question regarding a nested dictionary is there a way to merge a nested dictionary in one dictionary
Question:
I’m following a python course on runestone and i’m stuck with the following question:
Provided is a dictionary that contains pokemon go player data, where each player reveals the amount of candy each of their pokemon have. If you pooled all the data together, which pokemon has the highest number of candy? Assign that pokemon to the variable most_common_pokemon.
what i thought is to created a dictionary that merge the common keys (and their value or to make a comparison something like
if x>y
x=y
so I can get the pokemon with the highest number of candies
pokemon_go_data = {'bentspoon':
{'Rattata': 203, 'Pidgey':20, 'Drowzee': 89, 'Squirtle': 35, 'Pikachu': 3, 'Eevee': 34, 'Magikarp': 300, 'Paras': 38},
'Laurne':
{'Pidgey': 169, 'Rattata': 245, 'Squirtle': 9, 'Caterpie': 38, 'Weedle': 97, 'Pikachu': 6, 'Nidoran': 44, 'Clefairy': 15, 'Zubat': 79, 'Dratini': 4},
'picklejarlid':
{'Rattata': 32, 'Drowzee': 15, 'Nidoran': 4, 'Bulbasaur': 3, 'Pidgey': 56, 'Weedle': 21, 'Oddish': 18, 'Magmar': 6, 'Spearow': 14},
'professoroak':
{'Charmander': 11, 'Ponyta': 9, 'Rattata': 107, 'Belsprout': 29, 'Seel': 19, 'Pidgey': 93, 'Shellder': 43, 'Drowzee': 245, 'Tauros': 18, 'Lapras': 18}}
pokemon=[]
for i,k in pokemon_go_data.items():
b=k.keys()
b=list(b)
pokemon.append(b)
print (pokemon)
poke=[]
for i in pokemon:
for j in i:
if j not in poke:
poke.append(j)
else:
continue
print(poke)
d={}
n=0
count=[]
total=0
most_common_pokemon=""
for players in pokemon_go_data:
for pokemon in pokemon_go_data[players]:
if pokemon==poke[n]:
count.append(pokemon_go_data[players][pokemon])
counts=sum(count)
print (count)
print(counts)
d[poke[n]]=counts
print (d)
by doing so it prints a dictionary: {'Rattata': 587}
but if i add a counter like n+=1 i got the following
{'Rattata': 203, 'Pidgey': 372, 'Drowzee': 387}
if instead of creating a dictionary something like
if count>total:
total=count
most_common_pokemon=poke[n]
n+1=n
i got a out of range error message
i placed the counter everywhere but it doesn’t work…and also when i reset count
thanks any suggestion is more than welcome
Answers:
This should do it:
pokemon_total = {}
for player, dictionary in pokemon_go_data.items():
for pokemon, candy_count in dictionary.items():
if pokemon in pokemon_total.keys():
pokemon_total[pokemon] += candy_count
else:
pokemon_total[pokemon] = candy_count
most_common_pokemon = max(pokemon_total, key=pokemon_total.get)
print(most_common_pokemon)
out = {}
for k,v in [[k2,p[k1][k2]] for k1 in p for k2 in p[k1]]:
if k in out.keys():
out[k] = out[k] + v
else:
out[k] = v
print(max(out, key=out.get))
basically same as above answer in principle but slightly diff implementation
OR
from itertools import groupby
out = sorted([[k2,p[k1][k2]] for k1 in p for k2 in p[k1]])
result = {a:sum(c for _, c in b) for a, b in groupby(out, key=lambda x:x[0])}
print(max(result,key=result.get))
OR
out = sum(map(Counter, p.values()), Counter())
print(max(out,key=result.get))
pokemon_total = {}
for player, dictionary in pokemon_go_data.items():
for pokemon, candy_count in dictionary.items():
if pokemon in pokemon_total.keys():
pokemon_total[pokemon] += candy_count
else:
pokemon_total[pokemon] = candy_count
max_occurence = 0
for pokemon in pokemon_total:
if pokemon_total[pokemon] > max_occurence :
max_occurence = pokemon_total[pokemon]
