python - 破解这个嵌套列表列表？

Question

我正在尝试使用以下代码分解一组高度嵌套的列表，其中包含我需要的数据:

url = 'http://www.whoscored.com/stagestatfeed/9155/stageteams/'
                    url = str(''.join(url[0:3]))
                    params = {
            'against': '0',            
            'field': '0',
            'teamId': '-1',
            'type': '8'
            }

                    headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/36.0.1985.125 Safari/537.36',
           'X-Requested-With': 'XMLHttpRequest',
           'Host': 'www.whoscored.com',
           'Referer': 'http://www.whoscored.com/'}

                    responser = requests.get(url, params=params, headers=headers)

                    responser = json.loads(responser.text.replace("'", '"').decode('cp1252'))

                    results = defaultdict(int)
                    for match in responser: 
                        for num_events, team, events in match:
                            print "w = ", num_events
                            print "x = ", team

                            for y in events: 
                                print "y = ", events

示例输出如下:

w =  13
x =  Arsenal
y =  [[[[u'goal', u'openplay', u'header', [1]], [u'goal', u'openplay', u'leftfoot', [1]], 
[u'goal', u'openplay', u'rightfoot', [3]], [u'goal', u'owngoal', u'rightfoot', [1]], [u'miss', 
u'corner', u'header', [2]], [u'miss', u'corner', u'leftfoot', [3]], [u'miss', u'corner',
u'rightfoot', [2]], [u'miss', u'crossedfreekick', u'rightfoot', [2]], [u'miss', 
u'directfreekick',
u'leftfoot', [1]], [u'miss', u'openplay', u'header', [2]], [u'miss', u'openplay', 
u'leftfoot', [16]], [u'miss', u'openplay', u'rightfoot', [23]]]]]
...
...
...


 w =  171
 x =  Queens Park Rangers
 y =  [[[[u'goal', u'openplay', u'leftfoot', [1]], [u'miss', u'corner', u'header', [5]], [u'miss',
 u'crossedfreekick', u'header', [1]], [u'miss', u'directfreekick', u'rightfoot', [2]], 
 [u'miss', u'openplay', u'header', [1]], [u'miss', u'openplay', u'leftfoot', [4]], [u'miss', 
 u'openplay', u'rightfoot', [23]], [u'miss', u'throwin', u'header', [1]]]]]

我想要得到的是每个团队的输出如下:

w =  13
x =  Arsenal
y = [u'goal', u'openplay', u'header', [1]]
y = [u'goal', u'openplay', u'leftfoot', [1]]
y = [u'goal', u'openplay', u'rightfoot', [3]]
...
...
...
y = [u'miss', u'openplay', u'rightfoot', [23]]

在此示例中，我将迭代代码中包含变量 y 的每个嵌套列表，并依次打印每个嵌套列表。我已经尝试了各种循环和列表理解，但无法得到有效的东西。

有人可以提出答案吗？

谢谢

Answer 1

只需使用索引进行访问并迭代子列表:

for y in events[0]:
   for sub in y:
       print ("y = ", sub)

w =  162
x =  Crystal Palace
y =  [u'goal', u'corner', u'rightfoot', [1]]
y =  [u'goal', u'crossedfreekick', u'header', [1]]
y =  [u'goal', u'openplay', u'rightfoot', [1]]
y =  [u'miss', u'corner', u'header', [6]]
y =  [u'miss', u'corner', u'rightfoot', [1]]
y =  [u'miss', u'crossedfreekick', u'header', [2]]
y =  [u'miss', u'crossedfreekick', u'leftfoot', [1]]
y =  [u'miss', u'crossedfreekick', u'rightfoot', [3]]
y =  [u'miss', u'openplay', u'header', [2]]
y =  [u'miss', u'openplay', u'leftfoot', [9]]
y =  [u'miss', u'openplay', u'rightfoot', [14]]
w =  175
x =  West Bromwich Albion
y =  [u'goal', u'corner', u'header', [2]]
y =  [u'goal', u'openplay', u'rightfoot', [3]]
y =  [u'goal', u'penalty', u'rightfoot', [1]]
y =  [u'miss', u'corner', u'header', [1]]
y =  [u'miss', u'directfreekick', u'rightfoot', [3]]
y =  [u'miss', u'openplay', u'leftfoot', [12]]
y =  [u'miss', u'openplay', u'rightfoot', [21]]