循环通过字典列表 [英] loop through list of dictionaries

问题描述

我有一个字典列表。列表中有几个点，一些是多个。当有多个条目时，我想计算这一点的x和y的平均值。我的问题是，我不知道如何循环通过字典列表来比较ids的点！

当我使用这样的东西：

 为列表中的列表：
为列表中的j：
如果我['id'] == j ['id']：
 point = getPoint（i ['geom']）
 .... 
  

对不起，形成有点棘手...第二个循环是在第一个...
i认为它比较列表的第一个条目，所以是一样的。所以我必须从第二个循环开始，第二个条目，但我不能这样做与i-1，因为我是孔字典...
有人有个想法？
感谢提前！

 为范围内的j（1，len（NEWPoint）））
 if i ['gid'] == j ['gid']：
 allsamePoints.append（j）
 for all inamePoints：
 for l in range（1，len（allsamePoints）） ：
如果k ['gid'] == l ['gid']：
 Point1 = k ['geom'] 
 Point2 = l ['geom'] 
 X =（Point1.x（）+ Point2.x（））/ 2 
 Y =（Point1.y（）+ Point2.y（））/ 2 
 AVPoint = QgsPoint（X，Y）
 NEWReturnList.append（{'gid'：j ['gid']，'geom'：AVPoint}）
 del l 
在NEWReturnList中的m：
对于范围内的n（ 1，len（NEWReturnList））：
如果m ['gid'] == n ['gid']：
 Point1 = m ['geom'] 
 Point2 = n ['geom '] 
 X =（Point1.x（）+ Point2.x（））/ 2 
 Y =（Point1.y（）+ Point2.y（））/ 2 
 AV Point = QgsPoint（X，Y）
 NEWReturnList.append（{'gid'：j ['gid']，'geom'：AVPoint}）
 del n 
 else：
 pass 
  

好的，我想...在这个更令人困惑的时刻：）...

解决方案

一种方式将改变你存储积分的方式，因为你已经注意到，很难得到你想要的东西的。

一个更有用的结构将是一个dict，其中<​​code> id 映射到一个列表：

 从集合导入defaultdict 
 points_dict = defaultdict（list）
 
＃使新的dict 
 for point in point_list：
 id = point [id] 
 points_dict [id] .append（point ['geom']）
 
 def avg（lst） ：
lst的平均值
返回1.0 * sum（lst）/ len（lst）
 
＃现在简单得到平均
 for point in points_dict：
 print id，avg（points_dict [id]）
  

i have a list of dictionaries. there are several points inside the list, some are multiple. When there is a multiple entry i want to calculate the average of the x and the y of this point. My problem is, that i don't know how to loop through the list of dictionaries to compare the ids of the points!

when i use something like that:

for i in list:
  for j in list:
    if i['id'] == j['id']:
      point = getPoint(i['geom'])
      ....

sorry, the formating is a little bit tricky... the second loop is inside the first one... i think it compares the first entry of the list, so it's the same... so i have to start in the second loop with the second entry, but i can't do that with i-1 because i is the hole dictionary... Someone an idea? thanks in advance!

 for j in range(1, len(NEWPoint)):
      if i['gid']==j['gid']:
         allsamePoints.append(j)
      for k in allsamePoints:
         for l in range(1, len(allsamePoints)):
            if k['gid']==l['gid']:
                Point1 = k['geom']
                Point2=l['geom']
                X=(Point1.x()+Point2.x())/2
                Y=(Point1.y()+Point2.y())/2
                AVPoint = QgsPoint(X, Y)
                NEWReturnList.append({'gid': j['gid'], 'geom': AVPoint})
                del l
      for m in NEWReturnList:
          for n in range(1, len(NEWReturnList)):
              if m['gid']==n['gid']:
                 Point1 = m['geom']
                 Point2=n['geom']
                 X=(Point1.x()+Point2.x())/2
                 Y=(Point1.y()+Point2.y())/2
                 AVPoint = QgsPoint(X, Y)
                 NEWReturnList.append({'gid': j['gid'], 'geom': AVPoint})
                 del n
              else:
                 pass

ok, i think... at the moment thats more confusing :)...

解决方案

One way would be changing the way you store your points, because as you already noticed, it's hard to get what you want out of it.

A much more useful structure would be a dict where the id maps to a list of points:

from collections import defaultdict
points_dict = defaultdict(list)

# make the new dict
for point in point_list:
    id = point["id"]
    points_dict[id].append(point['geom'])

def avg( lst ):
    """ average of a `lst` """
    return 1.0 * sum(lst)/len(lst)

# now its simple to get the average
for id in points_dict:
    print id, avg( points_dict[id] )

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