Python-如何修复“；ValueError：没有足够的值来解包（预期为2，得到1）和#x201D；

我需要编写一个函数

add_to_dict（d，key\u-value\u-pairs）

，将每个给定的键/值对添加到python字典中。参数

key\u value\u pairs

将是一个元组列表，格式为

（key，value）

该函数应返回所有已更改的键/值对（及其原始值）的列表

我总是出错

ValueError:没有足够的值来解包（预期值为2，实际值为1）

如何解决此错误？

使用

items（）

解决此错误，如：

d = {"foo": "bar"}

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

如何调试代码 如果您想知道如何正确地遍历

dict

对象，请继续阅读剩余部分

---

遍历

dict

对象的不同方法 Python3.x 以下部分演示如何在Python 3.x中遍历

dict

迭代密钥集 迭代键值对集 迭代值集

---

Python2.x 以下部分演示如何在Python2.x中遍历

dict

迭代密钥集

keys（）

返回字典

d

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

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

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

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

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

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

iterkeys（）

返回字典

d

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

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

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

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

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

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

迭代键值对集

values（）

返回字典

d

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

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

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

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

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

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

itervalues（）

返回字典

d

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

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

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

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

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

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

迭代值集

values（）

返回字典

d

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

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

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

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

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

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

itervalues（）

返回字典

d

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

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

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

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

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

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

参考：

---

如果不迭代dict（100万个条目），而只迭代可能的更改列表，并查看它是否更改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

---

您还可以对您的参数进行一些验证-如果任何参数错误，将不会执行任何操作，并且会出现说话错误：

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

因此，您可以获得更清晰的错误消息：

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

---

请添加调用

add_to_dict（）

的脚本，该脚本不会对dct中的x执行

操作，只需在键上进行迭代即可？您可以指示错误发生在哪一行，并以可运行的形式提供调用函数的实例吗。i、 e将
d
更改为
d.items（）
@ulrichswarz:是的，确实如此。因此出现了错误。