most_common_pokemon = pokemon
print(most_common_pokemon)
T did this just now worked for me and gives the correct answer ‘Rattata’
pokemon_go_data = {'bentspoon':
{'Rattata': 203, 'Pidgey': 120, 'Drowzee': 89, 'Squirtle': 35, 'Pikachu': 3, 'Eevee': 34,
'Magikarp': 300, 'Paras': 38},
'Laurne':
{'Pidgey': 169, 'Rattata': 245, 'Squirtle': 9, 'Caterpie': 38, 'Weedle': 97, 'Pikachu': 6,
'Nidoran': 44, 'Clefairy': 15, 'Zubat': 79, 'Dratini': 4},
'picklejarlid':
{'Rattata': 32, 'Drowzee': 15, 'Nidoran': 4, 'Bulbasaur': 3, 'Pidgey': 56, 'Weedle': 21,
'Oddish': 18, 'Magmar': 6, 'Spearow': 14},
'professoroak':
{'Charmander': 11, 'Ponyta': 9, 'Rattata': 107, 'Belsprout': 29, 'Seel': 19, 'Pidgey': 93,
'Shellder': 43, 'Drowzee': 245, 'Tauros': 18, 'Lapras': 18}}
count_d={}
pokemon_main_lst=pokemon_go_data.keys()
#print(pokemon_main_lst)
for main_keys in pokemon_main_lst:
pokemon_sub_lst=pokemon_go_data[main_keys].keys()
#print(sub_lst)
for pokemon in pokemon_sub_lst:
if pokemon not in count_d:
count_d[pokemon]=0
count_d[pokemon]+=pokemon_go_data[main_keys][pokemon]
#print(count_d)
most_common_pokemon=sorted(count_d,key=lambda k:count_d[k])[-1]
print(most_common_pokemon)
This may help
d={}
for name,val in pokemon_go_data.items():
for key in val:
if key not in d:
d[key]=0
d[key]+=pokemon_go_data[name][key]
most_common_pokemon=sorted(d.keys(), key= lambda k:d[k])[-1]
pcandy={}
for v in pokemon_go_data.values():
for j in v:
pcandy[j]=pcandy.get(j,0)+v[j]
s_sorted=sorted(pcandy, key= lambda x: pcandy[x], reverse=True)
most_common_pokemon=s_sorted[0]
print(most_common_pokemon)
d={}
for pokemons in pokemon_go_data.values():
for name,value in pokemons.items():
d[name]=d.get(name,0)+value
most_common_pokemon=sorted(d.keys(), key= lambda k:d[k])[-1]
I’m following a python course on runestone and i’m stuck with the following question:
Provided is a dictionary that contains pokemon go player data, where each player reveals the amount of candy each of their pokemon have. If you pooled all the data together, which pokemon has the highest number of candy? Assign that pokemon to the variable most_common_pokemon.
what i thought is to created a dictionary that merge the common keys (and their value or to make a comparison something like
if x>y
x=y
so I can get the pokemon with the highest number of candies
pokemon_go_data = {'bentspoon':
{'Rattata': 203, 'Pidgey':20, 'Drowzee': 89, 'Squirtle': 35, 'Pikachu': 3, 'Eevee': 34, 'Magikarp': 300, 'Paras': 38},
'Laurne':
{'Pidgey': 169, 'Rattata': 245, 'Squirtle': 9, 'Caterpie': 38, 'Weedle': 97, 'Pikachu': 6, 'Nidoran': 44, 'Clefairy': 15, 'Zubat': 79, 'Dratini': 4},
'picklejarlid':
{'Rattata': 32, 'Drowzee': 15, 'Nidoran': 4, 'Bulbasaur': 3, 'Pidgey': 56, 'Weedle': 21, 'Oddish': 18, 'Magmar': 6, 'Spearow': 14},
'professoroak':
{'Charmander': 11, 'Ponyta': 9, 'Rattata': 107, 'Belsprout': 29, 'Seel': 19, 'Pidgey': 93, 'Shellder': 43, 'Drowzee': 245, 'Tauros': 18, 'Lapras': 18}}
pokemon=[]
for i,k in pokemon_go_data.items():
b=k.keys()
b=list(b)
pokemon.append(b)
print (pokemon)
poke=[]
for i in pokemon:
for j in i:
if j not in poke:
poke.append(j)
else:
continue
print(poke)
d={}
n=0
count=[]
total=0
most_common_pokemon=""
for players in pokemon_go_data:
for pokemon in pokemon_go_data[players]:
if pokemon==poke[n]:
count.append(pokemon_go_data[players][pokemon])
counts=sum(count)
print (count)
print(counts)
d[poke[n]]=counts
