A pile of small dictionaries like these:
born_place = {"David":"New York", "Juan":"Dallas"}
childhood_place = {"David":"Los Angeles"}
school_place = {"Mike":"New York", "David":"Houston"}
work_place = {"Kate":"Phoenix"}
university_place = {"Mike":"Chicago"}
I want to check, in a list of cities, all the names related (regardless what 'xx_place" is):
New York, Los Angeles, Chicago, Houston, Phoenix, Dallas
By doing the lines as below, I am able to check the "born_place" of 1 city:
New_York = []
Chicago = []
Houston = []
Phoenix = []
Dallas = []
try:
ny = list(born_place.keys())[list(born_place.values()).index('New York')] # reverse search in a dictionary
New_York.append(ny)
except:
pass
print (New_York)
The block of code is only for 1 xx_place for 1 city. 6 cities with 5 places require 30 blocks of these codes. (if the list of cities and and xx_place are longer...)
CodePudding user response:
Without using defaultdict you can utilise setdefault as follows:
born_place = {"David":"New York", "Juan":"Dallas"}
childhood_place = {"David":"Los Angeles"}
school_place = {"Mike":"New York", "David":"Houston"}
work_place = {"Kate":"Phoenix"}
university_place = {"Mike":"Chicago"}
result = {}
for d in born_place, childhood_place, school_place, work_place, university_place:
for k, v in d.items():
result.setdefault(v, []).append(k)
print(result)
Output:
{'New York': ['David', 'Mike'], 'Dallas': ['Juan'], 'Los Angeles': ['David'], 'Houston': ['David'], 'Phoenix': ['Kate'], 'Chicago': ['Mike']}
CodePudding user response:
The dictionaries can be combined together and traversed
from collections import defaultdict
born_place = {"David":"New York", "Juan":"Dallas"}
childhood_place = {"David":"Los Angeles"}
school_place = {"Mike":"New York", "David":"Houston"}
work_place = {"Kate":"Phoenix"}
university_place = {"Mike":"Chicago"}
res = defaultdict(list)
for item in [born_place, childhood_place, school_place, work_place, university_place]:
for k, v in item.items():
res[v].append(k)
print(dict(res))
# {'New York': ['David', 'Mike'], 'Dallas': ['Juan'], 'Los Angeles': ['David'], 'Houston': ['David'], 'Phoenix': ['Kate'], 'Chicago': ['Mike']}
CodePudding user response:
IIUC, you can use defaultdict(list)
and insert name
of dict
to key
of city
.
if the list of cities and xx_place are longer. if all city_name exist in dict
s and you don't want to check. You don't need to create a list of cities and You can create a list of dict
for xx_place
.
from collections import defaultdict
res = defaultdict(list)
# You can create a list of dict
# places = [{"David":"New York", "Juan":"Dallas"},
# {"David":"Los Angeles"},
# {"Mike":"New York", "David":"Houston"},
# {"Kate":"Phoenix"}, {"Mike":"Chicago"}]
# if you want to use 'list' of 'dict'
# for place in places:
for place in [born_place, childhood_place, school_place,
work_place, university_place]:
for k,v in place.items():
res[v].append(k)
print(res)
{
'New York': ['David', 'Mike'],
'Los Angeles': ['David'],
'Chicago': ['Mike'],
'Houston': ['David'],
'Phoenix': ['Kate'],
'Dallas': ['Juan']
}
CodePudding user response:
With if statement instead of defaultdict:
born_place = {"David":"New York", "Juan":"Dallas"}
childhood_place = {"David":"Los Angeles"}
school_place = {"Mike":"New York", "David":"Houston"}
work_place = {"Kate":"Phoenix"}
university_place = {"Mike":"Chicago"}
places = [born_place, childhood_place, school_place, work_place, university_place]
results = {}
for d in places:
for k, v in d.items():
if v not in results.keys():
results[v] = []
results[v].append(k)
print(results['New York'])