Python列表理解字典中的字典？

发布于 2024-09-08 19:31:36 字数 588 浏览 5 评论 0原文

我刚刚了解了列表理解，这是在一行代码中获取数据的快速方法。但有件事困扰着我。

在我的测试中，我的列表中有这种字典：

[{'y': 72, 'x': 94, 'fname': 'test1420'}, {'y': 72, 'x': 94, 'fname': 'test277'}]

列表理解 s = [ r for r in list if r['x'] > 92 且 r['x'] < 95、r['y']>； 70 且 r['y'] < 75] 完美地解决了这个问题（事实上，这是这一行的结果）

无论如何，我后来意识到我并没有真正在我的其他项目中使用列表，我正在使用字典。像这样：

{'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'}}

这样我就可以简单地使用 var['test1420'] = ... 编辑我的字典，

但是列表推导式对此不起作用！我无法以这种方式编辑列表，因为您无法分配这样的索引。

还有别的办法吗？

原文

I just learned about list comprehension, which is a great fast way to get data in a single line of code. But something's bugging me.

In my test I have this kind of dictionaries inside the list:

[{'y': 72, 'x': 94, 'fname': 'test1420'}, {'y': 72, 'x': 94, 'fname': 'test277'}]

The list comprehension s = [ r for r in list if r['x'] > 92 and r['x'] < 95 and r['y'] > 70 and r['y'] < 75 ] works perfectly on that (it is, in fact, the result of this line)

Anyway, I then realised I'm not really using a list in my other project, I'm using a dictionary. Like so:

{'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'}}

That way I can simply edit my dictionary with var['test1420'] = ...

But list comprehensions don't work on that!
And I can't edit lists this way because you can't assign an index like that.

Is there another way?

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你与清晨阳光 2024-09-15 19:31:36

您可以这样做：

s = dict([ (k,r) for k,r in mydict.iteritems() if r['x'] > 92 and r['x'] < 95 and r['y'] > 70 and r['y'] < 75 ])

这需要您指定的字典并返回“过滤”字典。

You can do this:

s = dict([ (k,r) for k,r in mydict.iteritems() if r['x'] > 92 and r['x'] < 95 and r['y'] > 70 and r['y'] < 75 ])

This takes a dict as you specified and returns a 'filtered' dict.

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萌辣 2024-09-15 19:31:36

如果 dct 是

{'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'},
 'test277': {'y': 72, 'x': 94, 'fname': 'test277'},}

也许你正在寻找类似的东西：

[ subdct for key,subdct in dct.iteritems() 
  if 92<subdct['x']<95 and 70<subdct['y']<75 ]

一点好处是 Python 允许你链接不等式：

92<dct[key]['x']<95

而不是

if r['x'] > 92 and r['x'] < 95

注意上面我已经写了一个列表理解，所以你会得到一个列表（在这种情况下，是字典）。

在 Python3 中，还有诸如 dict 推导式之类的东西：

{ n: n*n for n in range(5) } # dict comprehension
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

在 Python2 中，相当于

dict( (n,n*n) for n in range(5) )

I我不确定您是否正在寻找字典列表或字典的字典，但如果您理解上面的示例，则很容易修改我的答案以获得您想要的内容。

If dct is

{'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'},
 'test277': {'y': 72, 'x': 94, 'fname': 'test277'},}

Perhaps you are looking for something like:

[ subdct for key,subdct in dct.iteritems() 
  if 92<subdct['x']<95 and 70<subdct['y']<75 ]

A little nicety is that Python allows you to chain inequalities:

92<dct[key]['x']<95

instead of

if r['x'] > 92 and r['x'] < 95

Note also that above I've written a list comprehension, so you get back a list (in this case, of dicts).

In Python3 there are such things as dict comprehensions as well:

{ n: n*n for n in range(5) } # dict comprehension
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

In Python2 the equivalent would be

dict( (n,n*n) for n in range(5) )

I'm not sure if you are looking for a list of dicts or a dict of dicts, but if you understand the examples above, it is easy to modify my answer to get what you want.

