Using a lambda with sorted function

Question

I came across the following code and it is working fine.Though I looked for lambda functions in other questions , but did not find a relevant answer

In[9]: portfolio=[
                    {'name': 'IBM', 'shares': 100, 'price': 91.1},
                    {'name': 'IAM', 'shares': 100, 'price': 41.1},
                    {'name': 'IBM', 'shares': 100, 'price': 71.1} ,
                    {'name': 'IBM', 'shares': 100, 'price': 31.1} 
          ]
In [10]: s =  sorted(portfolio,key = lambda s : s['price'] )
Out[10]: s
[{'name': 'IBM', 'price': 31.1, 'shares': 100},
 {'name': 'IAM', 'price': 41.1, 'shares': 100},
 {'name': 'IBM', 'price': 71.1, 'shares': 100},
 {'name': 'IBM', 'price': 91.1, 'shares': 100}]

Questions:

Is lambda function called to return the price every time a dictionary element is called from the list ? lambda is called only once ?
If can anyone explain this whole of how the sorted works here, it will be very helpful

Possible duplicate of [Syntax behind sorted(key=lambda :)](https://stackoverflow.com/questions/8966538/syntax-behind-sortedkey-lambda) — Mike Scotty, Jun 17 '17 at 19:49

score 2 · Accepted Answer · answered Jun 17 '17 at 20:01

Well, let's try it:

portfolio = [
    {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'IAM', 'shares': 100, 'price': 41.1},
    {'name': 'IBM', 'shares': 100, 'price': 71.1} ,
    {'name': 'IBM', 'shares': 100, 'price': 31.1} 
]

def key_fn(s):
    print("called key_fn({})  ->  {}".format(s, s['price']))
    return s['price']

s = sorted(portfolio, key=key_fn)

which produces

called key_fn({'shares': 100, 'price': 91.1, 'name': 'IBM'})  ->  91.1
called key_fn({'shares': 100, 'price': 41.1, 'name': 'IAM'})  ->  41.1
called key_fn({'shares': 100, 'price': 71.1, 'name': 'IBM'})  ->  71.1
called key_fn({'shares': 100, 'price': 31.1, 'name': 'IBM'})  ->  31.1

Conclusion: the key function is called once per item being sorted.

This explanation helped me. Thanks – Joe_12345 Jun 17 '17 at 23:36 — Joe_12345, Jun 17 '17 at 23:36

Using a lambda with sorted function

1 Answers1