英文:
Is there any method to find the next possible element following the list's pattern?
问题
我想要一个方法，以列表的模式获取列表中的下一个可能元素作为继续。
比如，有一个列表 ls，
ls = [1,2,2,3,2,2,1,2,2,3]
而我想要获取列表的下一个可能元素.. 在这种情况下，是"2"。
英文:
I want a method to get the next possible element as a continuation of the list following the pattern on the list.
Say, there is a list ls,
ls = [1,2,2,3,2,2,1,2,2,3]
And I want to get the next possible element of the list.. In this case, "2".
答案1
得分: 1
你可以在遍历列表时跟踪最短循环的长度。
cycle = 1
for i, x in enumerate(ls):
if x != ls[i % cycle]:
cycle = i + 1
print(ls[len(ls) % cycle]) # 2
英文:
You can keep track of the length of the shortest cycle while iterating over the list.
cycle = 1
for i, x in enumerate(ls):
if x != ls[i % cycle]:
cycle = i + 1
print(ls[len(ls) % cycle]) # 2
答案2
得分: 1
以下是代码部分的翻译：
建议"AI是必需的"当然是荒谬的。但是你的问题陈述不清楚。你提供的示例序列后面可以跟一个2，但也可能有其他解决方案，这里是[OEIS网站](https://oeis.org/search?q=1%2C2%2C2%2C3%2C2%2C2%2C1%2C2%2C2%2C3&language=english)上的一些示例。
然而，假设你正在寻找平凡的扩展算术进度（无限重复一定数量的简单算术步骤），这是一个解决方案：
def find_pattern(xs):
if len(xs) < 2:
return [0]
n = 1
while n < len(xs):
pattern = list(map(lambda x: x[1] - x[0], zip(xs[:n], xs[1:n + 1])))
for i, (x, y) in enumerate(zip(xs, xs[1:])):
if y - x != pattern[i % n]:
n = i + 1
break
else:
return pattern
sample = [1, 2, 2, 3, 2, 2, 1, 2, 2, 3]
pattern = find_pattern(sample)
print(pattern)
使用这个示例模式，你可以扩展序列：
def extend_pattern(xs, pattern, n):
x = xs[0]
for i in range(n):
yield x
if i < len(xs) and xs[i] != x:
raise ValueError(f'序列{xs}在索引{i}不匹配模式{pattern}。')
x = x + pattern[i % len(pattern])
print(list(extend_pattern(sample, pattern, 15)))
当结合使用时，输出如下：
[1, 0, 1, -1, 0, -1]
[1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2]
但是，作为这个方法出现严重错误的示例：
sample = [2, 3, 5, 7, 11, 13, 17, 19]
pattern = find_pattern(sample)
print(pattern)
print(list(extend_pattern(sample, pattern, 15)))
结果：
[1, 2, 2, 4, 2, 4, 2]
[2, 3, 5, 7, 11, 13, 17, 19, 20, 22, 24, 28, 30, 34, 36]
对于人来说，很明显`[2, 3, 5, 7, 11, 13, 17, 19]`是质数列表，但这并不是该函数找到的内容。
英文:
The suggestion that "AI is required" is of course silly. But your problem statement is unclear. The series you provide as an example can be followed by a 2, but other solutions are possible as well, here's a few examples on the OEIS website.
However, assuming that you're looking for the trivial extended arithmetic progression (repeating a fixed number of simple arithmetic steps indefinitely), this is a solution:
def find_pattern(xs):
if len(xs) < 2:
return [0]
n = 1
while n < len(xs):
pattern = list(map(lambda x: x[1] - x[0], zip(xs[:n], xs[1:n + 1])))
for i, (x, y) in enumerate(zip(xs, xs[1:])):
if y - x != pattern[i % n]:
n = i + 1
break
else:
return pattern
sample = [1, 2, 2, 3, 2, 2, 1, 2, 2, 3]
pattern = find_pattern(sample)
print(pattern)
And with that sample pattern, you could extend the series:
def extend_pattern(xs, pattern, n):
x = xs[0]
for i in range(n):
yield x
if i < len(xs) and xs[i] != x:
raise ValueError(f'The series {xs} does not match the pattern {pattern} at index {i}.')
x = x + pattern[i % len(pattern)]
print(list(extend_pattern(sample, pattern, 15)))
When combined, the output:
[1, 0, 1, -1, 0, -1]
[1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2]
However, as an example that this gets horribly wrong:
sample = [2, 3, 5, 7, 11, 13, 17, 19]
pattern = find_pattern(sample)
print(pattern)
print(list(extend_pattern(sample, pattern, 15)))
Result:
[1, 2, 2, 4, 2, 4, 2]
[2, 3, 5, 7, 11, 13, 17, 19, 20, 22, 24, 28, 30, 34, 36]
To a person, it's fairly obvious that [2, 3, 5, 7, 11, 13, 17, 19] is a list of primes, but that's not what the function finds of course.
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