【发布时间】:2016-04-06 16:26:15
【问题描述】:
我创建了一个用于遍历多级字典的函数,并执行第二个函数ssocr,它需要四个参数:坐标、背景、前景、类型(它们是我的键的值)。这是我的字典,取自 json 文件。
def parse_image(self, d):
bg = d['background']
fg = d['foreground']
results = {}
for k, v in d['boxes'].iteritems():
if 'foreground' in d['boxes']:
myfg = d['boxes']['foreground']
else:
myfg = fg
if k != 'players_home' and k != 'players_opponent':
results[k] = MyAgonism.ssocr(v['coord'], bg, myfg, v['type'])
results['players_home'] = {}
for k, v in d['boxes']['players_home'].iteritems():
if 'foreground' in d['boxes']['players_home']:
myfg = d['boxes']['players_home']['foreground']
else:
myfg = fg
if k != 'background' and k != 'foreground':
for k2, v2 in d['boxes']['players_home'][k].iteritems():
if k2 != 'fouls':
results['players_home'][k] = {}
results['players_home'][k][k2] = MyAgonism.ssocr(v2['coord'], bg, myfg, v2['type'])
return results
在最后的迭代中,我只为 name 键获得了正确的值。 score 键没有出现。在我的 results['players_home'] 字典中看起来像 name 覆盖 score
输出:... "player4": {"name": 9}, "player5": {"name": 24} ...
我想要... "player4": {"name": 9, "score": value}, "player5": {"name": 24, "score": value} ...之类的东西
我做错了什么?这是完整的代码以防万一:Full Code
【问题讨论】:
标签: python json python-2.7 dictionary