CODEWITHSUNDEEP
pythonlistdictionary
MrKay
MrKay

Reputation: 63

Delete multiple dictionaries in a list

I have been trying to delete multiple dictionaries in a list but I can only delete one at a time.

Below is the main code I am working on. Records is the list of dictionaries. I want to delete dictionaries that have 0 in them.

Records = [{'Name':'Kelvin','Price': 0},{'Name': 'Michael','Price':10}]

I want to delete dictionaries with Prices of 0

def deleteUnsold(self):
    for d in records:
        for key, value in d.items():
            if d['Price'] == 0:
                records.remove(d)

Upvotes: 6

Views: 1246

Answers (3)

tglaria
tglaria

Reputation: 5866

Okay, generally speaking you shouldn't remove items from a list you're iterating for, because that will probable make you miss some items of the list.

Now, about some other answer spoken here, yes, they work, but strictly speaking they're not removing/deleting items from the list: they're creating a new list and replacing the old variable with a new list.

What could be done:

for d in list(records):
    if d['Price'] == 0:
        records.remove(d)

for d in reversed(records):
    if d['Price'] == 0:
        records.remove(d)

for idx in range(len(records)-1,-1,-1):
    if records[idx]['Price'] == 0:
        records.pop(idx)

I like this one, though:

for d in records[::-1]:
    if d['Price'] == 0:
        records.remove(d)

Upvotes: 0

Bhargav Rao
Bhargav Rao

Reputation: 52181

Use a list comprehension with an if condition

>>> Records = [{'Name':'Kelvin','Price': 0},{'Name': 'Michael','Price':10}]
>>> [i for i in Records if i['Price'] != 0]
[{'Price': 10, 'Name': 'Michael'}]

Check out if/else in Python's list comprehension? to learn more about using a conditional within a list comprehension.

Note that [as mentioned below] you can also leave out checking for the value 0. However this also works if Price is None, hence you may use the first alternative if you are not sure of the data type of the value of Price

>>> [i for i in Records if i['Price']]
[{'Price': 10, 'Name': 'Michael'}]

Upvotes: 12

Ahasanul Haque
Ahasanul Haque

Reputation: 11164

You can use filter:

print filter(lambda x:x['Price']!=0,Records)

Upvotes: 2

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