Question

如何检查列表中某项的频率，然后如果该项的频率为4，则删除所有匹配项？

上下文：

尝试用python创建围棋游戏，我需要能够检查玩家的手牌是否具有四个匹配数字，如果玩家的手牌确实存在，那么我需要删除所有四个匹配项，并在其中将得分提高{ {1}}

输入

char ch = ptr[3]; // 'd' in "World"

输出

玩家的手是一个数字列表。

这是游戏文件： ''' https://github.com/StarSpace-Interactive/GoFish/tree/master/GoFish

Answer 1

这是一个解决方案：

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)

输出为：

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

Answer 2

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)

Answer 3

您可以通过Counter来实现自己的目标。例如，

from collections import Counter

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

然后，counter是

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

然后，您可以编写一个python函数来更新得分和计数器。

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

它将返回新的当前分数和更新的计数器。

Answer 4

就我个人而言，我发现pandas value_count函数比上面建议的numpy.histogram更用户友好。您可以像这样使用它，假设您的手是一张牌（当然，如果手是系列，则此解决方案会更简单）：

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

如何计算列表中某项的频率？

4 个答案: