如何从json中提取值并递增

Question

示例 json 如下。我想将id已完成的（假和真）保存到单独的字典中todos = [{'userId': 1, 'id': 1, 'title': 'delectus aut autem', 'completed': False}, {'userId': 1, 'id': 2, 'title': 'quis ut nam facil ', 'completed': False}, {'userId': 1, 'id': 1, 'title': 'fugiat veniam minus', 'completed': False}, {'userId': 1, 'id': 2, 'title': 'et porro tempora', 'completed': True}, {'userId': 1, 'id': 1,'title': 'laprovident illum', 'completed': False}]预期结果如下todos_by_user_true = {1:0,2:1}todos_by_user_false = {1:3,2:1}代码在下面？为什么我的代码不起作用。我得到空白词典todos_by_user_true = {}todos_by_user_false = {}# Increment complete TODOs count for each user.for todo in todos:    if todo["completed"]==True:        try:            # Increment the existing user's count.            todos_by_user_true[todo["id"]] += 1        except KeyError:            # This user has not been seen. Set their count to 1.            todos_by_user_true[todo["id"]] = 0    elif todo["completed"]==False:        try:            # Increment the existing user's count.            todos_by_user_false[todo["id"]] += 1        except KeyError:            # This user has not been seen. Set their count to 1.            todos_by_user_false[todo["id"]] = 0我的字典不对我的输出如下all_by_user_false{1: 2, 2: 0}all_by_user_true{2: 0}免责声明：我还需要处理异常

Answer 1

查看您的输入数据，它是这样的： userId 1, id 1 has 0 true, and 3 false userId 1, id 2 has 1 true, and 1 false

给定所需的输出，看起来你真的想使用id而不是userId在你的查找中。id除此之外，第一次在生成的字典中插入时，会出现会计问题。我会这样修复它：

todos_by_user_true = {}

todos_by_user_false = {}

# Increment complete TODOs count for each user.

for todo in todos:

    if todo["completed"]==True:

        try:

            # Increment the existing user's count.

            todos_by_user_true[todo["id"]] += 1

        except KeyError:

            # This user has not been seen. Set their count to 1.

            todos_by_user_true[todo["id"]] = 1

    elif todo["completed"]==False:

        try:

            # Increment the existing user's count.

            todos_by_user_false[todo["id"]] += 1

        except KeyError:

            # This user has not been seen. Set their count to 1.

            todos_by_user_false[todo["id"]] = 1

其中（顺便说一句）已经是您的评论中的内容。

就个人而言，我会在插入之前检查字典中的键，而不是使用try..except，如下所示：

todos_by_user_true = {}

todos_by_user_false = {}

# Increment complete TODOs count for each user.

for todo in todos:

    key = todo["id"]

    if todo["completed"]:  # true case

            # If `id` not there yet, insert it to 0

            if key not in todos_by_user_true:

               todos_by_user_true[key] = 0

            # increment

            todos_by_user_true[key] += 1

    else:  # false case

            # If `id` not there yet, insert it to 0

            if key not in todos_by_user_false:

               todos_by_user_false[key] = 0

            # increment

            todos_by_user_false[key] += 1

这给出了：

todos_by_user_true = {2:1}

todos_by_user_false = {1:3,2:1}

逻辑是这样的，你不能有： todos_by_user_true = {1:0}

当你找到它时，你就计算它的价值；而不是id从单独的列表中迭代。