Question

我有一个嵌套字典 inputlist。每个列表有 2 个元组。我想按类 0 和 1 计算每个 tuple 第一个值的总和。

这是我的代码：

inputlist={1:{0: [(5.0, 3.6), (20.0, 0.0)],
              1: [(4.0, 0.0), (22.0, 0.0)]},
           2:{0: [(2.0, 0.5), (21.0, 0.5)],
              1: [(5.0, 0.0), (22.0, 0.0)]}
          }
      

sum_result={} 

for k1, v1 in inputlist.items():
   
    for (k2, v2) in v1.items():
        sum_result[k2]=1
        for i in range(len(v2)):
            (value1, value2) = v2[i]
            sum_result[k2] += value1
print(sum_result)

输出应如下所示：

{0: 48.0, 1: 53.0}

Answer 1

这是一个您可以尝试的解决方案。

from collections import defaultdict

sum_ = defaultdict(int)

for k, v in input_list.items():
    for ki, vi in v.items():
        sum_[ki] += sum(li[0] for li in vi)

print(sum_)

defaultdict(<class 'int'>, {0: 48.0, 1: 53.0})

Answer 2

你可以试试这个方法：

inputlist={1:{0: [(5.0, 3.6), (20.0, 0.0)],
              1: [(4.0, 0.0), (22.0, 0.0)]},
           2:{0: [(2.0, 0.5), (21.0, 0.5)],
              1: [(5.0, 0.0), (22.0, 0.0)]
              }
   }
sum_l={}
all_list=[]
for i in inputlist.values():
    for j in i.values():
        t_sum=0
        for k in j:
            t_sum+=k[0]
        all_list.append(t_sum)
j={0:sum(all_list[::2]),1:sum(all_list[1::2])}
print(j)

你也可以试试这个：

sum_l={}
all_list=[]
sum_={}
for a, b in inputlist.items():
    for c, d in b.items():
        try:
            sum_[c] += sum(li[0] for li in d)
        except KeyError:
            sum_[c] = sum(li[0] for li in d)
print(sum_)

Answer 3

这是代码 我希望你现在能理解代码。 0 和 1 的两个独立变量将使您的工作更轻松。当外循环完全执行时，您只需使用循环更新的值初始化 sum_result 字典即可。

  inputlist={1:{0: [(5.0, 3.6), (20.0, 0.0)],
              1: [(4.0, 0.0), (22.0, 0.0)]},
          2:{0: [(2.0, 0.5), (21.0, 0.5)],
              1: [(5.0, 0.0), (22.0, 0.0)]
              }}      
  sum_result={}
  sum_of_tuples_for_0=0
  sum_of_tuples_for_1=0
  for key,val in inputlist.items():
    for k1,v1 in val.items():
      if k1==0:
        for j1,j2 in v1:
          sum_of_tuples_for_0+=j1
        
      else :
        for j1,j2 in v1:
          sum_of_tuples_for_1+=j1   
  sum_result[0]=sum_of_tuples_for_0
  sum_result[1]=sum_of_tuples_for_1
  print(sum_result)

Answer 4

解决它只使用索引。

inputlist = {1: {0: [(5.0, 3.6), (20.0, 0.0)],
                 1: [(4.0, 0.0), (22.0, 0.0)]},
             2: {0: [(2.0, 0.5), (21.0, 0.5)],
                 1: [(5.0, 0.0), (22.0, 0.0)]
                 }
             }

sum_result_0 = {}
sum_result = {}

for i in inputlist.items():
    if i[0] == 1:
        sum_result_0[0] = i[1][0][0][0] + i[1][0][1][0]
        sum_result_0[1] = i[1][1][0][0] + i[1][1][1][0]
    if i[0] == 2:
        sum_result[0] = i[1][0][0][0] + i[1][0][1][0]
        sum_result[1] = i[1][1][0][0] + i[1][1][1][0]

sum_result = {0: sum_result[0] + sum_result_0[0], 1: sum_result[1] + sum_result_0[1]}

print(sum_result)

Answer 5

你可以试试这个：

sum_result=dict.fromkeys(range(2),0)


for v in inputlist.values():
    for k,i in v.items():
        sum_result[k] += i[0][0] + i[1][0]

print(sum_result)

或者这个

sum_result=dict.fromkeys(range(2),0)


for v in inputlist.values():
    sum_result[0] += v[0][0][0] + v[0][1][0]
    sum_result[1] += v[1][0][0] + v[1][1][0]

print(sum_result)

为每个元组添加第一个值

5 个答案: