CODEWITHSUNDEEP
pythonlistdictionary
Chris Clouten
Chris Clouten

Reputation: 1075

Find out if no items in a list are keys in a dictionary

I have this list:

source = ['sourceid', 'SubSourcePontiflex', 'acq_source', 'OptInSource', 'source',
      'SourceID', 'Sub-Source', 'SubSource', 'LeadSource_295', 'Source',
      'SourceCode', 'source_code', 'SourceSubID']

I am iterating over XML in python to create a dictionary for each child node. The dictionary varies in length and keys with each iteration. Sometimes the dictionary will contain a key that is also an item in this list. Sometimes it wont. What I want to be able to do is, if a key in the dictionary is also an item in this list then append the value to a new list. If none of the keys in the dictionary are in list source, I'd like to append a default value. I'm really having a brain block on how to do this. Any help would be appreciated.

Upvotes: 4

Views: 6148

Answers (5)

Sven Marnach
Sven Marnach

Reputation: 601779

In Python 3, you can perform set operations directly on the object returned by dict.keys():

in_source_and_dict = mydict.keys() & source
in_dict_not_source = mydict.keys() - source
in_source_not_dict = source - mydict.keys()

This will also work in Python 2.7 if you replace .keys() by .viewkeys().

Upvotes: 2

Dietrich Epp
Dietrich Epp

Reputation: 213368

You can do this with the any() function

dict = {...}
keys = [...]
if not any(key in dict for key in keys):
    # no keys here

Equivalently, with all() (DeMorgan's laws):

if all(key not in dict for key in keys):
    # no keys here

Upvotes: 1

Andrew Boyle
Andrew Boyle

Reputation: 41

my_dict = { some values }
values = []
for s in sources:
    if my_dict.get(s):
        values += [s]
if not values:
    values += [default]

You can loop through the sources array and see if there is a value for that source in the dictionary. If there is, append it to values. After that loop, if values is empty, append the default vaule.

Note, if you have a key, value pair in your dictionary (val, None) then you will not append the None value to the end of the list. If that is an issue you will probably not want to use this solution.

Upvotes: 1

Shashank
Shashank

Reputation: 13869

Just use the in keyword to check for membership of some key in a dictionary.

The following example will print [3, 1] since 3 and 1 are keys in the dictionary and also elements of the list.

someList = [8, 9, 7, 3, 1]
someDict = {1:2, 2:3, 3:4, 4:5, 5:6}
intersection = [i for i in someList if i in someDict]
print(intersection)

You can just check if this intersection list is empty at every iteration. If the list is empty then you know that no items in the list are keys in the dictionary.

Upvotes: 7

000
000

Reputation: 27247

in_source_and_dict = set(mydict.keys()).intersection(set(source))
in_dict_not_source = set(mydict.keys()) - set(source)
in_source_not_dict = set(source) - set(mydict.keys())

Iterate over the result of which one you want. In this case I guess you'll want to iterate over in_source_not_dict to provide default values.

Upvotes: 4

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