从列表中删除dict的子集-Removingasubsetofadictfromwithinalist

This is really only easy to explain with an example, so to remove the intersection of a list from within a dict I usually do something like this:

这只是一个例子很容易解释,所以要从dict中删除列表的交集我通常会这样做:

a = {1:'', 2:'', 3:'', 4:''}
exclusion = [3, 4, 5]

# have to build up a new list or the iteration breaks
toRemove = []
for var in a.iterkeys():
    if var in exclusion:
        toRemove.append(var)

for var in toRemove:
    del a[var]

This might seem like an unusual example, but it's surprising the number of times I've had to do something like this. Doing this with sets would be much nicer, but I clearly want to retain the 'values' for the dict.

这似乎是一个不寻常的例子,但令人惊讶的是我必须做这样的事情的次数。使用集合执行此操作会更好,但我显然希望保留dict的“值”。

This method is annoying because it requires two loops and an extra array. Is there a cleaner and more efficient way of doing this.

这种方法很烦人,因为它需要两个循环和一个额外的数组。是否有更清洁,更有效的方法。

4 个解决方案

for key in exclusion:
     a.pop(key, None)

a = dict((key,value) for (key,value) in a.iteritems() if key not in exclusion)

exclusion = set([3, 4, 5])

for key in exclusion.intersection(a):
    del a[key]