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Python 将元组列表转换为嵌套字典列表

转载 作者:行者123 更新时间:2023-12-01 05:46:44 25 4
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我有一个几乎让我发疯的问题。

我正在用 python 编程,我有一个元组列表:

my_tuples = [
('csrf_token', u'5bc7224ef8f71c8142b4bef1597883b049b83fee'),
('fields-team', u'dsada'),
('fields-year_from', u'dsa'),
('fields-year_to', u'dsa'),
('fields-description', u'dsa') ,
('fields-team', u''),
('fields-year_from', u''),
('fields-year_to', u''),
('fields-description', u'')
]

我必须将这个元组列表转换为:

 form_data = {
'csrf_token':u'5bc7224ef8f71c8142b4bef1597883b049b83fee',
'fields':[
{
'team': u'dsada',
'year_from': u'dsa',
'year_to': u'dsa',
'description': u'dsa'
},
{
'team': u'',
'year_from': u'',
'year_to': u'',
'description': u''
}
]
}

然后我必须删除第二个字典,因为每个键都有空值。所以最终的结果将是:

 form_data = {
'csrf_token':u'5bc7224ef8f71c8142b4bef1597883b049b83fee',
'fields':[
{
'team': u'dsada',
'year_from': u'dsa',
'year_to': u'dsa',
'description': u'dsa'
}
]
}

最佳答案

我的两分钱:

import collections

def transform(loft):
form_data = collections.defaultdict(list)
internal = {}
numfield = 0
for t in my_tuples:
keys = t[0].split("-")
if len(keys) == 1:
form_data[keys[0]] = t[1]
else:
internal[keys[1]] = t[1]
if numfield == 3:
numfield = -1
form_data[keys[0]].append(internal)
internal = {}
numfield += 1
return form_data

if __name__ == "__main__":

my_tuples = [
('csrf_token', u'5bc7224ef8f71c8142b4bef1597883b049b83fee'),
('fields-team', u'dsada'),
('fields-year_from', u'dsa'),
('fields-year_to', u'dsa'),
('fields-description', u'dsa') ,
('fields-team', u''),
('fields-year_from', u''),
('fields-year_to', u''),
('fields-description', u'')
]
trans = transform(my_tuples)
# cleanup
trans['fields'] = [f for i, f in enumerate(trans['fields'])
if trans['fields'][i]['team'] != ''
and trans['fields'][i]['year_from'] != ''
and trans['fields'][i]['year_to'] != ''
and trans['fields'][i]['description'] != '']
print dict(trans)

关于Python 将元组列表转换为嵌套字典列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15840285/

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