在python中使用lambda和filter从字典列表中过滤字典值

Question

I have a list of dictionaries as below:

l = [{'abc': 'def'}, {'ghi': 'jul'}, {'Name': 'my-name'}]

I want to get the value of the dictionary that has the key 'Name' Initially I wrote a function as below:

def get_json_val(l, key):
    for item in l:
        for k, v in item.iteritems():
            if k == key:
                return v

But I want to do it using lambdas and filter in a single line. So, I tried as below to filter out the dictionary:

name_lst = filter(lambda item: (k == 'Name' for k in item.iteritems()), l)

It is showing the whole list again. Is there a way to get only the value (not dictionary) using lambdas and filter? Thanks in advance

Edit 1: In the first function l is list and 'Name' is passed as key.

Answer 1

Why are you iterating over your dict? That defeats the purpose. Just do

[d for d in l if key in d]

Or if you feel some compulsion to use lambda and filter:

filter(lambda d: key in d, l)

Note, these return lists instead of the corresponding value. If you want the value, you'll have to index into the list. Simply out, using a loop is probably the most reasonable approach, just don't iterate over your dict:

def f(key, l):
    for d in l:
        if key in d:
            return d[key]

Answer 2

Seeing your initial code, this should do the same thing

>>> l = [{'abc': 'def'}, {'ghi': 'jul'}, {'Name': 'my-name'}]
>>> [i["Name"] for i in l if "Name" in i][0]
'my-name'

Your code returns only the first occurence.Using my approach, and probably also if you would use filter, you'll get (first) a list of all occurrences than only get the first (if any). This would make it less "efficient" IMHO, so I would probably change your code to be something more like this:

def get_json_val(l, key):
    for item in l:
        if not key in item: continue
        return item[key]

Answer 3

l = [{'abc': 'def'}, {'ghi': 'jul'}, {'Name': 'my-name', "no-name": "bbb"}]

I was able to do it in this way,

print filter(lambda item: item, map(lambda item: item.get("Name"), l))

if I break this down.

First check if item is in the dict or not?

fna = lambda item: item.get("Name")

then map list of dict to above function.

map(fna, l)

for above list it will print

[None, None, 'my-name']

Now we need to eliminate all the None. I am using filter with following function

fnb = lambda item: item

Now if you put all together you have your code in single line.

Answer 4

Here is an easy way to do it without lambdas or filter or any of that stuff. Just combine the list of dictionaries to one dictionary, then use that.

l = [{'abc': 'def'}, {'ghi': 'jul'}, {'Name': 'my-name'}]
k = {m.keys()[0] : m[m.keys()[0]] for m in l}['Name']

Note that this does not change the structure of l.

Answer 5

Don't underestimate the power of dict.get() . What follows is a simplification of a previous answer. In this case, there is no need for map . For the list:

l = [{'abc': 'def'}, {'ghi': 'jul'}, {'Name': 'my-name'}]

The following produces a list of dictionaries that have the key 'Name' :

filtered = list(filter(lambda x: x.get('Name'), l))

From here, you can retrieve the name values for all items in the list:

values = [x['Name'] for x in filtered]

If there is only one target, as in your example, wrap this all up in one line like this:

name = list(filter(lambda x: x.get('Name'), l))[0]['Name']