示例 json 如下。我想將id已完成的（假和真）保存到單獨(dú)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}]預(yù)期結(jié)果如下todos_by_user_true = {1:0,2:1}todos_by_user_false = {1:3,2:1}代碼在下面？為什么我的代碼不起作用。我得到空白詞典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我的字典不對我的輸出如下all_by_user_false{1: 2, 2: 0}all_by_user_true{2: 0}免責(zé)聲明：我還需要處理異常
1 回答
忽然笑
TA貢獻(xiàn)1806條經(jīng)驗(yàn) 獲得超5個贊
查看您的輸入數(shù)據(jù)，它是這樣的： userId 1, id 1 has 0 true, and 3 false userId 1, id 2 has 1 true, and 1 false
給定所需的輸出，看起來你真的想使用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時，會出現(xiàn)會計(jì)問題。我會這樣修復(fù)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（順便說一句）已經(jīng)是您的評論中的內(nèi)容。
就個人而言，我會在插入之前檢查字典中的鍵，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
這給出了：
todos_by_user_true = {2:1}
todos_by_user_false = {1:3,2:1}
邏輯是這樣的，你不能有： todos_by_user_true = {1:0}
當(dāng)你找到它時，你就計(jì)算它的價值；而不是id從單獨(dú)的列表中迭代。
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