Question

我有一个如下列表：

[(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)]

我已经尝试了groupby，但找不到办法为范围工作。任何帮助都会很棒。谢谢

Answer 1

将itertools.groupby与记住最后一项的关键功能一起使用，并使用当前项目进行检查。

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（参见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

使用https://pypi.python.org/pypi/cluster/1.1.0b1

>>> import cluster
>>> cl = cluster.HierarchicalClustering(data, lambda x,y: abs(x[1]-y[1]))
>>> cl.getlevel(20)
[
 [(220926340, 6768), (220926341, 6769), (220926344, 6772), (220926342, 6770), 
  (220926343, 6771)], 

 [(220927052, 7480), (220927053, 7481), (220927056, 7484), 
  (220927054, 7482), (220927055, 7483), (220927069, 7497), (220927071, 7499)], 

 [(220921998, 2426), (220921999, 2427), (220922000, 2428), (220922001, 2429)], 

 [(220922575, 3003), (220922563, 2991), (220922564, 2992), (220922565, 2993), 
  (220922566, 2994)], 

 [(220924912, 5340)], 

 [(220923958, 4386)], 

 [(220924161, 4589), (220924170, 4598), (220924171, 4599), (220924172, 4600), 
  (220924173, 4601)]
]

Answer 3

x=[(220921998, 2426), (220921999, 2427), ....    (220927071, 7499)]

start=0
flag=False
num=0
res=[]
for n,t in enumerate(x):
    #if not flag:start=x[n][1]
    if (x[n][1]-start)<20:
        res.append(t+('project%s' %num,))
        flag=True
    else:

        flag=False
        start=x[n][1]
        num+=1
        res.append(t+('project%s' %num,))
print res

返回

[(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, 'project2'), 
(220923958, 4386, 'project3'),
 (220924161, 4589, 'project4'), 
(220924170, 4598, 'project4'), 
(220924171, 4599, 'project4'), 
(220924172, 4600, 'project4'),
 (220924173, 4601, 'project4'), 
(220924912, 5340, 'project5'),
 (220926340, 6768, 'project6'),
 (220926341, 6769, 'project6'), 
(220926342, 6770, 'project6'), 
(220926343, 6771, 'project6'), 
(220926344, 6772, 'project6'), 
(220927052, 7480, 'project7'), 
(220927053, 7481, 'project7'),
 (220927054, 7482, 'project7'), 
(220927055, 7483, 'project7'), 
(220927056, 7484, 'project7'), 
(220927069, 7497, 'project7'), 
(220927071, 7499, 'project7')]

Answer 4

以下简单解决方案如何：

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))

Answer 5

尝试循环浏览数据：

prev = 0
currentProject = 1;
newx = []
for t[1] in x:
    if t - prev <= 20:
        pass
    else:
        currentProject += 1
    newx.append((t[0],t[1],"project"+currentProject))

编辑抱歉，我刚才意识到prev=None不起作用。我本可以检查prev是否None，但我认为对于给出的数据，只需执行prev=0就可以了。谢谢你指出我的错误！

根据python中的数字范围对值进行分组

5 个答案: