modifying list element in Python dictionary

Question

inventory = {'A':['Toy',3, 1000], 'B':['Toy',8, 1100], 
              'C':['Cloth',15, 1200], 'D':['Cloth',9, 1300], 
               'E':['Toy',11, 1400], 'F':['Cloth', 18, 1500], 'G':['Appliance', 300, 50]}

The alphabets are name of merchandise, the first field in the [] brackets are category of the merchandise, the second field in the [] brackets are price, the third are numbers sold.

I would like the price to be increased by 1 so the result would look like.

inventory = {'A':['Toy',4, 1000], 'B':['Toy',9, 1100], 
              'C':['Cloth',16, 1200], 'D':['Cloth',10, 1300], 
               'E':['Toy',12, 1400], 'F':['Cloth', 19, 1500], 'G':['Appliance', 301, 50]}

Then what would be a good way to loop through to find any item whose price is $19.

I am not good at lambda function. Could you please give me some explanation of your code so that I could manipulate for future use? Thank you

Finding items whose price is $19 is not as efficient. Lookups for key in dict is very fast (O(log N) or even O(1), but searching for a value you need to read through all elements (O(N)). — skyking
– skyking, Commented Aug 10, 2015 at 8:28
Not really an answer to the question, but it sounds like you don't really want a list for each key. If you have a fixed set of things that each object has it's easier and less error prone to create a class for it and access things by name rather than just using list indices (eg x.category, x.price, x.number_sold). — Holloway
– Holloway, Commented Aug 10, 2015 at 9:29

M. Shaw · Accepted Answer · 2015-08-10 08:28:09Z

2

Try this:

for k, v in inventory.iteritems():
    v[1] += 1

Then to find matches:

price_match = {k:v for (k,v) in inventory.iteritems() if v[1] == 19}

A lambda to find matches:

find_price = lambda x: {k:v for (k,v) in inventory.iteritems() if v[1] == x}
price_match = find_price(19)

answered Aug 10, 2015 at 8:28

M. Shaw

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2 Comments

skyking Over a year ago

Note that in that example the lambda looks like an excuse for using lambda. You could use an ordinary function instead.

M. Shaw Over a year ago

@skyking OP said I am not good at lambda function.. I'm assuming he wants a lambda function of some sort (?). But anyways it is trivial to change it into an ordinary function.

skyking · Accepted Answer · 2015-08-10 08:32:47Z

0

If you wan't to modify the data "in-place" you would loop through the dict:

for k, v in inventory.iteritems():
    v[1] += 1

if you're using python3 you'd use items instead of iteritems. The solution with using dict comprehension {k:[m,p+1,n] for k,(m,p,n) in inventory.items()} means that you will replace the entire dict with a new obejct (which is fine in some scenarios, but not that good in others).

answered Aug 10, 2015 at 8:32

skyking

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Community · Accepted Answer · 2017-05-23 12:14:37Z

0

You can use a dict comprehension :

>>> new={k:[m,p+1,n] for k,(m,p,n) in inventory.items()}
{'G': ['Appliance', 301, 50], 'E': ['Toy', 12, 1400], 'D': ['Cloth', 10, 1300], 'B': ['Toy', 9, 1100], 'A': ['Toy', 4, 1000], 'C': ['Cloth', 16, 1200], 'F': ['Cloth', 19, 1500]}
>>>

And for finding a special items :

>>> {k:[m,p,n] for k,(m,p,n) in new.items() if p==19}
{'F': ['Cloth', 19, 1500]}

edited May 23, 2017 at 12:14

CommunityBot

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answered Aug 10, 2015 at 8:27

Kasravnd

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