遍历字典列表时如何获取前一个字典的modificationDate值

我有如下字典列表：

history_changes = [
{'userName': 'recg', 'modificationDate': '2022-07-14T04:01:39+00:00', 'changeComment': 'more details about L2'},
{'userName': 'artf', 'modificationDate': '2022-07-15T04:01:39+00:00', 'changeComment': 'more details about L1'},
{'userName': 'zrt',  'modificationDate': '2022-07-16T04:01:39+00:00', 'changeComment': 'more details about L3'}]

我正在遍历该列表并尝试将数据插入数据库，当前代码如下：

for record in history_changes[1:]:
   comment = record['changeComment']
   to_time = record['modificationDate']

我还需要获取start_time信息，即前一个字典中的modificationDate值。预期输出如下：

comment = 'more details about L1'
 start_time = '2022-07-14T04:01:39+00:00'
 to_time = '2022-07-15T04:01:39+00:00'

 comment = 'more details about L3'
 start_time = '2022-07-15T04:01:39+00:00'
 to_time = '2022-07-16T04:01:39+00:00'

请问遍历该列表时，如何获取前一个字典中的modificationDate值？

方法1：通过索引遍历 #

直接遍历列表的索引（从1开始），利用索引差i-1获取前一个元素：

for i in range(1, len(history_changes)):
    current_record = history_changes[i]
    prev_record = history_changes[i-1]
    
    comment = current_record['changeComment']
    start_time = prev_record['modificationDate']
    to_time = current_record['modificationDate']
    
    # 此处执行插入数据库的逻辑
    print(f" comment = '{comment}'")
    print(f" start_time = '{start_time}'")
    print(f" to_time = '{to_time}'\n")

方法2：用zip配对前后元素 #

将原列表与原列表从第二个元素开始的切片配对，循环时可同时拿到前一个和当前元素，写法更简洁：

for prev_record, current_record in zip(history_changes, history_changes[1:]):
    comment = current_record['changeComment']
    start_time = prev_record['modificationDate']
    to_time = current_record['modificationDate']
    
    # 此处执行插入数据库的逻辑
    print(f" comment = '{comment}'")
    print(f" start_time = '{start_time}'")
    print(f" to_time = '{to_time}'\n")

两种方法都能满足需求，zip的方式更符合Pythonic风格，适合这类连续元素配对的场景。

内容的提问来源于stack exchange，提问作者coLby