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暖伴 2024-09-15 19:31:36

听起来你想要类似的东西：

my_dict = {'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'},
           'test277' : {'y': '072', 'x': '094', 'fname': 'test277'}}


new_dict = dict((k,v) for k,v in my_dict.items() 
                    if 92 < int(v['x']) < 95 and 70 < int(v['y']) < 75)

关于此代码的一些注释：

我正在使用生成器表达式
而不是列表推导式
Python 可以让你组合不等式，
测试为 low <值< high
dict() 构造函数接受一个可迭代对象
键/值元组来创建
字典

Sounds like you want something like:

my_dict = {'test1420': {'y': '060', 'x': '070', 'fname': 'test1420'},
           'test277' : {'y': '072', 'x': '094', 'fname': 'test277'}}


new_dict = dict((k,v) for k,v in my_dict.items() 
                    if 92 < int(v['x']) < 95 and 70 < int(v['y']) < 75)

Some notes on this code:

I'm using a generator expression
instead of a list comprehension
Python lets you combine inequality
tests as low < value < high
The dict() constructor takes an iterable
of key/value tuples to create a
dictionary

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凉栀 2024-09-15 19:31:36

在 Python 3 中，您可以使用字典理解，这可能是一个更短的解决方案：

{key_expression(item) : value_expression(item) for item in something if condition}

如果您想像原始问题一样过滤字典：

mydict = {'test1': {'y': 60},'test2': {'y': 70},'test3': {'y': 80}}
s = {k : r for k,r in mydict.items() if r['y'] < 75}
> {'test1': {'y': 60}, 'test2': {'y': 70}}

或者我们甚至可以在列表或范围之外创建一些内容。例如，如果我们想要一个包含所有奇数平方数的字典：
```
{i : i**2 for i in range(11) if i % 2 == 1}
> {1:1、3:9、5:25、7:49、9:81}
```

In Python 3 you can use dict comprehension which can be an even shorter solution:

{key_expression(item) : value_expression(item) for item in something if condition}

In case you want to filter a dictionary as in the original question:

mydict = {'test1': {'y':  60},'test2': {'y':  70},'test3': {'y':  80}}
s = {k : r for k,r in mydict.items() if r['y'] < 75 }
> {'test1': {'y': 60}, 'test2': {'y': 70}}

Or we can even create something out of a list or range. E.g. if we want a dictionary with all odd square numbers:
```
{i : i**2 for i in range(11) if i % 2 == 1}
> {1: 1, 3: 9, 5: 25, 7: 49, 9: 81}
```

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你的笑 2024-09-15 19:31:36

您可以使用 d.values() 获取字典 d 的值列表。你的列表理解应该使用它来工作，尽管我有点不清楚你到底想要输出是什么。

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長街聽風 2024-09-15 19:31:36

还有别的办法吗？

为什么不考虑使用一些轻量级的对象呢？

您仍然可以使用列表推导式来收集或过滤对象，并在清晰度/可扩展性方面获得很大的提升。

>>> class Item(object):
...     def __init__(self, x, y, name):
...         self.x = x
...         self.y = y
...         self.name = name
... 
>>> list_items = []
>>> list_items.append(Item(x=70, y=60, name='test1420'))                        
>>> list_items.append(Item(x=94, y=72, name='test277'))                         
>>> items_matching = [item for item in list_items 
                      if 92 < item.x < 95 and 70 < item.y < 75]
>>> for item in items_matching:
...     print item.name
... 
test277
>>> first_item = items_matching[0]
>>> first_item.x += 50
>>> first_item.x
144

Is there another way?

Why not consider the use of some lightweight objects?

You can still use list comprehensions for gathering or filtering the objects, and gain a lot in clarity / extensibility.

>>> class Item(object):
...     def __init__(self, x, y, name):
...         self.x = x
...         self.y = y
...         self.name = name
... 
>>> list_items = []
>>> list_items.append(Item(x=70, y=60, name='test1420'))                        
>>> list_items.append(Item(x=94, y=72, name='test277'))                         
>>> items_matching = [item for item in list_items 
                      if 92 < item.x < 95 and 70 < item.y < 75]
>>> for item in items_matching:
...     print item.name
... 
test277
>>> first_item = items_matching[0]
>>> first_item.x += 50
>>> first_item.x
144

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