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Es gibt zwei Wörterbücher a und b. Sie haben beide eine gemeinsame ID 1. Das Zusammenführen von Diktaten ist dem Ausführungsergebnis in MySQL sehr ähnlich
a = {
"id": "1",
"MUT": "1500",
}
b = {
"id": "1",
"neighbor": [2]
}
# result = addfunction(a,b)
result = {
"id": "1",
"MUT": "1500",
"neighbor": [2]
}
select a.id,a.MUT,b.neighbor from a full join b on a.id = b.id
Bedenken Sie Komplikationen:
a = [
{
"id": "1",
"MUT": "1500",
},
{
"id": "2",
"MUT": "1500",
}
]
b = [
{
"id": "1",
"neighbor": [2]
},
{
"id": "3",
"neighbor": [2]
}
]
# result = addfunction(a,b)
result = [
{
"id": "1",
"MUT": "1500",
"neighbor": [2]
},
{
"id": "2",
"MUT": "1500",
},
{
"id": "3",
"neighbor": [2]
}
]
PHP中文网2017-06-14 10:55:21
l_a = len(a)
l_b = len(b)
b_map = {}
result = []
for _ in range(l_b):
i = b.pop()
b_map[i['id']] = i
for _ in range(l_a):
i = a.pop()
if i['id'] in b_map:
i.update(b_map.pop(i['id']))
result.append(i)
result.extend(b_map.values())
print(result)
[{'MUT': '1500', 'id': '2'}, {'MUT': '1500', 'id': '1', 'neighbor': [2]}, {'id': '3', 'neighbor': [2]}]
某草草2017-06-14 10:55:21
from collections import defaultdict
def combineListDict(l1,l2,joinKeys = ["id"]):
d = defaultdict(dict)
for l in (l1, l2):
for elem in l:
if len(joinKeys) == 1:
joinKeysStr = elem[joinKeys[0]]
else:
joinKeysStr = reduce((lambda x, y: str(elem[x]) + str(elem[y])),joinKeys)
d[joinKeysStr].update(elem)
return d.values()
因为join on 条件可能时多个值,我在stack上找了个代码改了一下。