我有一个值列表，我想将 append 键值对迭代到 python 字典

Question

我有一个值列表，我想将 append 键值对迭代到 python 字典

lst = ['12233223','23232423','23453443']

我正在寻找 output 像这样：

new_list = [{'id':'12233223', 'new_id':'d2233223'},{'id':'13232423','new_id':'d3232423'}]

但我明白了

[{'id':13453443,'new_id':13453443},{'id':13453443,'new_id':13453443}]

这是我的代码

lst = [{'id':'12233223'},{'id':'13232423',} {'id':'23453443'}]
d={}
new_list=[]
for each in lst:
    d['id']= each
    if len(each)==8 and each[0]==1:
    new_id = each.replace('1','d',1)
    d['new_id'] = new_id
    new_list.append(d)

相同的值被添加到列表中两次

Answer 1

你不能有一个所有键都具有相同值的字典。 但是你可以有一个字典列表

new_list = [{'id': each} for each in lst]

Answer 2

您可以通过嵌套for的字典列表使用对 go 的列表理解来获取值。 然后应用过滤条件并从符合条件的字典/值构建新字典：

lst = [{'id':'12233223'},{'id':'13232423'}, {'id':'23453443'}]


new_list = [ {**each,'new_id':"d"+n[1:]} for each in lst 
             for n in each.values() if len(n)==8 and n[0]=='1' ]

print(new_list)
              
[{'id': '12233223', 'new_id': 'd2233223'}, 
 {'id': '13232423', 'new_id': 'd3232423'}]

如果lst中的字典有其他键，则可以使嵌套for具体以仅获取“id”键的值：

new_list = [ {**each,'new_id':"d"+n[1:]} for each in lst 
             for n in [each['id']] if len(n)==8 and n[0]=='1' ]