CODEWITHSUNDEEP
pythonpython-2.7
user2242044
user2242044

Reputation: 9253

eliminate entries in dictionary based on a certain value

I would like to eliminate entries in dictionary a and key-values for MyList if the second value is 8. I am missing something though.

My code:

a = {
   'Mylist' : [[1,2,3],[4,5,6],[7,8,9]]
    }
a = [x for x in a if a['Mylist'][1] != 8]
print a['MyList']

Desired Ouput:

[[1,2,3],[4,5,6]]

Upvotes: 0

Views: 54

Answers (3)

Shahriar
Shahriar

Reputation: 13814

If you want to apply this in whole dictionary, you can do this:

for k in a:
    a[k] = [ l for l in a[k] if l[1]!=8]

print a['Mylist']

Upvotes: 0

John Hopkins
John Hopkins

Reputation: 11

a = {
   'Mylist' : [[1,2,3],[4,5,6],[7,8,9]]
    }
a['Mylist'] = filter(lambda x: x[1] != 8, a['Mylist'])
print(a)

Result: {'Mylist': [[1, 2, 3], [4, 5, 6]]}

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1124758

You are looping over the keys of a, not the sublists of a['Mylist'].

The following corrects just the one value:

[sublist for sublist in a['Mylist'] if sublist[1] != 8]

To do so for all keys in the dictionary, nest this in a dictionary comprehension:

{key: [sublist for sublist in value if sublist[1] != 8]
 for key, value in a.iteritems()}

Demo:

>>> a = {
...    'Mylist' : [[1,2,3],[4,5,6],[7,8,9]]
...     }
>>> [sublist for sublist in a['Mylist'] if sublist[1] != 8]
[[1, 2, 3], [4, 5, 6]]
>>> {key: [sublist for sublist in value if sublist[1] != 8]
...  for key, value in a.iteritems()}
{'Mylist': [[1, 2, 3], [4, 5, 6]]}

Upvotes: 4

Related Questions