如何提取与字典重复的元组的一部分，并将元组的第二部分作为值？ [英] How do I extract part of a tuple that's duplicate as key to a dictionary, and have the second part of the tuple as value?

问题描述

我很熟悉Python和Qgis，现在我只是运行脚本，但我的最终目的是创建一个插件。

这是代码的一部分我有问题：

  import math 
 
 layer = qgis.utils .iface.activeLayer（）
 iter = layer.getFeatures（）
 dict = {} 
 
 #iterate over features 
为iter中的功能：
 #print feature.id（）
 geom = feature.geometry（）
 coord = geom.asPolyline（）
 points = geom.asPolyline（）
 #get Endpoints 
 first = points [0] 
 last = points [-1] 
 
 #Assemble特征
 dict [feature.id（）] = [first，last] 
 
 print dict 
  

这是我的结果：
{0L：[（ 355277,6.68901e + 06），（35538​​5,6.68906e + 06）]，1L：[（355238,6.68909e + 06），（355340,6.68915e + 06）]，2L：[（355340,6.68915e + 06 ），3L：[（355340,6.68915e + 06），（355364,6.6891e + 06）]，4L：[（355364,6.6891e + 06），（35538​​5,6.68906），（355452,6.68921e + 06） E + 06）]， 5L：[（355261,6.68905e + 06），（355364,6.6891e + 06）]，6L：[（355364,6.6891e + 06），（355481,6.68916e + 06）]，7L：[（35538​​5， 6.68906e + 06），（355501,6.68912e + 06）]}

如你所见，很多行都有一个共同的终点：（35538​​5,6.68906e +06）由7L，4L和0L共享。

我想创建一个新的字典，获取共享点作为关键字，并将第二个点作为值。

例如：{（35538​​5,6.68906e + 06）：[（355277,6.68901e + 06），（355364,6.6891e + 06），（355501,6.68912 e + 06）]}

我一直在寻找列表理解教程，但没有太多的成功：大多数人都在寻找删除重复，而我想使用它们作为键（具有唯一ID）。我在思考set（）中是否正确仍然有用？

我将非常感谢任何帮助，谢谢提前。

解决方案

也许这是你需要的？

  dictionary = {} 
 for i in dict：
 for j in dict：
c = set（dict [i]）。intersection（set（dict [j]））
 if len（c）== 1：
＃ok，so现在我们知道，同时在两个
＃集中都存在一个元组，但是这一个将是新字典
＃的关键，我们需要从集合中的第二个元组成为这个值新的键
＃所以我们可以从集合中减去键元组来获取另一个元组
d = set（dict [i]）。difference（c）
＃现在我们需要得到元组通过执行列表（c）从集合
＃返回，我们得到列表
＃，我们的元组是列表中的第一个元素，因此列表（c）[0] 
c = list ）[0] 
词典[c] =列表（d）[0] 
 else：pass 
  

此代码只将一个元组附加到字典中的键。如果你想为每个键多个值，你可以修改它，以便每个键都有一个值列表，这可以通过简单地修改：

 ＃some_value不能是一个集合，它可以通过c = list（c）[0] 
 key = some_value 
 dictionary.setdefault（key，[]）
词典[key] .append（value）
  

所以，正确的答案是：

  dictionary = {} 
 for a in a：
 for j in a： 
c = set（a [i]）。intersection（set（a [j]））
 if len（c）== 1：
d = set（a [i] （c）
c = list（c）[0] 
 value = list（d）[0] 
如果字典中的c和值不在字典[c]中：
 dictionary [c] .append（value）
 elif c不在字典中：
 dictionary.setdefault（c，[]）
词典[c] .append（value）
 
 else：pass 
  

I'm pretty new to Python and Qgis, right now I'm just running scripts but I my end-goal is to create a plugin.

Here's the part of the code I'm having problems with:

import math

layer = qgis.utils.iface.activeLayer()
iter = layer.getFeatures()
dict = {}

#iterate over features
for feature in iter:
    #print feature.id()
    geom = feature.geometry()
    coord = geom.asPolyline()
    points=geom.asPolyline()
#get Endpoints
    first = points[0]
    last = points[-1]

#Assemble Features
dict[feature.id() ]= [first, last]

print dict

This is my result : {0L: [(355277,6.68901e+06), (355385,6.68906e+06)], 1L: [(355238,6.68909e+06), (355340,6.68915e+06)], 2L: [(355340,6.68915e+06), (355452,6.68921e+06)], 3L: [(355340,6.68915e+06), (355364,6.6891e+06)], 4L: [(355364,6.6891e+06), (355385,6.68906e+06)], 5L: [(355261,6.68905e+06), (355364,6.6891e+06)], 6L: [(355364,6.6891e+06), (355481,6.68916e+06)], 7L: [(355385,6.68906e+06), (355501,6.68912e+06)]}

As you can see, many of the lines have a common endpoint:(355385,6.68906e+06) is shared by 7L, 4L and 0L for example.

I would like to create a new dictionary, fetching the shared points as a key, and having the second points as value.

eg : {(355385,6.68906e+06):[(355277,6.68901e+06), (355364,6.6891e+06), (355501,6.68912e+06)]}

I have been looking though list comprehension tutorials, but without much success: most people are looking to delete the duplicates, whereas I would like use them as keys (with unique IDs). Am I correct in thinking set() would still be useful?

I would be very grateful for any help, thanks in advance.

解决方案

Maybe this is what you need?

dictionary = {}
for i in dict:
    for j in dict:
        c = set(dict[i]).intersection(set(dict[j]))
        if len(c) == 1:
            # ok, so now we know, that exactly one tuple exists in both
            # sets at the same time, but this one will be the key to new dictionary
            # we need the second tuple from the set to become value for this new key
            # so we can subtract the key-tuple from set to get the other tuple
            d = set(dict[i]).difference(c)
            # Now we need to get tuple back from the set
            # by doing list(c) we get list
            # and our tuple is the first element in the list, thus list(c)[0]
            c = list(c)[0]
            dictionary[c] = list(d)[0]
        else: pass

This code attaches only one tuple to the key in dictionary. If you want multiple values for each key, you can modify it so that each key would have a list of values, this can be done by simply modifying:

# some_value cannot be a set, it can be obtained with c = list(c)[0]
key = some_value
dictionary.setdefault(key, [])
dictionary[key].append(value)

So, the correct answer would be:

dictionary = {}
for i in a:
        for j in a:
            c = set(a[i]).intersection(set(a[j]))
            if len(c) == 1:
                d = set(a[i]).difference(c)
                c = list(c)[0]
                value = list(d)[0]
                if c in dictionary and value not in dictionary[c]:
                    dictionary[c].append(value)
                elif c not in dictionary:
                    dictionary.setdefault(c, [])
                    dictionary[c].append(value)

            else: pass

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