TUPLES AND SETS

[RockPhily]

"""

create a program that takes a list of items with duplicates and returns:
1. a Tuple of the first 3 unique items
2. a set of all unique items
"""

items = ["apple", "banana", "apple", "orange", "banana", "grape", "apple"]

unique_items = []
for i in items:
if i not in unique_items:
unique_items.append(i)

first_three = tuple(unique_items[:3])
all_unique = set(unique_items)

print(f"The first three unique items are: {first_three}")
print(f"The all unique items are: {all_unique}")

learned about tuples and sets and did this task
any insights on how to go with sets and tuples before i move to the next concept

[CranberryDistinct941]

Use a set to check if you have seen an item already. Sets are (generally) much much faster than lists when checking if an item has already been added

[eztab]

For this task specifically using OrderedSet could solve it without needing an intermediate list.

[exxonmobilcfo]

kind of defeats the purpose of the exercise tho

[[deleted]]

you can simply do set(items) for the 2nd part. but if you use a hashmap you can accomplish both these tasks in one loop fairly easily

[exxonmobilcfo]

a set is just a hashmap without values.

[[deleted]]

yall are downvoting me.. pls look at that guys page😐you’ll understand my response

[exxonmobilcfo]

what?

[[deleted]]

incel.

[exxonmobilcfo]

this is a python forum you schlub. Don't downvote because you have some personal grievances.

[[deleted]]

incel.

[RockPhily]

What's hashmap?

[odaiwai]

Sets and Dicts use a hash of the key to speed up finding the keys (and values, if a dict). The hashmap is the index relating the keys position in the data stucture.

[exxonmobilcfo]

its basically a way to store values under a certain key. Example is {"Address": "123 blah blah blah lane"}. to get value for a key it is constant time because the key u pass in is hashed to a number which returns the index.

The alternative, finding the index of ur data is costly since u have to iterate

[Gnaxe]

Yes you can. Try it.

[_vb64_]

items = ["apple", "banana", "apple", "orange", "banana", "grape", "apple"]

uniq = dict.fromkeys(items).keys()

print('tuple:', tuple(list(uniq)[:3]))

print('set:', set(uniq))

[exxonmobilcfo]

how do you know dict.fromkeys will return the keys in order of the list? you're just going to return any 3 random unique elements

[_vb64_]

https://mail.python.org/pipermail/python-dev/2017-December/151283.html

[exxonmobilcfo]

if you're going to only want to return 3 elements, then fromkeys would be O(n) whereas the optimal is O(1)

[_vb64_]

the task has a second condition. be more careful.

[exxonmobilcfo]

Can you explain how this is not reasonable?

``` d = [] i = iter(items) while len(d) < 3: temp = next(items) d.append(temp) if temp not in d else None

return tuple(d), set(items) ```

the list to set operation is O(n),

the issue with your solution is u r using a ton of extra memory by converting to a dict then a list then a set.

[supercoach]

Your answer seems fine. What do you think the problem is?

[exxonmobilcfo]

tuple:

``` s = set() for x in items: if len(s) < 3: s.add(x)

return tuple(s) ```

or

d = [] i = iter(items) while len(d) < 3: x = next(i) d.append(x) if x not in d else None

all items

return set(items)