Python list comprehension for dictionaries in dictionaries?

Question

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?

Solution 1

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.

Solution 2

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.

Solution 3

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:

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

Solution 4

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}

1. 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}}
   

2. 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}
   

Solution 5

You can get a list of the values of a dictionary d with d.values(). Your list comprehension should work using that, although I'm a little unclear what exactly you want the output to be.

View more solutions