没有足够的值可解压缩（预期2，得到1

编写一个函数add_to_dict(d, key_value_pairs)，将每个给定的键/值对添加到给定的字典中。参数key_value_pairs将是形式为元组的列表(key, value)。

该函数应返回所有已更改的键/值对的列表（及其原始值）。

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

我不断收到错误消息

ValueError：没有足够的值可解包（预期2，得到1）

我如何完成这个问题？

'''
@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

如果您想知道如何dict正确地遍历对象，请继续阅读其余部分。

Python 3.x

以下各节演示了如何dict在Python 3.x中遍历a 。

迭代键集

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)

迭代键值对的集合

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

迭代一组值

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

Python 2.x

以下各节演示了如何dict在Python 2.x中遍历a 。

迭代键集

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

keys()返回字典键集的列表d

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

iterkeys()返回字典键集的迭代器d

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

迭代键值对的集合

values()返回字典的键值对集合的列表d

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

itervalues()返回字典的键值对集合的迭代器d

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

迭代一组值

values() 返回字典值集的列表 d

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

itervalues() 返回字典值集的迭代器 d

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

参考：

Python中的列表和迭代器有什么区别？

用于items()解决，例如：

d = {"foo": "bar"}

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