I have a grid of parameters which is composed of 234 dictionaries, everyone having the same keys. Then I have a list of weights through which I want to compute a weighted average of such dictionaries' values. In other words I need to obtin just one dictionary that has the same keys of the initial 243, but as values attached to each keys a weighted average of values, using the 243 weights.
我有一个参数网格,由234个字典组成,每个人都有相同的键。然后我有一个权重列表,我想通过它来计算这些词典值的加权平均值。换句话说,我需要只使用一个具有初始243的相同键的字典,但是使用243个权重作为每个键附加值的加权平均值。
I tried to use Counter
in order to cumulate results, but it returns very small values that do not make sense to me. w[0]
is the list of 243 weights, each of them related to 243 dictionaries inside grid
我尝试使用Counter来累积结果,但它返回的值非常小,对我来说没有意义。 w [0]是243个权重的列表,每个权重与网格内的243个词典相关
from collections import Counter
def avg_grid(grid, w, labels=["V0", "omega", "kappa", "rho", "theta"]):
avgHeston = Counter({label: 0 for label in labels})
for i in range(len(grid)):
avgHeston.update(Counter({label: grid[i][label] * w[0][i] for label in labels}))
totPar = dict(avgHeston)
return totPar
Is there an easier way to implement it?
有没有更简单的方法来实现它?
1 个解决方案
#1
0
You might want to use a defaultdict
instead:
您可能希望使用defaultdict:
from collections import defaultdict
def avg_grid(grid, wgts, labels=["V0", "rho"]):
avgHeston = defaultdict(int)
for g,w in zip(grid, wgts):
for label in labels:
avgHeston[label] += g[label] * w
return avgHeston
weights = [4,3]
grd = [{'V0':4,'rho':3}, {'V0':1,'rho':2}]
print(avg_grid(grd, weights))
Output:
defaultdict(<class 'int'>, {'V0': 19, 'rho': 18})
Notes:
I have changed w
so that you need to pass in a straight list. You might want to call the function like this: avg_grid(grids, w[0])
我已经改变了w,所以你需要传递一个直接列表。您可能想要调用这样的函数:avg_grid(grid,w [0])
Also this doesn't produce an average as such. you may want to do a divide by len(grid)
at some point.
这也不会产生平均值。你可能想在某个时候通过len(网格)进行除法。
Also for g,w in zip(grid, wgts):
is a more pythonic iteration
同样对于g,w in zip(grid,wgts):是一个更加pythonic迭代
#1
0
You might want to use a defaultdict
instead:
您可能希望使用defaultdict:
from collections import defaultdict
def avg_grid(grid, wgts, labels=["V0", "rho"]):
avgHeston = defaultdict(int)
for g,w in zip(grid, wgts):
for label in labels:
avgHeston[label] += g[label] * w
return avgHeston
weights = [4,3]
grd = [{'V0':4,'rho':3}, {'V0':1,'rho':2}]
print(avg_grid(grd, weights))
Output:
defaultdict(<class 'int'>, {'V0': 19, 'rho': 18})
Notes:
I have changed w
so that you need to pass in a straight list. You might want to call the function like this: avg_grid(grids, w[0])
我已经改变了w,所以你需要传递一个直接列表。您可能想要调用这样的函数:avg_grid(grid,w [0])
Also this doesn't produce an average as such. you may want to do a divide by len(grid)
at some point.
这也不会产生平均值。你可能想在某个时候通过len(网格)进行除法。
Also for g,w in zip(grid, wgts):
is a more pythonic iteration
同样对于g,w in zip(grid,wgts):是一个更加pythonic迭代