我应该如何从列表中删除所有不包含其值之一的字典？

Question

Suppose I have a list like so:

[{'name': 'Blah1', 'age': x}, {'name': 'Blah2', 'age': y}, {'name': None, 'age': None}]

It is guaranteed that both 'name' and 'age' values will either be filled or empty.

I tried this:

for person_dict in list:
    if person_dict['name'] == None:
        list.remove(person_dict)

But obviously that does not work because the for loop skips over an index sometimes and ignores some blank people.

I am relatively new to Python, and I am wondering if there is a list method that can target dicts with a certain value associated with a key.

EDIT: Fixed tuple notation to list as comments pointed out

Answer 1

You can use list comprehension as a filter like this

[c_dict for c_dict in dict_lst if all(c_dict[key] is not None for key in c_dict)]

This will make sure that you get only the dictionaries where all the values are not None .

Answer 2

Just test for the presence of None in the dict's values to test ALL dict keys for the None value:

>>> ToD=({'name': 'Blah1', 'age': 'x'}, {'name': 'Blah2', 'age': 'y'}, {'name': None, 'age': None})
>>> [e for e in ToD if None not in e.values()]
[{'age': 'x', 'name': 'Blah1'}, {'age': 'y', 'name': 'Blah2'}]

Or, use filter:

>>> filter(lambda d: None not in d.values(), ToD)
({'age': 'x', 'name': 'Blah1'}, {'age': 'y', 'name': 'Blah2'})

Or, if it is a limited test to 'name':

>>> filter(lambda d: d['name'], ToD)
({'age': 'x', 'name': 'Blah1'}, {'age': 'y', 'name': 'Blah2'})

Answer 3

for index,person_dict in enumerate(lis):
    if person_dict['name'] == None:
        del lis[index]

you can also try

lis=[person_dict  for person_dict in lis if person_dict['name'] != None]  

never use List as variable

Answer 4

You can create new list with accepted data. If you have tuple then you have to create new list.

List comprehension could be faster but this version is more readable for beginners.

data = ({'name': 'Blah1', 'age': 'x'}, {'name': 'Blah2', 'age': 'y'}, {'name': None, 'age': None})

new_data = []

for x in data:
    if x['name']: # if x['name'] is not None and x['name'] != ''
        new_data.append(x)

print new_data