Python: sort dictionary with tuples as key by the value

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How do I sort a dictionary by value? 38 answers

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This gives what you want, sorted. But it's not in a dictionary at the end, since you can't sort a dictionary.

d={('w1','u1'):3,('w1','u2'):8,('w2','u1'):1,('w1','u3'):11,('w2','u3'):6}

d2 = {}

for (w,u) , value in d.items():
    if w not in d2:
        d2[w] = [(u,value)]
    else:
        d2[w].append((u, value))


for key, values in d2.items():
    print key, ":t", sorted(values, key=lambda x: -x[1]), "n"

which gives:

w2 :    [('u3', 6), ('u1', 1)] 

w1 :    [('u3', 11), ('u2', 8), ('u1', 3)] 

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Given the final dict you are after, it doesn't seem to me you need to sort. Just iterate through the key-value pairs, and rearrange them into a new dict:

d = {('w1', 'u1'): 3, ('w1', 'u2'): 8, ('w2', 'u1'): 1, ('w1', 'u3'): 11,
     ('w2', 'u3'): 6}
result = {}
for (w, u), val in d.iteritems():
    result.setdefault(w, {})[u] = val
print(result)

yields

{'w2': {'u1': 1, 'u3': 6}, 'w1': {'u1': 3, 'u3': 11, 'u2': 8}}

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