28 дек. 2011 г. · This answer returns a list with all matches for 'Pam' in people, alternatively we could get a list of all the people that are not 'Pam' by changing the ...

9 окт. 2010 г. · Here's one way to do it: if not any(d['main_color'] == 'red' for d in a): # does not exist The part in parentheses is a generator expression that returns True ...

18 дек. 2018 г. · It's not possible to do this without iteration. However, but you can transform you dictionary into a different data structure, such as a dictionary where names ...

28 янв. 2022 г. · If all of the keys are unique, you can flatten the list of dictionaries into a dictionary with a straightforward dictionary comprehension.

10 авг. 2016 г. · If you are searching for a single item then this is the "best" approach def search(name): for p in people: if p['name'] == name: return p

27 мар. 2023 г. · If you are looking for a specific date and sensor you can make a loop to iterate over every dictionary in the list and save the index of matching item in the ...

8 дек. 2010 г. · A simple readable version is def find(lst, key, value): for i, dic in enumerate(lst): if dic[key] == value: return i return -1

29 дек. 2021 г. · By using dict_list[-1][0], you are trying to access a list with a list, which you do not have. You have a list with a dict key within a list.

24 окт. 2018 г. · I need to iterate over list_of_dict and find the keys that match with the element in the list and append the values to a an empty defaultdict of list items.

9 мар. 2018 г. · You can just fetch the index of the first item that has the key, eg, like this: idx = next((i for i,d in enumerate(listofpeople) if "Jack" in d), None)