list to dictionary conversion with multiple values per key?
PythonListDictionaryType ConversionPython Problem Overview
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?
Python Solutions
Solution 1 - Python
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.
Solution 2 - Python
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.
Solution 3 - Python
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.
Solution 4 - Python
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',)}
Solution 5 - Python
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)
Solution 6 - Python
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']}