'itertools._grouper‘对象没有属性'user’
'itertools._grouper‘对象没有属性'user’
提问于 2019-05-21 04:29:28
为什么我不能将groupby中的循环groupby转换为list？目前，我正在使用Django==2.2.1，当我将下面的data = [...]尝试到python中时，它运行得很好。
from itertools import groupby
from operator import itemgetter
@login_required
def list(request, template_name='cart/list.html'):
# I also try with this dummy data
test_data = [{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':7.0}]
print(type(data)) # a list
sorted_totals = sorted(test_data, key=itemgetter('total_order'))
for agent_name, group in groupby(sorted_totals, key=lambda x: x['agent_name']):
print(agent_name, list(group)) # I stopped here when converting the `group` as list.但是，当我在Django中的视图中尝试这个错误时，会遇到这样的错误。
我也用defaultdict试过
from collections import defaultdict
@login_required
def list(request, template_name='cart/list.html'):
test_data = [{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':7.0}]
grouped = defaultdict(list)
for data_total in test_data:
grouped[data_total['agent_name']].append(data_total) # stoped here
grouped_out = []
for agent_name, group in grouped.items():
total_order = 0
total_pcs = 0
total_kg = 0
if isinstance(group, list):
for data_total in group:
total_order += data_total.get('total_order')
total_pcs += data_total.get('total_pcs')
total_kg += data_total.get('total_kg')
grouped_out.append({
'agent_name': agent_name,
'total_order': total_order,
'total_pcs': total_pcs,
'total_kg': total_kg
})但是我发现的错误被包装器视图阻止了。如果我们按照前面的问题，它引用了这个_wrapped_view
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-05-22 01:53:37
最后，我使用dict手动修复了它。
test_data = [{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':5.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agentbeli','total_pcs':1,'total_kg':6.0},{'total_order':1,'agent_name':'agent123','total_pcs':1,'total_kg':7.0}]
grouped = {}
for data_total in test_data:
agent_name = data_total.get('agent_name')
if agent_name in grouped:
new_data = grouped[agent_name] # dict
new_data['total_order'] += data_total.get('total_order')
new_data['total_pcs'] += data_total.get('total_pcs')
new_data['total_kg'] += data_total.get('total_kg')
grouped[agent_name].update(**new_data)
else:
grouped[agent_name] = data_totalgrouped的结果如下所示：
{'agent123': {'agent_name': 'agent123',
'total_kg': 18.0,
'total_order': 3,
'total_pcs': 3},
'agentbeli': {'agent_name': 'agentbeli',
'total_kg': 17.0,
'total_order': 3,
'total_pcs': 3}}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56231089
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