列表的python dict

Question

I have following dict:

items_temp = dict(fruits=["apple", "orange"], vegetables=["carrot", "potato"], animals=["dog", "cat"])

and following list to be verified what kind of things it contains.

check = ["orange", "dog", "apple"]

is there any clever pythonic way to obtain following dict from data above?:

output = dict(fruits = ["orange", "apple"], animals=["dog"])

Answer 1

I think you should be able to do the following:

check = set(['orange', 'dog', 'apple'])
output = {k: check.intersection(v) for k, v in items_temp.items() if check.intersection(v)}

Basically, I'm checking the intersection between check and the values of your dictionary. If there is an intersection, we add it to the output.

This will give you a dictionary with sets as values, but you can convert that pretty easily.

---

Note that we're doing the intersection check twice. That's a bit annoying (and we definitely don't need to) if we add an extra step into the processing pipeline...

check = set(['orange', 'dog', 'apple'])
keys_intersect = ((k, check.intersection(v)) for k, v in items_temp.iteritems())
output = {k: intersect for k, intersect in keys_intersect if intersect}

Answer 2

There isn't a clean, single step way to get what you want, but you can do it in two:

>>> output = {k:[v for v in vs if v in check] for k,vs in items_temp.items()}
>>> output
{'vegetables': [], 'animals': ['dog'], 'fruits': ['apple', 'orange']}

Next, we just need to filter out the empty lists:

>>> output = {k:vs for k,vs in output.items() if vs}
>>> output
{'animals': ['dog'], 'fruits': ['apple', 'orange']}

If you've got a lot of items to check, you can speed things up considerably by turning check into a set , but premature optimization is the root of all evil .

Edit: I suppose you can do it in one:

>>> output = {k:[v for v in vs if v in check] for k,vs in items_temp.items() if any(v in check for v in vs)}
>>> output
{'animals': ['dog'], 'fruits': ['apple', 'orange']}

But that's over-complicating things with redundant testing.

You could also do

{k:vs for k,vs in ((k,[v for v in vs if v in check]) for k,vs in items_temp.items()) if vs}

To do it in one step without redundant membership checking, but now we're getting a little silly.

Answer 3

Why don't you use a dictionary?

a={"Mobin":"Iranian","Harry":"American"}

and you can get that with this:

print a.get("Mobin")

when you run this code,you can see 'Iranian' in screen