Francis - 1 month ago 6x

Python Question

This is a question about Python. I have the following list of dictionaries:

`listA = [`

{"t": 1, "tid": 2, "gtm": 3, "c1": 4, "id": "111"},

{"t": 3, "tid": 4, "gtm": 3, "c1": 4, "c2": 5, "id": "222"},

{"t": 1, "tid": 2, "gtm": 3, "c1": 4, "c2": 5, "id": "333"},

{"t": 5, "tid": 6, "gtm": 3, "c1": 4, "c2": 5, "id": "444"}

]

and a dictionary I wanted to compare with:

`dictA = {"t": 1, "tid": 2, "gtm": 3}`

I wanted to create a list of dicts that match all the items in

`listB = [`

{"t": 1, "tid": 2, "gtm": 3, "id": "111"},

{"t": 1, "tid": 2, "gtm": 3, "id": "333"}

]

I tried doing this:

`for k in listA:`

for key, value in k.viewitems() & dictA.viewitems():

print key, value

But it's matching any item in

Answer

You would need to check the *length* of the intersection, just checking `if dct.viewitems() & dictA.viewitems()`

would evaluate to True for any intersection :

```
[dct for dct in listA if len(dct.viewitems() & dictA.viewitems()) == len(dictA)]
```

Or just check for a subset, if the items from dictA are a subset of each dict:

```
[dct for dct in listA if dictA.viewitems() <= dct.viewitems()]
```

Or reverse the logic looking for a superset:

```
[dct for dct in listA if dct.viewitems() >= dictA.viewitems()]
```

Source (Stackoverflow)

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