filterfunc = lambda: True # take every line
result = []
sub_ = []
for line in data:
if filterfunc():
sub_.append(line)
else:
if sub_:
result.append(sub_)
sub_ = []
return result
# Example filterfunc
def example_filter(take=1, leave=1):
"""example_filter is a less-fancy version of itertools.cycle"""
while True:
for _ in range(take):
yield True
for _ in range(leave):
yield False
# Your example
final = get_data(range(1, 33), example_filter(take=3, leave=2))
As alluded to in the docstring of example_filter, the filterfunc
for get_data is really just expecting a True
or False
based on a call. 正如example_filter的文档字符串中提到的
filterfunc
用于get_data的filterfunc
实际上只是期望基于调用的True
或False
。 You could change this easily to be of the signature: 您可以轻松地将其更改为签名:
def filterfunc(some_data: object) -> bool:
So that you can determine whether to take or leave based on the value (or even the index), but it currently takes no arguments and just functions as a less magic itertools.cycle
(since it should return its value on call, not on iteration) 这样您就可以基于值(甚至是索引)来确定是否接受或离开,但是它目前不接受任何参数,而只是充当一个不太魔术的
itertools.cycle
(因为它应该在调用时返回其值,而不是在迭代时返回) )
from itertools import islice
def grouper(iterable, n, min_chunk=1):
it = iter(iterable)
while True:
chunk = list(islice(it, n))
if len(chunk) < min_chunk:
return
yield chunk
def pick_skip_seq(seq, pick, skip, skip_first=False):
if skip_first:
ret = [ x[skip:] for x in grouper(seq, pick+skip, skip+1) ]
else:
ret = [ x[:pick] for x in grouper(seq, pick+skip) ]
return ret
pick_skip_seq(range(1,33), 3, 2)
gives required list. pick_skip_seq(range(1,33), 3, 2)
给出所需的列表。
In pick_skip_seq(seq, pick, skip, skip_first=False)
, seq
is sequence to pick/skip from, pick
/ skip
are no. 在
pick_skip_seq(seq, pick, skip, skip_first=False)
, seq
是从中进行拾取/跳过的顺序,而pick
/ skip
是no。 of elements to pick/skip, skip_first
is to be set True
if such behavior is desired. 元素的选择/跳过,如果需要这种行为,将
skip_first
设置为True
。
grouper
returns chunks of n elements, it ignores last group if it has less than min_chunk elements. grouper
返回n个元素的块,如果少于min_chunk个元素,它将忽略最后一个组。 It is derived from stuff given in https://stackoverflow.com/a/8991553/1921546 . 它源自https://stackoverflow.com/a/8991553/1921546中提供的内容 。
Demo: 演示:
# pick 3 skip 2
for i in range(30,35):
print(pick_skip_seq(range(1,i), 3, 2))
# skip 3 pick 2
for i in range(30,35):
print(pick_skip_seq(range(1,i), 3, 2, True))
An alternative implementation of pick_skip_seq
: pick_skip_seq
的替代实现:
from itertools import chain,cycle,repeat,compress
def pick_skip_seq(seq, pick, skip, skip_first=False):
if skip_first:
c = cycle(chain(repeat(0, skip), repeat(1, pick)))
else:
c = cycle(chain(repeat(1, pick), repeat(0, skip)))
return list(grouper(compress(seq, c), pick))
All things used are documented here: https://docs.python.org/3/library/itertools.html#itertools.compress 此处记录了所有使用的东西: https : //docs.python.org/3/library/itertools.html#itertools.compress
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