压缩两个字典列表

Question

我是Python新手，试图从json响应中创建一个新结构。两个json响应是来自2个环境的测试，但长度和顺序相同，只是不同的结果，为简洁起见，我简化了示例。

response1.json

[{"qa":"o"}, {"qa":"o"}] 

response2.json

[{"prod":"x"}, {"prod": "x"}] 

create.py

with open('response1.json') as data_file:  
    data1 = json.load(data_file) 

with open('response2.json') as data_file:  
    data2 = json.load(data_file) 

#i want to be able to create a structure like this: 
# [{"qa":"o", "prod":"x"},{"qa":"o", "prod":"x"}] 

list = [] 

#This is wrong but was thinking that logic would be close to this. 
for i in range(0,len(data1)): 
    list[i]['qa'] = data1[i]['qa'] 

for i in range(0,len(data2)): 
    list[i]['prod'] = data[i]['prod'] 

Answer 1

json_list = [dict(list(x[0].items()) + list(x[1].items())) for x in zip(data1,data2)] 
print(json_list) 

结果：

[{'qa': 'o', 'prod': 'x'}, {'qa': 'o', 'prod': 'x'}] 

这里的一个更优雅但可能不太有效的解决方案：

json_list = [dict(sum([list(y.items()) for y in x], [])) 
      for x in zip(data1,data2)] 

Answer 2

1）的Python 3.5运营商使用zip()功能溶液和字典拆包**：

data1 = [{"qa":"o"},{"qa":"o"}] 
data2 = [{"prod":"x"}, {"prod": "x"}] 

new_struct = [{**x, **y} for x,y in zip(data1, data2)] 
print(new_struct) 

输出：

[{'qa': 'o', 'prod': 'x'}, {'qa': 'o', 'prod': 'x'}] 

---

2）的Python < 3.5使用dict.update()方法解决：

new_struct = [] 
for x,y in zip(data1, data2): 
    x.update(y) 
    new_struct.append(x) 

print(new_struct) # will give the same output