print (d)
by doing so it prints a dictionary: {'Rattata': 587}
but if i add a counter like n+=1 i got the following
{'Rattata': 203, 'Pidgey': 372, 'Drowzee': 387}
if instead of creating a dictionary something like
if count>total:
total=count
most_common_pokemon=poke[n]
n+1=n
i got a out of range error message
i placed the counter everywhere but it doesn’t work…and also when i reset count
thanks any suggestion is more than welcome
This should do it:
pokemon_total = {}
for player, dictionary in pokemon_go_data.items():
for pokemon, candy_count in dictionary.items():
if pokemon in pokemon_total.keys():
pokemon_total[pokemon] += candy_count
else:
pokemon_total[pokemon] = candy_count
most_common_pokemon = max(pokemon_total, key=pokemon_total.get)
print(most_common_pokemon)
out = {}
for k,v in [[k2,p[k1][k2]] for k1 in p for k2 in p[k1]]:
if k in out.keys():
out[k] = out[k] + v
else:
out[k] = v
print(max(out, key=out.get))
basically same as above answer in principle but slightly diff implementation
OR
from itertools import groupby
out = sorted([[k2,p[k1][k2]] for k1 in p for k2 in p[k1]])
result = {a:sum(c for _, c in b) for a, b in groupby(out, key=lambda x:x[0])}
print(max(result,key=result.get))
OR
out = sum(map(Counter, p.values()), Counter())
print(max(out,key=result.get))
pokemon_total = {}
for player, dictionary in pokemon_go_data.items():
for pokemon, candy_count in dictionary.items():
if pokemon in pokemon_total.keys():
pokemon_total[pokemon] += candy_count
else:
pokemon_total[pokemon] = candy_count
max_occurence = 0
for pokemon in pokemon_total:
if pokemon_total[pokemon] > max_occurence :
max_occurence = pokemon_total[pokemon]
most_common_pokemon = pokemon
print(most_common_pokemon)
T did this just now worked for me and gives the correct answer ‘Rattata’
pokemon_go_data = {'bentspoon':
{'Rattata': 203, 'Pidgey': 120, 'Drowzee': 89, 'Squirtle': 35, 'Pikachu': 3, 'Eevee': 34,
'Magikarp': 300, 'Paras': 38},
'Laurne':
{'Pidgey': 169, 'Rattata': 245, 'Squirtle': 9, 'Caterpie': 38, 'Weedle': 97, 'Pikachu': 6,
'Nidoran': 44, 'Clefairy': 15, 'Zubat': 79, 'Dratini': 4},
'picklejarlid':
{'Rattata': 32, 'Drowzee': 15, 'Nidoran': 4, 'Bulbasaur': 3, 'Pidgey': 56, 'Weedle': 21,
'Oddish': 18, 'Magmar': 6, 'Spearow': 14},
'professoroak':
{'Charmander': 11, 'Ponyta': 9, 'Rattata': 107, 'Belsprout': 29, 'Seel': 19, 'Pidgey': 93,
'Shellder': 43, 'Drowzee': 245, 'Tauros': 18, 'Lapras': 18}}
count_d={}
pokemon_main_lst=pokemon_go_data.keys()
#print(pokemon_main_lst)
for main_keys in pokemon_main_lst:
pokemon_sub_lst=pokemon_go_data[main_keys].keys()
#print(sub_lst)
for pokemon in pokemon_sub_lst:
if pokemon not in count_d:
count_d[pokemon]=0
count_d[pokemon]+=pokemon_go_data[main_keys][pokemon]
#print(count_d)
most_common_pokemon=sorted(count_d,key=lambda k:count_d[k])[-1]
print(most_common_pokemon)
This may help
d={}
for name,val in pokemon_go_data.items():
for key in val:
if key not in d:
d[key]=0
d[key]+=pokemon_go_data[name][key]
most_common_pokemon=sorted(d.keys(), key= lambda k:d[k])[-1]
pcandy={}
for v in pokemon_go_data.values():
for j in v:
pcandy[j]=pcandy.get(j,0)+v[j]
s_sorted=sorted(pcandy, key= lambda x: pcandy[x], reverse=True)
most_common_pokemon=s_sorted[0]
print(most_common_pokemon)
d={}
for pokemons in pokemon_go_data.values():
for name,value in pokemons.items():
d[name]=d.get(name,0)+value
most_common_pokemon=sorted(d.keys(), key= lambda k:d[k])[-1]