如何合并列表列表？答案

【问题标题】：How to union a list of lists?如何合并列表列表？
【发布时间】：2014-02-27 23:11:29
【问题描述】：

我编写了这个程序来计算一组点的 3d 距离。

points = [(472765.09, 6191522.78, 13.0), (472764.82, 6191524.09, 9.0), (472763.8, 6191525.68, 8.0), (472764.07, 6191524.39, 16.0)]

def dist3d((x0, y0, z0), (x1, y1, z1)):
    return math.sqrt((x0-x1)**2+(y0-y1)**2+(z0-z1)**2)

def dist_3d(obs):
    dist_list = list()
    while len(obs) != 1:
        obs_g = [(obs[0], x) for x in obs[1:]]
        dist_list.append([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])
        obs.pop(0)
    return dist_list

结果是一个距离列表：

test = dist_3d(points)
print test
[[4.217700795331081, 5.922339064664832, 3.554222840244929], [2.1374049685457694, 7.046453008421205], [8.107835716151763]]

我希望得到的结果如下：

[4.217700795331081, 5.922339064664832, 3.554222840244929, 2.1374049685457694, 7.046453008421205, 8.107835716151763]

P.S.：代码没有优化，因为函数返回后的“points”列表只有一个元素

points = [(472765.09, 6191522.78, 13.0), (472764.82, 6191524.09, 9.0), (472763.8, 6191525.68, 8.0), (472764.07, 6191524.39, 16.0)]
test = dist_3d(points)
points
[(472764.07, 6191524.39, 16.0)]

【问题讨论】：

标签： python list optimization coding-style

【解决方案1】：

要解决您的第一个问题，请在此行使用list.extend：

dist_list.append([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])

如：

          ######
dist_list.extend([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])
          ######

Here 说明了为什么会有帮助。

您的第二个问题来自这一行：

obs.pop(0)

因为obs和points都引用了同一个列表对象，所以调用obs.pop和直接调用points.pop是一样的。换句话说，每次执行该行时，都会从points 弹出一个项目。最后，这将导致points 仅包含一项。

要解决此问题，请在dist_3d 内制作points 的浅表副本：

copy = obs[:]

然后，在副本上调用pop。这样做不会影响points。

修改后的代码如下所示：

import math

points = [(472765.09, 6191522.78, 13.0), (472764.82, 6191524.09, 9.0), (472763.8, 6191525.68, 8.0), (472764.07, 6191524.39, 16.0)]

def dist3d((x0, y0, z0), (x1, y1, z1)):
    return math.sqrt((x0-x1)**2+(y0-y1)**2+(z0-z1)**2)

def dist_3d(obs):
    dist_list = list()
    copy = obs[:]
    while len(copy) != 1:
        obs_g = [(copy[0], x) for x in copy[1:]]
        dist_list.extend([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])
        copy.pop(0)
    return dist_list

这是一个示例运行：

>>> import math
>>> points = [(472765.09, 6191522.78, 13.0), (472764.82, 6191524.09, 9.0), (472763.8, 6191525.68, 8.0), (472764.07, 6191524.39, 16.0)]
>>> def dist3d((x0, y0, z0), (x1, y1, z1)):
...     return math.sqrt((x0-x1)**2+(y0-y1)**2+(z0-z1)**2)
...
>>> def dist_3d(obs):
...     dist_list = list()
...     copy = obs[:]
...     while len(copy) != 1:
...         obs_g = [(copy[0], x) for x in copy[1:]]
...         dist_list.extend([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])
...         copy.pop(0)
...     return dist_list
...
>>> test = dist_3d(points)
>>> print test
[4.217700795331081, 5.922339064664832, 3.554222840244929, 2.1374049685457694, 7.046453008421205, 8.107835716151763]
>>> print points
[(472765.09, 6191522.78, 13.0), (472764.82, 6191524.09, 9.0), (472763.8, 6191525.68, 8.0), (472764.07, 6191524.39, 16.0)]
>>>

【讨论】：

感谢 ICodez。测试中的一个元素的问题仍然存在。查看帖子中的更新

【解决方案2】：

import itertools

def dist_3d(obs):
    dist_list = list()
    while len(obs) != 1:
        obs_g = [(obs[0], x) for x in obs[1:]]
        dist_list.append([dist3d(obs_g[i][0], obs_g[i][1]) for i in xrange(len(obs_g))])
        obs.pop(0)
    return list(itertools.chain(*dist_list))

【讨论】：