Python: How to access the dictionary that has specific key-values in a list

Question

I have a list of dictionaries:

[
{"START":"Denver", "END":"Chicago", "Num":0},
{"START":"Dallas", "END":"Houston", "Num":3},
{"START":"Virginia", "END":"Boston", "Num":1},
{"START":"Washington", "END":"Maine", "Num":7}
]

How do I access the dictionary in this list that has "START":"Virginia", "END":"Boston" in most pythonic way?

Do you guarantee uniqueness? Can the ordering change? – thegrinner Jan 16 '13 at 19:20 — thegrinner, Jan 16 '13 at 19:20
Yes I guarantee uniqueness, and ordering can't change. – alwbtc Jan 16 '13 at 19:22 — alwbtc, Jan 16 '13 at 19:22
i'm trying to change "Num" values – alwbtc Jan 16 '13 at 19:25 — alwbtc, Jan 16 '13 at 19:25

Mark Reed · Accepted Answer · 2013-01-16T22:33:19.313

The most Pythonic way is probably a list comprehension:

[ d for d in dict_list if d["START"] == "Virginia" and d["END"] == "Boston" ]

As mgilson pointed out, if you are assuming that there's only one item in the list with that pair of locations, you can use next with the same generator expression, instead of the brackets. That will return just the matching dict, rather than a one-item list containing it:

trip = next( d for d in dict_list 
               if d["START"] == "Virginia" and d["END"] == "Boston" )

Either way, the result references the same dict object(s) as the original list, so you can make changes:

trip["Num"] = trip["Num"] + 1

And those changes will be there when accessed through the original list:

print(dict_list[2]["Num"])    # 2

As Ashwini Chaudhary indicated in his answer, your search may itself be specified as a dict. In that case, the if condition in the generator expression is a little different, but the logic is otherwise the same:

search = { "START": "Virginia", "END": "Boston" }
trip = next(d for d in dict_list if all(i in d.items() for i in search.items()))

The latter can also be written as `...if set(d.items()) & set(search.items())` — georg, Jan 16 '13 at 19:30
after accessing this dict, i will change the value of its "Num" key, how to do it? — alwbtc, Jan 16 '13 at 19:34
You could also do `next( d for d in lis if d['START'] == 'Virginia' ...)` — mgilson, Jan 16 '13 at 19:34

Ashwini Chaudhary · Answer 2 · 2013-01-16T19:43:55.317

use all() with dict.items():

In [66]: lis=[                                      
   ....: {"START":"Denver", "END":"Chicago", "Num":0},
   ....: {"START":"Dallas", "END":"Houston", "Num":3},
   ....: {"START":"Virginia", "END":"Boston", "Num":1},
   ....: {"START":"Washington", "END":"Maine", "Num":7}
   ....: ]

In [67]: for x in lis:                              
   ....:         if all(y in x.items() for y in search.items()):
   ....:                 x['Num']="foobar"        #change Num here
   ....:         

In [68]: lis
Out[68]: 
[{'END': 'Chicago', 'Num': 0, 'START': 'Denver'},
 {'END': 'Houston', 'Num': 3, 'START': 'Dallas'},
 {'END': 'Boston', 'Num': 'foobar', 'START': 'Virginia'},
 {'END': 'Maine', 'Num': 7, 'START': 'Washington'}]

using a list comprehension:

In [58]: [x for x in lis if all(y in x.items() for y in search.items())]
Out[58]: [{'END': 'Boston', 'Num': 1, 'START': 'Virginia'}]

Or in comprehension form, `[x for x in lis if all(y in x.items() for y in search.items())]` — Eric, Jan 16 '13 at 19:25
after accessing this dict, i will change the value of its "Num" key, how to do it? — alwbtc, Jan 16 '13 at 19:39

Facundo Casco · Answer 3 · 2013-01-16T20:15:49.737

2

If you are going to do the search a lot of times and you know there is one and only one dict for every pair of locations you could create a dict from the list like this:

d = dict(((x['START'], x['END']), x) for x in list_of_dicts)

And then just do lookups in the new dict like this:

found_dict = d[('Virginia', 'Chicago')]

You could change found_dict then however you need.

edited Jan 16 '13 at 20:15

answered Jan 16 '13 at 19:44

Facundo Casco

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or: `{','.join([x['START'], x['END']]): x for x in list_of_dicts}` – Eric Jan 16 '13 at 19:59
@Eric, sure for Python >= 2.7 – Facundo Casco Jan 16 '13 at 20:00
why join them? better to use a `(start, end)` tuple as the key – Eevee Jan 16 '13 at 20:12

score 1 · Answer 4 · answered Jan 16 '13 at 19:25

If each item is unique, then you could do this:

def get_dict(items, start, end):
    for dict in items:
        if dict['START'] == start and dict['END'] == end:
            return dict

And then:

>>> items = [
    {"START":"Denver", "END":"Chicago", "Num":0},
    {"START":"Dallas", "END":"Houston", "Num":3},
    {"START":"Virginia", "END":"Boston", "Num":1},
    {"START":"Washington", "END":"Maine", "Num":7}
]
>>> get_dict(items, 'Virginia', 'Boston')
{"START":"Virginia", "END":"Boston", "Num":1}

It's fairly straightforward, but I thought I'd post it for the sake of completeness.

`return None` is completely redundant, since a function returns none by default. — Eric, Jan 16 '13 at 19:26

Python: How to access the dictionary that has specific key-values in a list

4 Answers4