如何使用python删除列表中的括号和单个婴儿床

问题描述 投票:0回答:1

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
python-3.x
1个回答
0
投票

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'
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