我有两套，两者都被导入到Python和当前已被放置在列表如下两个单独导入文件获取的数据的。

I have two sets of data taken from two separate import files which are both being imported into python and have currently been placed in lists as follows.

表1的格式为：

（参考号，x坐标，y坐标）

(reference number, x coordinate, y coordinate)

实施例列表1：[[1，0,0]，[2,0，10]，[3,0，20]，[4,0，30]，[5,0，40]]

Example list 1: [[1, 0, 0], [2, 0, 10], [3, 0, 20], [4, 0, 30], [5, 0, 40]]

表2的格式为：

（x坐标，y坐标，温度）

(x coordinate, y coordinate, temperature)

实施例列表2：[[0，0，100]，[0，10，110]，[0，20，120]，[0，30，130]，[0，40，140]]

Example list 2: [[0, 0, 100], [0, 10, 110], [0, 20, 120], [0, 30, 130], [0, 40, 140]]

我需要使用x和y坐标来比较两个列表，如果他们找到一个匹配产生含有相应的参考号码和温度的新列表。

I need to compare the two lists using the x and y coordinates and if they find a match produce a new list containing the corresponding reference number and temperature.

例如从上面的输出列表中两个名单将遵循以下形式：

for example from the two lists above the output list would follow the form:

（参考号，温度）

实施例输出列表：[[1，100]，[2，110]，[3，120]，[4，130]，[5，140]

Example Output list: [[1, 100], [2, 110], [3, 120], [4, 130], [5, 140]]

这是要与大数据量的完成，我真的在努力寻找解决办法，任何帮助将是非常美联社preciated。干杯

This is to be done with a large amount of data and I am really struggling to find a solution, any help would be really appreciated. Cheers

推荐答案

本作品 0（N ^ 2）但它是非常容易阅读和理解。

This works 0(n^2) but it is very easy to read and understand.

 result = []
 for reference, x, y in list1:
     for a, b, temperature in list2:
         if x == a and y == b:
             result.append([temperature, reference])

您可以减少复杂性， 0（N）通过循环列表和在商店负责协调字典如下：

You can reduce the complexity to 0(n) by iterating over the lists and store coordinates in a dict as follows:

 dict1 = {}
 for reference, x, y in list1:
     dict[(x, y)] = reference

 dict2 = {}
 for x, y, temperature in list2:
     dict2[(x, y)] = temperature

 result = []
 for coordinate, reference in dict1.iteritems():
     temperature = dict2.get(coordinate)
     if temperature:
         result.append([temperature, reference])