This topic created in 3192 days ago, the information mentioned may be changed or developed.
以下这段数组
[{'beifan': u'D551', 'id': 2L, 'shuliang': 1334L}, {'beifan': u'D552', 'id': 3L, 'shuliang': 4003L}, {'beifan': u'D550', 'id': 4L, 'shuliang': 5103L}]
怎么转换成为
<li>D551:1334</li><li>D552:4003</li><li>D550:5103</li>
这样的字符串?
22 replies • 2017-09-13 00:57:19 +08:00
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1
linboki Sep 10, 2017 1
```python a = [{'beifan': u'D551', 'id': 2L, 'shuliang': 1334L}, {'beifan': u'D552', 'id': 3L, 'shuliang': 4003L}, {'beifan': u'D550', 'id': 4L, 'shuliang': 5103L}] str = ''.join(map(lambda x: '<li>{}:{}</li>'.format(x['beifan'], x['shuliang']), a)) print(str) ```
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4
billion Sep 10, 2017
>>> x = [{'beifan': 'D551', 'id': 2, 'shuliang': 1334}, {'beifan': 'D552', 'id': 3, 'shuliang': 4003}, {'beifan': 'D550', 'id': 4, 'shuliang': 5103}] >>> ''.join([f'<li>{a["beifan"]}:{a["shuliang"]}</>' for a in x]) '<li>D551:1334</><li>D552:4003</><li>D550:5103</>'
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5
billion Sep 10, 2017
</li>写成</>了,不过不影响。你改一下就可以。
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6
billion Sep 10, 2017
上面的代码需要 Python 3.6 或者以上才能成功运行。
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11
Victor215 Sep 11, 2017 via Android
为什么 python 做前端的事?
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12
secondwtq Sep 11, 2017
建议老实用模板引擎之类的,功能比 builtin 的强大,还帮你处理转义问题
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13
CEBBCAT Sep 11, 2017 via Android
贴代码可以用 gist
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15
Terry05 Sep 11, 2017
用模板语言处理这个很容易呀
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17
RealLiuSha Sep 11, 2017
print([ "{beifan}:{shuliang}".format(**item) for item in [{'beifan': u'D551', 'id': 2L, 'shuliang': 1334L}, {'beifan': u'D552', 'id': 3L, 'shuliang': 4003L}, {'beifan': u'D550', 'id': 4L, 'shuliang': 5103L}] ])
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21
kisscucci Sep 11, 2017 via iPhone
按照一楼的方法解决了
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22
yucongo Sep 13, 2017
In [578]: d = [{'beifan': u'D551', 'id': 2, 'shuliang': 1334}, {'beifan': u'D552', 'id': 3, 'shuliang': 4003}, {'beifan': u'D550', 'id': 4, 'shuliang': 5103}] In [579]: ''.join(['<li>' + elm.get('beifan') + ':' + str(elm.get('shuliang')) + '</li>' for elm in d]) Out[579]: '<li>D551:1334</li><li>D552:4003</li><li>D550:5103</li>'
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