Python - 如何修复“ValueError：没有足够的值来解包（预期 2，得到 1）”[关闭]

Question

我需要编写一个函数add_to_dict(d, key_value_pairs)，它将每个给定的键/值对添加到 python 字典中。参数key_value_pairs 将是(key, value) 形式的元组列表。

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

ValueError: 没有足够的值来解包（预期 2，得到 1）

我该如何解决这个错误？

Answer 1

你应该如何调试你的代码

'''
@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

如果您想知道如何正确遍历dict 对象，请继续阅读剩余部分。

---

遍历dict对象的不同方式

Python 3.x

以下部分演示了如何在 Python 3.x 中遍历 dict。

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)

迭代一组键值对

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

迭代值集

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

---

Python 2.x

以下部分演示了如何在 Python 2.x 中遍历 dict。

迭代键集

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

keys()返回字典d的键集的列表

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

iterkeys()返回字典d的键集的迭代器

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

迭代一组键值对

values() 返回字典 d 的键值对集合的列表

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

itervalues() 返回字典 d 的键值对集的 迭代器

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

迭代值集

values()返回字典d的值集列表

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

itervalues()返回字典d的值集的迭代器

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

参考：

• What is the difference between list and iterator in Python?

Answer 2

使用items()解决，比如：

d = {"foo": "bar"}

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

Answer 3

如果您不迭代 dict（包含 100 万个条目）而只迭代 可能 更改的列表并查看它是否更改了 dict 中的任何内容，则可以避免此错误：

def add_to_dict(d, key_value_pairs):
    """Adds all tuples from key_value_pairs as key:value to dict d, 
    returns list of tuples of keys that got changed as (key, old value)"""
    newlist = []


    for item in key_value_pairs:

        # this handles your possible unpacking errors
        # if your list contains bad data 
        try:
            key, value = item
        except (TypeError,ValueError):
            print("Unable to unpack {} into key,value".format(item))

        # create entry into dict if needed, else gets existing
        entry = d.setdefault(key,value) 

        # if we created it or it is unchanged this won't execute
        if entry != value:
            # add to list
            newlist.append( (key, entry) )
            # change value
            d[key] = value

    return newlist



d = {}
print(add_to_dict(d, (  (1,4), (2,5) ) ))    # ok, no change
print(add_to_dict(d, (  (1,4), (2,5), 3 ) )) # not ok, no changes
print(add_to_dict(d, (  (1,7), (2,5), 3 ) )) # not ok, 1 change

输出：

[] # ok

Unable to unpack 3 into key,value
[] # not ok, no change

Unable to unpack 3 into key,value
[(1, 4)] # not ok, 1 change

---

您也可以对您的参数进行一些验证 - 如果任何参数有误，则不会执行任何操作并出现语音错误：

import collections 

def add_to_dict(d, key_value_pairs):
    """Adds all tuples from key_value_pairs as key:value to dict d, 
    returns list of tuples of keys that got changed as (key, old value)"""

    if not isinstance(d,dict):
        raise ValueError("The dictionary input to add_to_dict(dictionary,list of tuples)) is no dict")

    if not isinstance(key_value_pairs,collections.Iterable):
        raise ValueError("The list of tuples input to add_to_dict(dictionary,list of tuples)) is no list")  

    if len(key_value_pairs) > 0:
        if any(not isinstance(k,tuple) for k in key_value_pairs):
            raise ValueError("The list of tuples includes 'non tuple' inputs")        

        if any(len(k) != 2 for k in key_value_pairs):
            raise ValueError("The list of tuples includes 'tuple' != 2 elements")        

    newlist = []
    for item in key_value_pairs:            
        key, value = item

        # create entry into dict if needed, else gets existing
        entry = d.setdefault(key,value) 

        # if we created it or it is unchanged this won't execute
        if entry != value:
            # add to list
            newlist.append( (key, entry) )
            # change value
            d[key] = value

    return newlist

所以你会得到更清晰的错误信息：

add_to_dict({},"tata") 
# The list of tuples input to add_to_dict(dictionary,list of tuples)) is no list

add_to_dict({},["tata"])
# The list of tuples includes 'non tuple' inputs

add_to_dict({},[ (1,2,3) ])
# The list of tuples includes 'tuple' != 2 elements

add_to_dict({},[ (1,2) ])
# ok