Find the ONE key in a dict with a certain property

Question

{1: [], 2: [5], 3: [4], 4: [], 5: []}

This is an example for the occurency of numbers after rolling 5 dices
One can read it following:
2 ocurencies of 5
3 occurencies of 4

I am looking for an (again) ELEGANT way to find the one key > 2 with an not empty list and there the value.
In this example I want go get the 4 because it is the only number, which occurs at least 3 times.

Answer 1

Oma Falk , here is code to demonstrate how this could be done. Instead of randint(), choices is used to get 5 numbers in a range.
To count the generated dice values, a counter from collections module is used. The code is not put to maximum compactness, but for better readability. If necessary, this can be shortened. I hope i understood it right.

from random import choices as ch
from collections import Counter as co
dices = ch(range(1,6),k=5) # using choices from random: range 1-6 (= 1-5) k= number of values
print(dices) # for demo only
res = co(dices) # uses a counter from collections
out = [i for i in res.items() if i[1] >= 3]
if out:
print(f'{dict(out)} - dice {out[0][0]} occurs {out[0][1]} times')
else:
print('no value count >= 3 found')

Answer 2

Sven_m
Auch nicht schlecht...

Answer 3

After a very little not offical and cheating criticism of my last post:
Edit: and now to answer @KIIbo
Here is my attempt:
v = None
for i in range(3,6):
if dicdice[i]:
v=dicdice[i][0]
break
print(v)

Answer 4

Oma Falk, just a a short question. You mentioned in your post: "... the one key > 2 with ...." This could be read as "key greater then 2", or you just want to mention that "key 2 has the greatest value". As in the dict are 2 values that are greater than 2: 2:[5] and 3:[4]. From my understanding you are looking for the key, that has the greatest value ???

Answer 5

Oma Falk , using your latest code, i did the last line, that shows the same result as your code:

# find one value with occurency 3+
dicdice= {i:j for i in range(3,6) for j in range(1,7) if dice.count(j) == i or None}
print(dicdice)

# code lothar:
print({dice.count(i):i for i in set(dice) if dice.count(i) >= 3})

Answer 6

Lothar looks gooooood!

Answer 7

If I would want to get the values of the key, which should be greater than 3 of a dict, where the keys are the number of ocurencies, I would:

for k,v in d.items():
if k > 2 and len(v) >0:
print(v)

But I am not sure, if it is elegant.
Its more a kind of "solid".

A for iteration with the keys and values of a dict.

Its good readable (I guess), its possible to turn it in a function fast, if neeeded.

Its the best crafted solid and maybe not 100% ugly codesolution, I can offer for this problem atm.

But I guess, I know what you mean with the "elegant" codestyle.

Its not often. But sometimes I rework some of my old code to more elegant statements.

Good luck with your project.

Answer 8

Play about with this:-
mydict = {1: [], 2: [5], 3: [4], 4: [], 5: []}
print(*[b[0] for a, b in mydict.items() if b and a > 2])

Answer 9

Lothar
i have 5 dice.
So 1 or 0 value can occur 3 times or more.
I am looking for that special value if existing.

so key is 3 or greater (occurence)
and value is the value I am looking for.

Answer 10

I found a better dictionary
it now surely has 1/0 entries.

dice =[ri(1,6) for i in range(5)]

# find one value with occurency 3+
dicdice= {i:j for i in range(3,6) for j in range(1,7) if dice.count(j) == i or None}

Answer 11

A boring version:

d = {1: [], 2: [5], 3: [4], 4: [], 5: []}
res = None
for i in d:
if i < 3:
continue
if not res and d[i]:
res = i
elif res and d[i]:
res = None
break
print(res)

(Now will I find something more interesting?)

Answer 12

الوووو