[英]Average values in nested dictionary
我想创建一个新的值列表my_qty
,其中每个项目等于d[key]['qty']
中所有值的平均值,其中d[key]['start date']
与my_dates
中的值匹配。 我想我已经很近了,但是正挂在嵌套部分上。
import datetime
import numpy as np
my_dates = [datetime.datetime(2014, 10, 12, 0, 0), datetime.datetime(2014, 10, 13, 0, 0), datetime.datetime(2014, 10, 14, 0, 0)]
d = {
'ID1' : {'start date': datetime.datetime(2014, 10, 12, 0, 0) , 'qty': 12},
'ID2' : {'start date': datetime.datetime(2014, 10, 13, 0, 0) , 'qty': 34},
'ID3' : {'start date': datetime.datetime(2014, 10, 12, 0, 0) , 'qty': 35},
'ID4' : {'start date': datetime.datetime(2014, 10, 11, 0, 0) , 'qty': 40},
}
my_qty = []
for item in my_dates:
my_qty.append([np.mean(x for x in d[key]['qty']) if d[key]['start date'] == my_dates[item]])
print my_qty
所需输出:
[23.5,34,0]
要澄清每个请求的输出:
[average of d[key]['qty'] where d[key]['start date '] == my_dates[0], average of d[key]['qty'] where d[key]['start date '] == my_dates[1], average of d[key]['qty'] where d[key]['start date '] == my_dates[2],]
简单的方法是按日期将数量分组为字典:
import collections
quantities = collections.defaultdict(lambda: [])
for k,v in d.iteritems():
quantities[v["start date"]].append(v["qty"])
然后在该字典上运行以计算均值:
means = {k: float(sum(q))/len(q) for k,q in quantities.iteritems()}
给予:
>>> means
{datetime.datetime(2014, 10, 11, 0, 0): 40.0,
datetime.datetime(2014, 10, 12, 0, 0): 23.5,
datetime.datetime(2014, 10, 13, 0, 0): 34.0}
如果您想变得聪明,可以通过保持当前均值和所见值的数量相加来一次计算均值。 您甚至可以在一个类中抽象它:
class RunningMean(object):
def __init__(self, mean=None, n=0):
self.mean = mean
self.n = n
def insert(self, other):
if self.mean is None:
self.mean = 0.0
self.mean = (self.mean * self.n + other) / (self.n + 1)
self.n += 1
def __repr__(self):
args = (self.__class__.__name__, self.mean, self.n)
return "{}(mean={}, n={})".format(*args)
一遍您的数据将给您答案:
import collections
means = collections.defaultdict(lambda: RunningMean())
for k,v in d.iteritems():
means[v["start date"]].insert(v["qty"])
真正简单的方法是使用pandas
库,因为它是为类似这样的事情制作的。 这是一些代码:
import pandas as pd
df = pd.DataFrame.from_dict(d, orient="index")
means = df.groupby("start date").aggregate(np.mean)
给予:
>>> means
qty
start date
2014-10-11 40.0
2014-10-12 23.5
2014-10-13 34.0
单行答案:
mean_qty = [np.mean([i['qty'] for i in d.values()\
if i.get('start date') == day] or 0) for day in my_dates]
In [12]: mean_qty
Out[12]: [23.5, 34.0, 0.0]
or 0
的目的是,如果没有qty
,则返回0作为OP所需的值,因为空列表上的np.mean默认情况下返回nan
。
如果您需要速度,那么可以在jme的出色第二部分上进行构建(您可以这样做(我不重新计算均值直到被要求)将他的时间减少了3倍):
class RunningMean(object):
def __init__(self, total=0.0, n=0):
self.total=total
self.n = n
def __iadd__(self, other):
self.total += other
self.n += 1
return self
def mean(self):
return (self.total/self.n if self.n else 0)
def __repr__(self):
return "RunningMean(total=%f, n=%i)" %(self.total, self.n)
means = defaultdict(RunningMean)
for v in d.values():
means[v["start date"]] += (v["qty"])
Out[351]:
[RunningMean(mean= 40.000000),
RunningMean(mean= 34.000000),
RunningMean(mean= 23.500000)]
这是一些可以帮助您的工作代码:
for item in my_dates:
nums = [ d[key]['qty'] for key in d if d[key]['start date'] == item ]
if len(nums):
avg = np.mean(nums)
else:
avg = 0
print item, nums, avg
请注意, np.mean
不适用于空白列表,因此您必须检查要平均的数字的长度。
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