I performed this operation to retrieve a queryset:
我执行此操作以检索查询集：
Name.objects.values_list('name', flat=True)
And it returns these results:
它返回以下结果：
[u'accelerate', u'acute', u'bear', u'big']
The results are all in unicode (u'). How do I remove them all so that I get the result:
结果都是unicode（u'）。如何将它们全部删除以便得到结果：
['accelerate', 'acute', 'bear', 'big']
5 个解决方案
#1
14
If you want to encode in utf8, you can simply do:
如果你想用utf8编码，你可以简单地做：
definitions_list = [definition.encode("utf8") for definition in definitions.objects.values_list('title', flat=True)]
#2
9
You could call str on all the values (note that map is a bit lazy, list() added to immediately turn it back into an indexable object):
您可以在所有值上调用str（请注意，map有点懒，添加了list（）以立即将其转换回可索引对象）：
thingy = list(map(str, [u'accelerate', u'acute', u'bear', u'big']))
Or use a list comprehension:
或者使用列表理解：
[str(item) for item in [u'accelerate', u'acute', u'bear', u'big']]
In the end though, why would you require them to be str explicitly; added to a django template (like {{ value }}), the u's will disappear.
最后，为什么你要求他们明确表达;添加到django模板（如{{value}}），u将消失。
#3
2
I think the short way is using json.dumps()
我认为简短的方法是使用json.dumps（）
json.dumps(definitions.objects.values_list('title', flat=True))
so you will get the string result as json format like
所以你会得到json格式的字符串结果
'["accelerate", "acute", "bear", "big" ...]'
and if you want to change that into python variable, simply use eval as the function, so your code more like this
如果你想把它改成python变量，只需使用eval作为函数，所以你的代码更像这样
json_format_string = json.dumps(definitions.objects.values_list('title', flat=True))
my_list = eval(json_format_string) # ['accelerate', 'acute', 'bear', 'big' ...]
#4
1
in python 2.7 justing using map without list is fine as well
在python 2.7中使用没有列表的地图也很好
list_of_python_strings = map(str, [u'accelerate'])
#5
0
i used this
我用过这个
my_list = list(map(str, [u'accelerate', u'acute', u'bear', u'big'...]))
#1
14
If you want to encode in utf8, you can simply do:
如果你想用utf8编码，你可以简单地做：
definitions_list = [definition.encode("utf8") for definition in definitions.objects.values_list('title', flat=True)]
#2
9
You could call str on all the values (note that map is a bit lazy, list() added to immediately turn it back into an indexable object):
您可以在所有值上调用str（请注意，map有点懒，添加了list（）以立即将其转换回可索引对象）：
thingy = list(map(str, [u'accelerate', u'acute', u'bear', u'big']))
Or use a list comprehension:
或者使用列表理解：
[str(item) for item in [u'accelerate', u'acute', u'bear', u'big']]
In the end though, why would you require them to be str explicitly; added to a django template (like {{ value }}), the u's will disappear.
最后，为什么你要求他们明确表达;添加到django模板（如{{value}}），u将消失。
#3
2
I think the short way is using json.dumps()
我认为简短的方法是使用json.dumps（）
json.dumps(definitions.objects.values_list('title', flat=True))
so you will get the string result as json format like
所以你会得到json格式的字符串结果
'["accelerate", "acute", "bear", "big" ...]'
and if you want to change that into python variable, simply use eval as the function, so your code more like this
如果你想把它改成python变量，只需使用eval作为函数，所以你的代码更像这样
json_format_string = json.dumps(definitions.objects.values_list('title', flat=True))
my_list = eval(json_format_string) # ['accelerate', 'acute', 'bear', 'big' ...]
#4
1
in python 2.7 justing using map without list is fine as well
在python 2.7中使用没有列表的地图也很好
list_of_python_strings = map(str, [u'accelerate'])
#5
0
i used this
我用过这个
my_list = list(map(str, [u'accelerate', u'acute', u'bear', u'big'...]))