CODEWITHSUNDEEP
pythonlistpython-2.7dictionarykey
user2057574
user2057574

Reputation: 799

How to check if key exists in list of dicts in python?

Say I have a list of dicts that looks like this:

[{1: "a"}, {2: "b"}]

What is the pythonic way to indicate if a certain key is in one of the dicts in the list?

Upvotes: 63

Views: 92276

Answers (7)

Super Kai - Kazuya Ito
Super Kai - Kazuya Ito

Reputation: 1

You can check if a key exists in a list of dictionaries with <Key> in <Dictionary> and <Key> in <Dictionary>.keys() as shown below:

dict_list = [{1: "a"}, {2: "b"}]

# <Key> in <Dictionary>
print(any(1 in dict for dict in dict_list)) # True
print(any(3 in dict for dict in dict_list)) # False

# <Key> in <Dictionary>.keys()
print(any(1 in dict.keys() for dict in dict_list)) # True
print(any(3 in dict.keys() for dict in dict_list)) # False

Upvotes: 1

FastGTR
FastGTR

Reputation: 423

I was thrown aback by what was possible in python2 vs python3. I will answer it based on what I ended up doing for python3. My objective was simple: check if a json response in dictionary format gave an error or not. My dictionary is called "token" and my key that I am looking for is "error"

if ((token.get('error', None)) is None):
    do something

I am looking for key "error" and if it was not there, then setting it to value of None, then checking is the value is None, if so proceed with my code. An else statement to handle the if I do have the key "error

Upvotes: -4

Marky0
Marky0

Reputation: 2084

To search through deeply nested data structure I used this code to recursively look for keys in both lists and dictionaries 

def isKey(dictORlist, key):
# dictORlist is the data structure you want to search
# key is the keyword you want to search for
def checkList(List, key):
    if isinstance(List, list):
        for i in List:
             return isKey(i, key)

result = checkList(dictORlist, key)
if isinstance(dictORlist, dict):
    for k in dictORlist.keys():
        data = dictORlist[k]
        if k == key:
            return True
        elif isinstance(data, dict):
            result = isKey(data, key)
        else:
            result = checkList(data, key)

if result == None:
    result = False
return result

Upvotes: 0

Lyncean Patel
Lyncean Patel

Reputation: 2721

To see in single dictoray we use 'in' keyword:

key in dic_instance

To check in list of dictionary, iterate through dictionary list and use 'any' function, so if key found in any of the dictionary, it will not iterate the list further. 

dic_list = [{1: "a"}, {2: "b"}]
any(2 in d for d in dic_list)
True
any(4 in d for d in dic_list)
False

Upvotes: 6

Ranvijay sachan
Ranvijay sachan

Reputation: 2444

parsedData=[]
dataRow={}
if not any(d['url'] == dataRow['url'] for d in self.parsedData):
       self.parsedData.append(dataRow)

Upvotes: 2

DSM
DSM

Reputation: 353019

I'd probably write:

>>> lod = [{1: "a"}, {2: "b"}]
>>> any(1 in d for d in lod)
True
>>> any(3 in d for d in lod)
False

although if there are going to be a lot of dicts in this list you might want to reconsider your data structure.

If you want the index and/or the dictionary where the first match is found, one approach is to use next and enumerate:

>>> next(i for i,d in enumerate(lod) if 1 in d)
0
>>> next(d for i,d in enumerate(lod) if 1 in d)
{1: 'a'}
>>> next((i,d) for i,d in enumerate(lod) if 1 in d)
(0, {1: 'a'})

This will raise StopIteration if it's not there:

>>> next(i for i,d in enumerate(lod) if 3 in d)
Traceback (most recent call last):
  File "<ipython-input-107-1f0737b2eae0>", line 1, in <module>
    next(i for i,d in enumerate(lod) if 3 in d)
StopIteration

If you want to avoid that, you can either catch the exception or pass next a default value like None:

>>> next((i for i,d in enumerate(lod) if 3 in d), None)
>>>

As noted in the comments by @drewk, if you want to get multiple indices returned in the case of multiple values, you can use a list comprehension:

>>> lod = [{1: "a"}, {2: "b"}, {2: "c"}]
>>> [i for i,d in enumerate(lod) if 2 in d]
[1, 2]

Upvotes: 127

Rohit Jain
Rohit Jain

Reputation: 213223

Use any function with a generator:

>>> d = [{1: "a"}, {2: "b"}]
>>> any(1 in x for x in d)
True

any function returns True, if at least one element in the iterable passed to it is True. But you really need to consider, why are you not having all the key: value pairs in a single dict?

Upvotes: 4

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