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Delete item in list in dictionary

In a large dictionary like

d = {}                                                                                   
d['a']=[1,2,3,4]                                                                         
d['b']=[1,2,3,4,5,6]                                                                     
d['c']=[1,2]                                                                             
d['d']=[1,4]

How can I quickly delete "four" in lists?

EDIT

Is there a way to link fours in lists? As in, eliminating one excludes the others.

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5 answers


Go through the dictionary values ​​and remove 4 from each list:

for a in d.itervalues():
    try:
        a.remove(4)
    except ValueError:
        pass

This is not very efficient as removing an item from a list is an O (n) operation. To improve performance, use a different data type (for example, a set).



If the dictionary values ​​are sets, you can do

for a in d.itervalues():
    a.discard(4)
+8


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It looks like you can use the dictionary "Symmetric", you can do something like this:

def make_symmetric(D):
    for key, value in D.items():
        for v in value:
            D.setdefault(v, set()).add(key)

def add(D, a, b):
    D.setdefault(a, set()).add(b)
    D.setdefault(b, set()).add(a)

def remove(D, a):
    values = D.pop(a)
    for v in values:
        D[v].remove(a)

And use it something like this:



>>> d = {'a': set([1, 2, 3, 4]),
         'b': set([1, 2, 3, 4, 5, 6]),
         'c': set([1, 2]),
         'd': set([1, 4])}
>>> make_symmetric(d)
>>> d
{1: set(['a', 'c', 'b', 'd']),
 2: set(['a', 'c', 'b']),
 3: set(['a', 'b']),
 4: set(['a', 'b', 'd']),
 5: set(['b']),
 6: set(['b']),
 'a': set([1, 2, 3, 4]),
 'b': set([1, 2, 3, 4, 5, 6]),
 'c': set([1, 2]),
 'd': set([1, 4])}
>>> remove(d, 4)
>>> d
{1: set(['a', 'c', 'b', 'd']),
 2: set(['a', 'c', 'b']),
 3: set(['a', 'b']),
 5: set(['b']),
 6: set(['b']),
 'a': set([1, 2, 3]),
 'b': set([1, 2, 3, 5, 6]),
 'c': set([1, 2]),
 'd': set([1])}
>>> add(d, 'd', 4)
>>> d
{1: set(['a', 'c', 'b', 'd']),
 2: set(['a', 'c', 'b']),
 3: set(['a', 'b']),
 4: set(['d']),
 5: set(['b']),
 6: set(['b']),
 'a': set([1, 2, 3]),
 'b': set([1, 2, 3, 5, 6]),
 'c': set([1, 2]),
 'd': set([1, 4])}

I'm using a set here, but you can do something similar with lists. I wouldn't be surprised if there was some kind of implementation of a "symmetric" dictionary somewhere. Hopefully someone else can point you in the right direction if it exists.

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Try to remove all 4s:

for i in d:
    while 4 in d[i]:
       d[i].remove(4)

I don't know how you can eliminate all 4s at once.

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You haven't specified what you want if there are duplicates in the list. The solution I wrote (iterate and then delete ..) doesn't handle duplicates.

>>> d = {'a':[1,2,3,4],'b':[1,2,3,4,5,6],'c':[1,2],'d':[1,4],'e':[1,4,4]}
>>> for l in d.values():
...     if 4 in l: l.remove(4)
... 
>>> d
{'a': [1, 2, 3], 'c': [1, 2], 'b': [1, 2, 3, 5, 6], 'e': [1, 4], 'd': [1]}

It is not very efficient. if 4 in l

will iterate over the list once and l.remove()

repeat the list again.

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If the values ​​contain duplicates of 4, you can use something like this:

d = {}                                                                                   
d['a']=[1,2,3,4]                                                                         
d['b']=[1,2,3,4,4,5,6]                                                                     
d['c']=[1,2]                                                                             
d['d']=[1,4,4,4]

def remove_dup(x,n):
    while n in x:
        x.remove(n)
    return x

for ele in d.itervalues():
    try:
        remove_dup(ele,4)
    except ValueError:
        pass

Result:

>>> d
{'a': [1, 2, 3], 'c': [1, 2], 'b': [1, 2, 3, 5, 6], 'd': [1]}
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