Python – How to fix “ValueError: not enough values to unpack (expected 2, got 1)”

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I need to write a function add_to_dict(d, key_value_pairs) which adds each given key/value pair to a python dictionary. The argument key_value_pairs will be a list of tuples in the form (key, value).

The function should return a list of all of the key/value pairs which have changed (with their original values).

def add_to_dict(d, key_value_pairs):

   newlist = []

   for key,value in d:
       for x,y in key_value_pairs:
           if x == key:
              newlist.append(x,y)
            
   return newlist

I keep getting an error

ValueError: not enough values to unpack (expected 2, got 1)

How do I solve this error?


Solution

How you should debug your code

'''
@param d: a dictionary
@param key_value_pairs: a list of tuples in the form `(key, value)`
@return: a list of tuples of key-value-pair updated in the original dictionary
'''
def add_to_dict(d, key_value_pairs):

    newlist = []

    for pair in key_value_pairs:

        # As is mentioned by Mr Patrick
        # you might not want to unpack the key-value-pair instantly
        # to avoid possible corrupted data input from
        # argument `key_value_pairs`
        # if you can't guarantee its integrity
        try:
            x, y = pair
        except (ValueError):
            # unable to unpack tuple
            tuple_length = len(pair)
            raise RuntimeError('''Invalid argument `key_value_pairs`!
                Corrupted key-value-pair has ({}) length!'''.format(tuple_length))

        # Instead of using nesting loop
        # using API would be much more preferable
        v = d.get(x)

        # Check if the key is already in the dictionary `d`
        if v:
            # You probably mean to append a tuple
            # as `array.append(x)` takes only one argument
            # @see: https://docs.python.org/3.7/library/array.html#array.array.append
            #
            # Besides, hereby I quote
            # "The function should return a list of all of the key/value pairs which have changed (with their original values)."
            # Thus instead of using the following line:
            #
            # newlist.append((x, y,))
            #
            # You might want a tuple of (key, old_value, new_value)
            # Hence:
            newlist.append((x, v, y,))

        # I don't know if you want to update the key-value-pair in the dictionary `d`
        # take out the following line if you don't want it
        d[x] = y

    return newlist

Please keep reading the remaining part if you want to know how to traverse a dict object properly.


Different ways to traverse a dict object

Python 3.x

The following segments demonstrate how to traverse a dict in Python 3.x.

Iterate the set of keys

for key in d:
    value = d[key]
    print(key, value)

the code segment above has the same effect as the following one:

for key in d.keys():
    value = d[key]
    print(key, value)

Iterate the set of key-value-pairs

for key, value in d.items():
    print(key, value)

Iterate the set of values

for value in d.values():
    print(value)

Python 2.x

The following segments demonstrate how to traverse a dict in Python 2.x.

Iterate the set of keys

for key in d:
    value = d[key]
    print(key, value)

keys() returns a list of the key set of dictionary d

for key in d.keys():
    value = d[key]
    print(key, value)

iterkeys() returns an iterator of the key set of dictionary d

for key in d.iterkeys():
    value = d[key]
    print(key, value)

Iterate the set of key-value-pairs

values() returns a list of the key-value-pair set of dictionary d

for key, value in d.items():
    print(key, value)

itervalues() returns an iterator of the key-value-pair set of dictionary d

for key, value in d.iteritems():
    print(key, value)

Iterate the set of values

values() returns a list of the value set of dictionary d

for value in d.values():
    print(value)

itervalues() returns a iterator of the value set of dictionary d

for value in d.itervalues():
    print(value)

Reference:

Source: StackOverflow.com

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