问题
I have a dictionary that I want to group by the common values:
init_dict = {'00001': 'string1', '00002': 'string2', '00003': 'string1', '00004': 'string3', '00005': 'string2'}
I want to create a new dictionary that groups the values and lists the keys like this:
new_dict = {'string1': ['00001', '00003'], 'string2':['00002', '00004'], 'string3': ['00004']}
I tried many things and this is the closest I can get.
lookup = 'string1'
all_keys = []
for k, v in init_dict.items():
if v == lookup:
all_keys.append(k)
print(all_keys)
This produces the first list: ['00001', '00003']
so I thought I could somehow loop through a list of lookup
values but can't since I'm working with strings. Is there a way to do this and is there a way that is relatively efficient because my initial dictionary has 53,000 items in it. Any help would be much appreciated as I've been trying different things for hours.
回答1:
Use a defaultdict, specifying a list
as default argument, and append the corresponding values from the dictionary:
from collections import defaultdict
d = defaultdict(list)
for k,v in init_dict.items():
d[v].append(k)
print(d)
defaultdict(list,
{'string1': ['00001', '00003'],
'string2': ['00002', '00005'],
'string3': ['00004']})
回答2:
You can use defaultdict
result = defaultdict(list)
for k, v in init_dict.items():
result[v].append(k)
or itertools.groupby
result = {k: [x[0] for x in v] for k, v in
groupby(sorted(init_dict.items(), key=lambda kv: kv[1]), key=lambda kv: kv[1])}
回答3:
You can also use a normal dict
(instead of defaultdict
):
new_dict = {}
for key, val in init_dict.items():
if val in new_dict:
new_dict[val].append(key)
else:
new_dict[val] = []
new_dict[val].append(key)
Output:
new_dict = {'string1': ['00001', '00003'],
'string2': ['00002', '00005'],
'string3': ['00004']}
来源:https://stackoverflow.com/questions/61427141/how-to-reverse-dictionary-items-and-list-keys-grouped-by-common-values