Python dictionary: Get list of values for list of keys

Question

A list comprehension seems to be a good way to do this:

>>> [mydict[x] for x in mykeys]
[3, 1]

A couple of other ways than list-comp:

• Build list and throw exception if key not found: map(mydict.__getitem__, mykeys)
• Build list with None if key not found: map(mydict.get, mykeys)

Alternatively, using operator.itemgetter can return a tuple:

from operator import itemgetter
myvalues = itemgetter(*mykeys)(mydict)
# use `list(...)` if list is required

Note: in Python3, map returns an iterator rather than a list. Use list(map(...)) for a list.

A little speed comparison:

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[1]: l = [0,1,2,3,2,3,1,2,0]
In[2]: m = {0:10, 1:11, 2:12, 3:13}
In[3]: %timeit [m[_] for _ in l]  # list comprehension
1000000 loops, best of 3: 762 ns per loop
In[4]: %timeit map(lambda _: m[_], l)  # using 'map'
1000000 loops, best of 3: 1.66 µs per loop
In[5]: %timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
1000000 loops, best of 3: 1.65 µs per loop
In[6]: %timeit map(m.__getitem__, l)
The slowest run took 4.01 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 853 ns per loop
In[7]: %timeit map(m.get, l)
1000000 loops, best of 3: 908 ns per loop
In[33]: from operator import itemgetter
In[34]: %timeit list(itemgetter(*l)(m))
The slowest run took 9.26 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 739 ns per loop

So list comprehension and itemgetter are the fastest ways to do this.

Update

For large random lists and maps I had a bit different results:

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[2]: import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)

%timeit f(m)
1000 loops, best of 3: 1.14 ms per loop

%timeit list(itemgetter(*l)(m))
1000 loops, best of 3: 1.68 ms per loop

%timeit [m[_] for _ in l]  # list comprehension
100 loops, best of 3: 2 ms per loop

%timeit map(m.__getitem__, l)
100 loops, best of 3: 2.05 ms per loop

%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
100 loops, best of 3: 2.19 ms per loop

%timeit map(m.get, l)
100 loops, best of 3: 2.53 ms per loop

%timeit map(lambda _: m[_], l)
100 loops, best of 3: 2.9 ms per loop

So in this case the clear winner is f = operator.itemgetter(*l); f(m), and clear outsider: map(lambda _: m[_], l) .

Update for Python 3.6.4

import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)

%timeit f(m)
1.66 ms ± 74.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit list(itemgetter(*l)(m))
2.1 ms ± 93.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit [m[_] for _ in l]  # list comprehension
2.58 ms ± 88.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(m.__getitem__, l))
2.36 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
2.98 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(m.get, l))
2.7 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(lambda _: m[_], l)
3.14 ms ± 62.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

So, results for Python 3.6.4 is almost the same.

Here are three ways.

Raising KeyError when key is not found:

result = [mapping[k] for k in iterable]

Default values for missing keys.

result = [mapping.get(k, default_value) for k in iterable]

Skipping missing keys.

result = [mapping[k] for k in iterable if k in mapping]

Try This:

mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one','ten']
newList=[mydict[k] for k in mykeys if k in mydict]
print newList
[3, 1]

Try this:

mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one'] # if there are many keys, use a set

[mydict[k] for k in mykeys]
=> [3, 1]

new_dict = {x: v for x, v in mydict.items() if x in mykeys}