With list comprehensions:

myDict = [{'first': 'James', 'middle': 'Smith', 'last': 'Joule'}, 
      {'first': 'James', 'middle': 'Johnson', 'last': 'Watt'},
      {'first': 'Christian', 'middle': 'Edward', 'last': 'Doppler'},
      {'first': 'Robert', 'last': 'Antonio'}]

keys = {"middle", "last"}

l = [{k:v for k, v in i.items() if k in keys} for i in myDict]

But you can also use map for this:

myDict = [{'first': 'James', 'middle': 'Smith', 'last': 'Joule'}, 
      {'first': 'James', 'middle': 'Johnson', 'last': 'Watt'},
      {'first': 'Christian', 'middle': 'Edward', 'last': 'Doppler'},
      {'first': 'Robert', 'last': 'Antonio'}]

keys = {"middle", "last"}

l = list(map(lambda x: {k:v for k, v in x.items() if k in keys}, myDict))
print(l)

output:

[{'last': 'Joule', 'middle': 'Smith'}, {'last': 'Watt', 'middle': 'Johnson'}, {'last': 'Doppler', 'middle': 'Edward'}, {'last': 'Antonio'}]

Check if first value of the tuple key is 2, with dict comprehension:

{k: v for k, v in dct.items() if k[0]==2}

So:

In [11]: dct = {(0, 'DRYER'): [103.0, 131.0, 9.0, 1.24],
    ...:  (2, 'DRYER'): [106.0, 120.0, 5.0, 1.24],
    ...:  (2, 'WASHING'): [70.0, 90.0, 11.0, 0.19]}
    ...:  

In [12]: {k: v for k, v in dct.items() if k[0]==2}
Out[12]: 
{(2, 'DRYER'): [106.0, 120.0, 5.0, 1.24],
 (2, 'WASHING'): [70.0, 90.0, 11.0, 0.19]}

Constructing a new dict:

dict_you_want = {key: old_dict[key] for key in your_keys}

Uses dictionary comprehension.

If you use a version which lacks them (ie Python 2.6 and earlier), make it dict((key, old_dict[key]) for ...). It's the same, though uglier.

Note that this, unlike jnnnnn's version, has stable performance (depends only on number of your_keys) for old_dicts of any size. Both in terms of speed and memory. Since this is a generator expression, it processes one item at a time, and it doesn't looks through all items of old_dict.

Removing everything in-place:

unwanted = set(old_dict) - set(your_keys)
for unwanted_key in unwanted: del your_dict[unwanted_key]

If you first sort your list by "name", you can use itertools.groupby to group them, then use min with a lambda to find the minimum "price" in each group.

>>> from itertools import groupby
>>> sorted_orders = sorted(orders, key=lambda i: i["name"])
>>> [min(g, key=lambda j: j["price"]) for k,g in groupby(sorted_orders , key=lambda i: i["name"])]
[{'name': 'v', 'price': 123, 'location': 'Mars'},
 {'name': 'x', 'price': 124, 'location': 'Mars'},
 {'name': 'y', 'price': 456, 'location': 'Mars'},
 {'name': 'z', 'price': 123, 'location': 'Mars'}]