如何从 for 循环创建一个字典答案

【问题标题】：How to create one dictionary from for loop如何从 for 循环创建一个字典
【发布时间】：2021-09-21 23:35:06
【问题描述】：

我对编码很陌生，我正在尝试遍历一个列表并创建一个具有相同键的字典。我的代码目前正在使用相同的键创建大约 1000 个格式正确的字典，但我不知道如何将它们变成一个字典。

for salaries, match in matching_salaries_non_employees.items():

   final_match.update(matches)

   Power_match = (match['Status']['Power'])+ ": " + (salaries)
    
   Length_match = (match['Status']['Length'])+ ": " + (match['Break']['Temp'])
        
   duration_Status = (match ['Status'][str('duration')])
   duration_Break = (match ['Break']['duration'])
   duration_match = (str(duration_Status) + ': ' + str(duration_Break))
    
   freq_Status = (match ['Status'][str('freq')])
   freq_Break = (match ['Break']['freq'])
   freq_match = (str(freq_Status) + ": " + str(freq_Break))
    
   start_times = (match['Status']['start time']) + ": " + (match['Break']['start time'])

   matches = {'Power': [power_match], 'Length': [Length_match], 'Duration': [duration_match], 'Freq': [freq_match], 'Start Times': [start_times]}

我的部分输出示例如下所示；

{'Power': ['high: high'], 'Length': ['Long: Short'], 'Duration': ['2.22: 1.01'], 'Freq': ['0.081229: 0.079503'], 'Start Times': ['10.10.08: 11.09.16']}
{'Power': ['med-high: med-high'], 'Length': ['Long: Long'], 'Duration': ['1.16: 0.81'], 'Freq': ['0.0988: 0.565'], 'Start Times': ['17.08.98: 31:01:03']}

而我希望将它合并到一个字典中，这样我就可以将它保存到单个 csv 文件中

【问题讨论】：

标签： python dictionary for-loop

【解决方案1】：

最简单直接的解决方案是使用defaultdict

from collections import defaultdict

matches = defaultdict(list)

for salaries, match in matching_salaries_non_employees.items():

   Power_match = (match['Status']['Power'])+ ": " + (salaries)
    
   Length_match = (match['Status']['Length'])+ ": " + (match['Break']['Temp'])
        
   duration_Status = (match ['Status'][str('duration')])
   duration_Break = (match ['Break']['duration'])
   duration_match = (str(duration_Status) + ': ' + str(duration_Break))
    
   freq_Status = (match ['Status'][str('freq')])
   freq_Break = (match ['Break']['freq'])
   freq_match = (str(freq_Status) + ": " + str(freq_Break))
    
   start_times = (match['Status']['start time']) + ": " + (match['Break']['start time'])

   matches['Power'].append(Power_match)
   matches['Length'].append(Length_match)
   matches['Duration'].append(duration_match)
   matches['Freq'].append(freq_match)
   matches['Start Times'].append(start_times)

【讨论】：

【解决方案2】：

@Matias Cicero 的回答很棒，但是显式初始化也很简单：

keys = ['Power', 'Length', 'Duration', 'Freq', 'Start Times']
matches = {k: [] for k in keys}

将替换行

from collections import defaultdict
matches = defaultdict(list)

【讨论】：