Removing a subset of a dict from within a list
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:
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.
This method is annoying because it requires two loops and an extra array. Is there a cleaner and more efficient way of doing this.
Asked by: Sarah280 | Posted: 28-01-2022
Answer 1
Consider dict.pop:
for key in exclusion:
a.pop(key, None)
The None keeps pop from raising an exception when key isn't a key.
Answer 2
a = dict((key,value) for (key,value) in a.iteritems() if key not in exclusion)
Answer 3
Why not just use the keys method, instead of iterkeys? That way you can do it in one loop because it returns a list, not an iterator.
Answer 4
You could change your exclusion list to a set, then just use intersection to get the overlap.
exclusion = set([3, 4, 5])
for key in exclusion.intersection(a):
del a[key]
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