More pythonic way of delete a dict if a key matches something

Question

I have a list of dictionaries in the following format:

list = [
{'name':'bob','age':12},
{'name':'jill','age':34}
]

I want to remove the first dictionary where the name equals a value. Is there a better way of doing this?

value = 'bob'
for dict in list:
    if dict['name'] == value:
        list.remove(dict)
        break

Output:

list = [
{'name':'jill','age':34}
]

Answer 1

list1 = [
{'name':'bob','age':12},
{'name':'jill','age':34}
]

value = 'bob'

for i, item in enumerate(list1):
  if item['name'] == value:
    list1.pop(i)
    break

Answer 2

This is a perfect pandas problem. pandas is part of the PyData suite (I think) and it's widely accepted. I would use it for this type of problem fosho.

Python 3.5.2 |Anaconda custom (x86_64)| (default, Jul  2 2016, 17:52:12)
[GCC 4.2.1 Compatible Apple LLVM 4.2 (clang-425.0.28)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import pandas as pd
>>> query_list = [ {'name':'bob','age':12}, {'name':'jill','age':34} ]
>>> DF = pd.DataFrame(query_list)
>>> DF
     age  name
0   12   bob
1   34  jill
>>> mask = DF["name"] != "bob"
>>> mask
0    False
1     True
Name: name, dtype: bool
>>> DF2 = DF.loc[mask,:]
>>> DF2
     age  name
1   34  jill
>>> DF2.to_dict()
{'age': {1: 34}, 'name': {1: 'jill'}}

Answer 3

If you want a one liner, you could use a generator expression with next(), which short-circuits once you've found the first item to remove, and then use list.remove():

l.remove(next(d for d in l if d['name'] == value))

Example:

>>> l = [{'name':'bob','age':12},{'name':'jill','age':34}]
>>> value = 'bob'
>>> l.remove(next(d for d in l if d['name'] == value))
>>> l
[{'name': 'jill', 'age': 34}]

Note this will raise a StopIteration is the value is not found, which can be avoided, but it's a bit longer because although next() has a default argument, list.remove() does not:

>>> l = [{'name':'bob','age':12},{'name':'jill','age':34}]
>>> value = 'bob'
>>> value_to_remove = next((d for d in l if d['name'] == value), None)
>>> 'Value not in list' if value_to_remove is None else l.remove(value_to_remove)
>>> l
[{'name': 'jill', 'age': 34}]
>>> value_to_remove = next((d for d in l if d['name'] == value), None)
>>> 'Value not in list' if value_to_remove is None else l.remove(value_to_remove)
'Value not in list'

Answer 4

From python's documentation on dictionaries.

https://docs.python.org/3/library/stdtypes.html#dict

d.clear() will clear all the values of a dictionary.

value = 'bob'
for dict in list:
    if dict['name'] == value:
        #Empty dictionary of all values, will now return {}
        dict.clear()
        break

If the dictionary is in a list of others, you can simply pop it out of the list or delete it like this.

list2 = [x for x in list1 if x != {}]

Answer 5

In my opinion using filter is the most pythonic way:

new_list = filter(lambda x: x['name'] != value, original_list)

However this doesn't remove from the original list and creates a new one in memory. Assigning old_list = filter(...) mimics removing from the original one (in the local scope at least) however there is still the memory overhead. For small lists (and most cases) this won't play a role. If it does then the following might be more appropriate:

for item in original_list:
    if item['name'] == value:
        original_list.remove(item)
        break

Answer 6

I would create a separate utility function to do it. The following first determines which (if any) dictionaries meet the criteria, and then pops (deletes) the first one found.

from operator import itemgetter

people = [
    {'name': 'bob', 'age': 12},
    {'name': 'jill', 'age': 34},
    {'name': 'bob', 'age': 14},
]

def remove(name, seq):
    """Remove first dictionary in seq with 'name' entry equal to name."""
    matches = [i for i,n in enumerate(map(itemgetter('name'), seq)) if n == name]
    if matches:
        seq.pop(matches[0])


remove('bob', people)
print(people)  # -> [{'name': 'jill', 'age': 34}, {'name': 'bob', 'age': 14}]

Answer 7

list1 = [
     {'name': 'bob', 'age':12},
     {'name': 'jill', 'age':34}
]

map(lambda d: d.pop('name', 0) if d['name'] == 'bob' else d, list1)