如何覆盖python字典值？

Question

A simplification of my dictionary:

my_dict = {
    'DOC': [
        [('k', 'v'), ('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3'), ('k4', 'v4')],
        [('k5', 'v5'), ('k', 'v'), ('k1', 'v1'), ('k6', 'v6')]
    ],
    'DIC': [
        [('k7', 'v7'), ('k3', 'v3'), ('k8', 'v8')],
        [('k5', 'v5'), ('k3', 'v3'), ('k1', 'v1'), ('k2', 'v2')],
        [('k4', 'v4'), ('k9', 'v9')]
    ],
    'INFRA': [
        [('k5', 'v5'), ('k3', 'v3'), ('k1', 'v1'), ('k2', 'v2')],
        [('k', 'v'), ('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3'), ('k4', 'v4')]
    ]
}

Overwriting my dictionary values:

for k, v in my_dict.items():
     my_dict[k] = map(lambda x: dict(x), my_dict[k])

Returning...

my_dict 
{'DIC': [{'k3': 'v3', 'k7': 'v7', 'k8': 'v8'},
         {'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'k5': 'v5'},
         {'k4': 'v4', 'k9': 'v9'}],
 'DOC': [{'k': 'v', 'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'k4': 'v4'},
         {'k': 'v', 'k1': 'v1', 'k5': 'v5', 'k6': 'v6'}],
 'INFRA': [{'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'k5': 'v5'},
           {'k': 'v', 'k1': 'v1', 'k2': 'v2', 'k3': 'v3', 'k4': 'v4'}]}

When I ran my real code following the same logic I got the error:

ValueError: dictionary update sequence element #0 has length 6; 2 is required

I've tried to create another dictionary to receive the map() but the same error occurred. Someone could help me?

Answer 1

The input in the real testcase doesn't conform to the structure you expect. In particular, instead of all being tuples, at least one of the elements is a sequence of length 6. For example, a faulty input may look like:

{'DOC': [
    [('123456')],               # A string is a sequence, too
    [('123' '456')]             # Note the lack of a comma
    [('1','2','3','4','5','6')] # A tuple, but with 6 elements
    [['1','2','3','4','5','6']] # A list
]}

To solve this problem, check your input, like:

for k, v in my_dict.items():
    for l in v:
        for subl in v:
            for tup in subl:
                if not len(tup) == 2:
                    raise ValueError('In %r: value is not a tuple(2), but %r'
                                     % (k, tup))

Oh, by the way, when designing data structures, it's often a good idea to follow the Zen of Python , which includes:

Flat is better than nested.

Answer 2

The dict constructor expects a sequence of 2 item subsequences (lists or tuples). At least one of subsequences contains more then two items.

Perhaps you have missing comma (see last item):

>>> dict([('1', 'red'),('2', 'blue'),('3' 'purple')]) # missing comma in last item
Traceback (most recent call last):
  File "<pyshell#166>", line 1, in <module>
    dict([('1', 'red'),('2', 'blue'),('3' 'purple')]) 
ValueError: dictionary update sequence element #2 has length 7; 2 is required

Or maybe a single item instead of a tuple (again, see last item)

>>> dict([('1', 'red'),('2', 'blue'),('purple')]) # last "tuple" has one item
Traceback (most recent call last):
  File "<pyshell#167>", line 1, in <module>
    dict([('1', 'red'),('2', 'blue'),('purple')])
ValueError: dictionary update sequence element #2 has length 6; 2 is required

Answer 3

The problem is that in your "real code", the first element of one of your lists has more than two elements (6). dict requires exactly two to interpret them as key, value .

Here's an example that generates the error you describe:

my_dict = {
    'DOC': [
        [('k', 'v', 'extra1', 'extra2', 'extra3', 'extra4'), ('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3'), ('k4', 'v4')],
        [('k5', 'v5'), ('k', 'v'), ('k1', 'v1'), ('k6', 'v6')]
    ],
    'DIC': [
        [('k7', 'v7'), ('k3', 'v3'), ('k8', 'v8')],
        [('k5', 'v5'), ('k3', 'v3'), ('k1', 'v1'), ('k2', 'v2')],
        [('k4', 'v4'), ('k9', 'v9')]
    ],
    'INFRA': [
        [('k5', 'v5'), ('k3', 'v3'), ('k1', 'v1'), ('k2', 'v2')],
        [('k', 'v'), ('k1', 'v1'), ('k2', 'v2'), ('k3', 'v3'), ('k4', 'v4')]
    ]
}

Here's a simple way to drop in a print statement to see the problem element:

for k, v in my_dict.items():

    # Print out the first element of each list,
    # because one of them has six elements when it should only have two.
    for l in my_dict[k]:
        print l[0]

    my_dict[k] = map(lambda x: dict(x), my_dict[k])

Answer 4

I can see no problem with code and data in your question. Assuming there's nothing significantly different about the "real" code, you could do something like this to find out what the malformed data is that causes the error to occur:

def test(x):
    print x
    return dict(x)

for k, v in my_dict.items():
#     my_dict[k] = map(lambda x: dict(x), my_dict[k])
     my_dict[k] = map(test, my_dict[k])

Answer 5

How about this code which is simpler than yours:

temp_dict = {}
for k in my_dict():
    temp_dict[k] = {}
    for t in my_dict[k]:
        temp_dict[k][t[0]] = t[1]
my_dict = temp_dec
del temp_dec