基于相同值将字典列表与单个字典合并的实现方法

首先注意到你的dict1是数字对应扩展名的映射，而我们需要的是扩展名对应数字的反向映射，这样才能快速匹配每个字典的FileExtension值。

步骤1：反转字典映射 #

先把dict1反转成扩展名到数字的映射，方便后续快速查找：

dict1 = {21409: 'docx', 44334: 'xlsx', 33635: 'jpg'}
# 生成 {扩展名: 对应数字} 的反向映射
ext_to_count = {v: k for k, v in dict1.items()}

步骤2：遍历字典列表添加Count字段 #

循环遍历list_of_dicts中的每个字典，根据FileExtension从反向映射中取值，新增Count键：

list_of_dicts= [
    {"Customer":"test1","Field1":"yy","Field2":"bb","FileExtension":"jpg"},
    {"Customer":"test2","Field1":"aa","Field2":"bb","FileExtension":"docx"},
    {"Customer":"test3","Field1":"cc","Field2":"yy","FileExtension":"xlsx"}
]

for item in list_of_dicts:
    item['Count'] = ext_to_count[item['FileExtension']]

最终结果 #

执行后list_of_dicts会完全符合你的期望输出：

[{"Customer":"test1","Field1":"yy","Field2":"bb","FileExtension":"jpg","Count":33635},{"Customer":"test2","Field1":"aa","Field2":"bb","FileExtension":"docx","Count":21409},{"Customer":"test3","Field1":"cc","Field2":"yy","FileExtension":"xlsx","Count":44334}]

如果需要处理FileExtension不在dict1中的异常情况，可以改用get方法避免报错：

for item in list_of_dicts:
    # 不存在对应扩展名时，Count设为None（或其他默认值）
    item['Count'] = ext_to_count.get(item['FileExtension'], None)

内容的提问来源于stack exchange，提问作者Rusty cole