在列表中查找数量递增的组 [英] Finding groups of increasing numbers in a list

问题描述

目的是在给定整数列表的情况下找到递增/单调数的组.结果组中的每个项目必须比上一个项目增加+1

输入:

x = [7, 8, 9, 10, 6, 0, 1, 2, 3, 4, 5]

我需要找到数量递增的组并实现:

I need to find groups of increasing numbers and achieve:

increasing_numbers = [(7,8,9,10), (0,1,2,3,4,5)]

最终还有越来越多的数字:

And eventually also the number of increasing numbers:

len(list(chain(*increasing_numbers)))

还有小组的len:

increasing_num_groups_length = [len(i) for i in increasing_numbers]

我已经尝试了以下方法来获得越来越多的数字:

I have tried the following to get the number of increasing numbers:

>>> from itertools import tee, chain
>>> def pairwise(iterable): 
...     a, b = tee(iterable)
...     next(b, None)
...     return zip(a, b)
... 
>>> x = [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
>>> set(list(chain(*[(i,j) for i,j in pairwise(x) if j-1==i])))
set([1, 2, 3, 4, 5, 6, 8, 9, 10, 11])
>>> len(set(list(chain(*[(i,j) for i,j in pairwise(x) if j-1==i]))))
10

但是我无法保持顺序和数量递增的组.

But I'm unable to keep the order and the groups of increasing numbers.

如何获得整数元组的increasing_numbers组以及increasing_num_groups_length?

How can I achieve the increasing_numbers groups of integer tuples and also the increasing_num_groups_length?

也有这样/类似问题的名字吗?

我想出了这个解决方案，但是它太冗长了，我敢肯定有一种更简单的方法可以实现increasing_numbers输出:

I've came up with this solution but it's super verbose and I'm sure there's an easier way to achieve the increasing_numbers output:

>>> from itertools import tee, chain
>>> def pairwise(iterable): 
...     a, b = tee(iterable)
...     next(b, None)
...     return zip(a, b)
... 
>>> x = [8, 9, 10, 11, 7, 1, 2, 3, 4, 5, 6]
>>> boundary =  iter([0] + [i+1 for i, (j,k) in enumerate(pairwise(x)) if j+1!=k] + [len(x)])
>>> [tuple(x[i:next(boundary)]) for i in boundary]
[(8, 9, 10, 11), (1, 2, 3, 4, 5, 6)]

是否有更多使用pythonic/不太冗长的方法来做到这一点?

另一个输入/输出示例:

Another input/output example:

[输入]:

[17、17、19、20、21、22、0、1、2、2、4、5、6、7、8、9、10、11、12、13， 14、14、14、28、29、30、31、32、33、34、35、36、40]

[17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35, 36, 40]

[输出]:

[(19，20，21，22)，(0，1，2)，(4，5，6，7，8，9，10，11，12，13，14)， (28，29，30，31，32，33，34，35，36)]

[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36)]

推荐答案

使用itertools和numpy的几种不同方式:

A couple of different ways using itertools and numpy:

from itertools import groupby, tee, cycle

x = [17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35,
     36, 1, 2, 3, 4,34,54]


def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    for k, v in grps:
        if k:
            yield tuple(v) + (next((next(grps)[1])),)


print(list(sequences(x)))

[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]

或者使用python3并从中产生:

Or using python3 and yield from:

def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    yield from (tuple(v) + (next((next(grps)[1])),) for k,v in grps if k)

print(list(sequences(x)))

使用numpy.split在此处使用我的答案的变体:

Using a variation of my answer here with numpy.split :

out = [tuple(arr) for arr in np.split(x, np.where(np.diff(x) != 1)[0] + 1) if arr.size > 1]

print(out)

[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]

与ekhumoro的答案类似:

And similar to ekhumoro's answer:

def sequences(x):
    it = iter(x)
    prev, temp = next(it), []
    while prev is not None:
        start = next(it, None)
        if prev + 1 == start:
            temp.append(prev)
        elif temp:
            yield tuple(temp + [prev])
            temp = []
        prev = start

获取长度和元组:

def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, key=lambda j: j + 1 == next(x2))
    for k, v in grps:
        if k:
            t = tuple(v) + (next(next(grps)[1]),)
            yield t, len(t)


def sequences(l):
    x2 = cycle(l)
    next(x2)
    grps = groupby(l, lambda j: j + 1 == next(x2))
    yield from ((t, len(t)) for t in (tuple(v) + (next(next(grps)[1]),)
                                      for k, v in grps if k))



def sequences(x):
        it = iter(x)
        prev, temp = next(it), []
        while prev is not None:
            start = next(it, None)
            if prev + 1 == start:
                temp.append(prev)
            elif temp:
                yield tuple(temp + [prev]), len(temp) + 1
                temp = []
            prev = start

这三个输出都相同:

[((19, 20, 21, 22), 4), ((0, 1, 2), 3), ((4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), 11)
, ((28, 29, 30, 31, 32, 33, 34, 35, 36), 9), ((1, 2, 3, 4), 4)]

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