列表到字典转换，每个键有多个值？

Question

I have a Python list which holds pairs of key/value:

l = [[1, 'A'], [1, 'B'], [2, 'C']]

I want to convert the list into a dictionary, where multiple values per key would be aggregated into a tuple:

{1: ('A', 'B'), 2: ('C',)}

The iterative solution is trivial:

l = [[1, 'A'], [1, 'B'], [2, 'C']]
d = {}
for pair in l:
    if pair[0] in d:
        d[pair[0]] = d[pair[0]] + tuple(pair[1])
    else:
        d[pair[0]] = tuple(pair[1])

print(d)

{1: ('A', 'B'), 2: ('C',)}

Is there a more elegant, Pythonic solution for this task?

Answer 1

from collections import defaultdict

d1 = defaultdict(list)

for k, v in l:
    d1[k].append(v)

d = dict((k, tuple(v)) for k, v in d1.items())

d contains now {1: ('A', 'B'), 2: ('C',)}

d1 is a temporary defaultdict with lists as values, which will be converted to tuples in the last line. This way you are appending to lists and not recreating tuples in the main loop.

Answer 2

Using lists instead of tuples as dict values:

l = [[1, 'A'], [1, 'B'], [2, 'C']]
d = {}
for key, val in l:
    d.setdefault(key, []).append(val)

print(d)

Using a plain dictionary is often preferable over a defaultdict , in particular if you build it just once and then continue to read from it later in your code:

First, the plain dictionary is faster to build and access.

Second, and more importantly, the later read operations will error out if you try to access a key that doesn't exist, instead of silently creating that key. A plain dictionary lets you explicitly state when you want to create a key-value pair, while the defaultdict always implicitly creates them, on any kind of access.

Answer 3

This method is relatively efficient and quite compact:

reduce(lambda x, (k,v): x[k].append(v) or x, l, defaultdict(list))

In Python3 this becomes (making exports explicit):

dict(functools.reduce(lambda x, d: x[d[0]].append(d[1]) or x, l, collections.defaultdict(list)))

Note that reduce has moved to functools and that lambdas no longer accept tuples. This version still works in 2.6 and 2.7.

Answer 4

Are the keys already sorted in the input list? If that's the case, you have a functional solution:

import itertools

lst = [(1, 'A'), (1, 'B'), (2, 'C')]
dct = dict((key, tuple(v for (k, v) in pairs)) 
           for (key, pairs) in itertools.groupby(lst, lambda pair: pair[0]))
print dct
# {1: ('A', 'B'), 2: ('C',)}

Answer 5

I had a list of values created as follows:

performance_data = driver.execute_script('return window.performance.getEntries()')  

Then I had to store the data (name and duration) in a dictionary with multiple values:

dictionary = {}
    for performance_data in range(3):
        driver.get(self.base_url)
        performance_data = driver.execute_script('return window.performance.getEntries()')
        for result in performance_data:
            key=result['name']
            val=result['duration']
            dictionary.setdefault(key, []).append(val)
        print(dictionary)

Answer 6

My data was in a Pandas.DataFrame

myDict = dict()

for idin set(data['id'].values):
    temp = data[data['id'] == id]
    myDict[id] = temp['IP_addr'].to_list()
    
myDict

Gave me a Dict of the keys, ID, mappings to >= 1 IP_addr . The first IP_addr is Guaranteed. My code should work even if temp['IP_addr'].to_list() == []

{'fooboo_NaN': ['1.1.1.1', '8.8.8.8']}