我有一个列表字典如下:
defaultdict(
我需要一个字典,通过对列表的值求和来减少维度,如下所示
{0: [6.0, 24.0], 1: [15.0, 33.0]}
提示: 6.0 = [1.0+2.0 +3.0] 24.0 =[7.0+8.0+9.0]
答案 0 :(得分:1)
迭代键并对列表推导式中的每个子列表求和:
d = {0: [[1.0, 2.0, 3.0], [7.0, 8.0, 9.0]], 1: [[4.0, 5.0, 6.0], [10.0, 11.0, 12.0]]}
for k in d:
d[k] = [sum(l) for l in d[k]]
>>> d
{0: [6.0, 24.0], 1: [15.0, 33.0]}
也适用于 defaultdict
。
使用字典推导可以在不修改原始字典的情况下获得新字典:
new_d = {k:[sum(l) for l in d[k]] for k in d}
答案 1 :(得分:0)
递归是遍历和评估嵌套字典的有效方法
defaultDict={0: [[1.0, 2.0, 3.0], [7.0, 8.0, 9.0]], 1: [[4.0, 5.0, 6.0], [10.0, 11.0, 12.0]]}
def walk_the_tree(inputDict,parent_index=None):
if isinstance(inputDict,dict):
for key, value in inputDict.items():
if isinstance(value,list):
ret_index,type,result=walk_the_tree(value,key)
print(key, ret_index, type, result)
elif isinstance(inputDict,list):
sum=0
for index,alist in enumerate(inputDict):
#print(index,alist)
ret_index,type,value=walk_the_tree(alist,index)
if type=="Sum List":
print ("replace value",parent_index, index, value)
defaultDict[parent_index][index]=value
print(type,value)
if type=="Leaf":
sum+=value
return index,"Sum List",sum
else:
return 0,"Leaf", inputDict
return
walk_the_tree(defaultDict)
print(defaultDict)
output:
{0: [6.0, 24.0], 1: [15.0, 33.0]}