CODEWITHSUNDEEP
pythonlistfrequency
Nova 4026
Nova 4026

Reputation: 45

How do to Count the frequency of an item in a list?

How do I check the frequency of an item in a list and then if that item has a frequency of 4 remove all the matching items?

context:

trying to make a go fish game in python and I need to be able to check if a players hand has four matching numbers if the player's hand does then I need to remove all four of the matching items and increase there score by 1

input 

score = 0
[1,2,4,3,5,6,1,1,1]

output

[2,4,3,5,6]
score += 1

the player's hand is a list of numbers.

here is the file for the game: ''' https://github.com/StarSpace-Interactive/GoFish/tree/master/GoFish

Upvotes: 0

Views: 173

Answers (4)

Itamar Mushkin
Itamar Mushkin

Reputation: 2895

Personally, I find pandas value_count function more user-friendly than numpy.histogram suggested above. You could use it like this, assuming your hand is a List (of course this solution is simpler if the hand is a Series):

import pandas as pd

hand = [1,1,2,3,4,1,1]

cards_count = pd.Series.value_counts(hand) 
# count how many times each card appears
score += (cards_count>=4).sum() 
# add 1 to score for each card that repeats at least 4 times
hand = [card for card in hand if card not in cards_count.index[cards_count>=4]] 
# keeps only cards that did not appear >=4 times

Upvotes: 0

Chun Kit Yeung
Chun Kit Yeung

Reputation: 143

You can do achieve your goal by Counter. For example,

from collections import Counter

mylist = [1,2,4,3,5,6,1,1,1]
counter = Counter(mylist)

Then, the counter is

Counter({
    1: 4, 
    2: 1, 
    4: 1, 
    3: 1, 
    5: 1, 
    6: 1
})

Then, you can write a python function to update the score and the counter.

def update_score(counter, curr_score):
   remove_keys = list()

   # update score
   for key, value in counter.items():
       if value >= 4:
           curr_score += 1
           remove_keys.append(key)

   # remove key
   for key in remove_keys:
       del counter[key]

   return counter, curr_score

It will return the new current score and the updated counter back.

Upvotes: 0

Jae Yang
Jae Yang

Reputation: 539

from collections import defaultdict
score = 0

hand = [1,2,4,3,5,6,1,1,1] # current hand
# im guessing you are doing this in a loop
d= defaultdict( int )
for card in hand:
    d[card] += 1
fourList = []
for key, value in d.items():
  if value >= 4:
    fourList.append(key)

hand = [n for n in hand if n not in fourList]
score += len(fourList)
print(hand)
print(score)

Upvotes: 0

bleand
bleand

Reputation: 376

Here's a solution:

from collections import Counter

score = 0
hand = [1,2,4,3,5,6,1,1,1]

counts = Counter(hand)

for num, count in counts.items():
    if count >= 4:
        hand = list(filter((num).__ne__, hand))
        score += 1

print(hand)
print(score)

And the output is:

[2, 4, 3, 5, 6]
1

Upvotes: 1

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