根据python中的数字范围对值进行分组

Question

我有一个列表如下:

[(220921998, 2426),
(220921999, 2427),
(220922000, 2428),
(220922001, 2429),
(220922563, 2991),
(220922564, 2992),
(220922565, 2993),
(220922566, 2994),
(220922575, 3003),
(220923958, 4386),
(220924161, 4589),
(220924170, 4598),
(220924171, 4599),
(220924172, 4600),
(220924173, 4601),
(220924912, 5340),
(220926340, 6768),
(220926341, 6769),
(220926342, 6770),
(220926343, 6771),
(220926344, 6772),
(220927052, 7480),
(220927053, 7481),
(220927054, 7482),
(220927055, 7483),
(220927056, 7484),
(220927069, 7497),
(220927071, 7499)]

我想根据第二个数字在列表中添加一个字符串.如果列表中的第二个数字在其他第二个数字的20左右之内,则会为它们分配相同的"项目"名称.见下文:

[(220921998, 2426,project1),
(220921999, 2427,project1),
(220922000, 2428,project1),
(220922001, 2429,project1),
(220922563, 2991,project2),
(220922564, 2992,project2),
(220922565, 2993,project2),
(220922566, 2994,project2),
(220922575, 3003,project3),
(220923958, 4386,project4),
(220924161, 4589,project5),
(220924170, 4598,project5),
(220924171, 4599,project5),
(220924172, 4600,project5),
(220924173, 4601,project5),
(220924912, 5340,project6),
(220926340, 6768,project7),
(220926341, 6769,project7),
(220926342, 6770,project7),
(220926343, 6771,project7),
(220926344, 6772,project7),
(220927052, 7480,project8),
(220927053, 7481,project8),
(220927054, 7482,project8),
(220927055, 7483,project8),
(220927056, 7484,project8),
(220927069, 7497,project8),
(220927071, 7499,project8)]

我已经尝试过了groupby,但找不到一种方法来处理它的范围.任何帮助都会很棒.谢谢

Answer 1

使用itertools.groupby记住最后一项的键功能并使用当前项检查它.

lst = [(220921998, 2426),
       (220921999, 2427),
       (220922000, 2428),
       (220922001, 2429),
       (220922563, 2991),
       (220922564, 2992),
       (220922565, 2993),
       (220922566, 2994),
       (220922575, 3003),
       (220923958, 4386),
       (220924161, 4589),
       ....]

class Delta:
    def __init__(self, delta):
        self.last = None
        self.delta = delta
        self.key = 1
    def __call__(self, value):
        if self.last is not None and abs(self.last - value[1]) > self.delta:
            # Compare with the last value (`self.last`)
            # If difference is larger than 20, advance to next project
            self.key += 1
        self.last = value[1]  # Remeber the last value.
        return self.key

import itertools
for key, grp in itertools.groupby(lst, key=Delta(20)):
    for tup in grp:
        print(tup + ('project{}'.format(key),))

如果使用Python 3.x,则可以使用以下函数(请参阅参考资料nonlocal):

def Delta(delta):
    last = None
    key = 1
    def keyfunc(value):
        nonlocal last, key
        if last is not None and abs(last - value[1]) > delta:
            key += 1
        last = value[1]
        return key
    return keyfunc

Answer 2

下面的简单解决方案怎么样：

data = [(220921998, 2426),
        (220921999, 2427),
        (220922000, 2428),
        (220922001, 2429),
        ...
        (220922563, 2991),
        (220922564, 2992)]

ref = 0
cnt = 0
out = []
for dt in data:
    if dt[1]-ref > 20:
        cnt += 1
        ref = dt[1]
    out.append((dt[0],dt[1],'project%d'%cnt))