如果想要删除的项总是在dict迭代的“开始”，有一种方法可能是合适的

while mydict:
    key, value = next(iter(mydict.items()))
    if should_delete(key, value):
       del mydict[key]
    else:
       break

“开头”只保证在某些Python版本/实现中是一致的。例如，来自Python 3.7新提示

dict对象的插入顺序保存特性已被声明为Python语言规范的正式组成部分。

这种方法避免了许多其他答案所建议的字典副本，至少在Python 3中是这样。

在python3中，迭代dic.keys()将引发字典大小错误。你可以用这种替代方式:

使用python3测试，它工作正常，并且没有引发错误“迭代期间字典更改大小”:

my_dic = { 1:10, 2:20, 3:30 }
# Is important here to cast because ".keys()" method returns a dict_keys object.
key_list = list( my_dic.keys() )

# Iterate on the list:
for k in key_list:
    print(key_list)
    print(my_dic)
    del( my_dic[k] )


print( my_dic )
# {}

相反，迭代一个副本，例如items()返回的副本:

for k, v in list(mydict.items()):

您可以首先构建一个要删除的键的列表，然后遍历该列表删除它们。

dict = {'one' : 1, 'two' : 2, 'three' : 3, 'four' : 4}
delete = []
for k,v in dict.items():
    if v%2 == 1:
        delete.append(k)
for i in delete:
    del dict[i]

在迭代集合时不能修改集合。这种方式是疯狂的——最明显的是，如果你被允许删除和删除当前项，迭代器将不得不移动(+1)，而下一次调用next将超出(+2)，所以你最终会跳过一个元素(在你删除的元素后面的那个元素)。你有两个选择:

Copy all keys (or values, or both, depending on what you need), then iterate over those. You can use .keys() et al for this (in Python 3, pass the resulting iterator to list). Could be highly wasteful space-wise though. Iterate over mydict as usual, saving the keys to delete in a seperate collection to_delete. When you're done iterating mydict, delete all items in to_delete from mydict. Saves some (depending on how many keys are deleted and how many stay) space over the first approach, but also requires a few more lines.

你也可以分两步来做:

remove = [k for k in mydict if k == val]
for k in remove: del mydict[k]

我最喜欢的方法通常是创建一个新词典:

# Python 2.7 and 3.x
mydict = { k:v for k,v in mydict.items() if k!=val }
# before Python 2.7
mydict = dict((k,v) for k,v in mydict.iteritems() if k!=val)