Python基于键的求和值
问题内容：
如何从值是字符串的字典中求和。我的意思是在字典中乘法键相同的情况下如何求和。
字典
values = [
{
"rashod": 0,
"prihod": 230.0,
"prod_name": "r",
"prod_hola": "t"
},
{
"rashod": 0,
"prihod": 230.0,
"prod_name": "r",
"prod_hola": "t"
},
{
"rashod": 0.0,
"prihod": 0,
"prod_name": "c",
"prod_hola": "f"
},
{
"rashod": 0,
"prihod": 100.0,
"prod_name": "c",
"prod_hola": "f"
},
{
"rashod": 0.0,
"prihod": 0,
"prod_name": "a",
"prod_hola": "b"
},
{
"rashod": 0,
"prihod": 1500.0,
"prod_name": "a",
"prod_hola": "b"
}]
蟒蛇
dictf = reduce(lambda x, y: dict((k, v + y[k]) for k, v in x.iteritems()), values)
print dictf
但这将dict中的所有值相加，输出如下：
{'rashod': 1930.0, 'prihod': -17020.0, 'prod_name': 'abcfrtabcfrtabcfrt'}
我想要的是这样的输出：
[{'rashod': 1930.0, 'prihod': -17020.0, 'prod_name': 'a','prod_hola':'b},
{'rashod': 1930.0, 'prihod': -17020.0, 'prod_name': 'c','prod_hola':'f},
{'rashod': 1930.0, 'prihod': -17020.0, 'prod_name': 'r','prod_hola':'t},]
问题答案：
具体解决方案，不尝试巧妙：
def regroup(values):
groups = dict()
for d in values:
key = (d["prod_name"], d["prod_hola"])
if key in groups:
group = groups[key]
group["rashod"] += d["rashod"]
group["prihod"] += d["prihod"]
else:
groups[key] = d.copy()
return list(groups.values())
还有一个更通用的解决方案：
def generic_regroup(values, keys):
groups = dict()
valkeys = [k for k in values[0] if k not in key]
for d in values:
key = tuple(d[k] for k in keys)
if key in groups:
group = groups[key]
for k in valkeys:
group[k] += d[k]
else:
groups[key] = d.copy()
return list(groups.values())
results = generic_regroup(values, ("prod_name", "prod_hola"))
现在肯定有人会采用涉及itertools的更聪明的解决方案…
