我尝试过此代码
>>> do = [{'no': [0], 'name': ['MSI Afterburner 4.6.2'], 'version': ['4.6.2']}]
>>> flattened = [val for sublist in do for val in sublist]
>>> flattened
我的输出
['no', 'name', 'version']
所需的输出
no:0, name:MSI Afterburner 4.6.2, version: 4.6.2
解决方案不言自明:
", ".join(f"{k}: {v[0]}" for k, v in do[0].items())
#'no: 0, name: MSI Afterburner 4.6.2, version: 4.6.2'